Pathway to Japanese Literature

Author’s Preface

Within this book lies the “Arithmetical Restorations Competition,” its venues spanning from the First Venue through to the Thirtieth.
Let all who deem themselves worthy embark from that First Venue without reservation—there to assay their intellect and temper their resolve.

“What exactly are ‘Arithmetical Restorations’?”

Simply put, *Arithmetical Restorations* is a discipline within arithmetic mathematics where digits rendered illegible by insect damage are determined through the power of deduction.

However, though it is called a discipline, it is not some tedious pursuit that leaves your head throbbing with nothing but agony.
“Arithmetical Restorations are tremendously interesting and enjoyable; those who once engage with them will never forget the delightful taste of Arithmetical Restorations for the rest of their lives.”
I myself have loved these for twenty years, and at times even incorporated them into detective novels I wrote (Unno Juza’s *Cryptographic Numbers*).

This book also includes several examples of Cryptarithmetic—a relative of Arithmetical Restorations.
“Cryptarithmetic differs from Arithmetical Restorations in that instead of digits being eaten away by insects leaving holes, calculations are masked with letters or symbols—and you apprentice detectives must use deductive reasoning to pry off those masks and reveal the numbers beneath.”

I combined both types into this “Arithmetical Restorations Competition” presented here; while the initial venues are straightforward, the later ones grow increasingly challenging.
Instead, the allure of Arithmetical Restorations will gradually intensify, and you will likely regret how few venues remain as you approach the final Thirtieth Venue.
If you pass the First Venue, you may proclaim yourself a First-tier Selectee.
However, becoming a Thirtieth-tier Selectee is an exceedingly arduous endeavor.

Easy problems could be solved even by first-year middle school students. Even the most difficult problems might likely be solved by high school students. Moreover, these Arithmetical Restorations possessed such bewitching magic that they compelled even university professors to chew their pencils and groan in contemplation—a quality imbued with genuine mystique.

This book intentionally included blank spaces where readers were heartily encouraged to exercise their pencils for partial calculations and revisions—as was only proper.

Within this volume I had managed to compile a total of 132 problems; even gathering this number had required considerable effort over many years. Though I wished to discover more new problems, what I obtained during two decades was all contained here. Should any fellow enthusiasts know of others, I would gratefully welcome such contributions. Finally, I recorded that this work had referenced the following books, expressing both gratitude and respect.

F.C.Boon, *Puzzle Papers in Arithmetic*

G.C.Barnard, *An Elementary Puzzle Arithmetic*

F.F.Potter and F.C.Rice, *Common Sense Arithmetic*

H.E. Dudeney, *Modern Puzzles*

A.S.E. Ackermann, *Scientific Paradoxes and Problems*

Fujimura Kousaburou, *New Mathematical Puzzles*

January 7, 1946 (Showa 21)

Unno Juza

1. What Are Arithmetical Restorations?

Old ledgers, deeds, and account books had been eaten by bookworms called *shimi*, leaving holes where crucial numbers once were.
Oh dear—we were in quite a predicament now!
At such times, through deductive reasoning, you would determine that those digits reduced to holes must indeed be of such-and-such numbers.
This discipline of arithmetic was what came to be called *Arithmetical Restorations*.

Of course, if you have only a standalone numerical entry—such as “Fifty thousand yen in total borrowed under the aforementioned conditions”—and its digits have been eaten by bookworms, there would be no way to determine them; but if it were a set of arithmetic operations involving addition, subtraction, multiplication, or division—even with portions devoured—you could accurately deduce the missing digits through reasoning based on their contextual relationships. At times, even when all digits have been devoured by bookworms yet their arrangement remains known, you might apply deductive reasoning to correctly determine every digit.

Furthermore, there exists something similar to *Arithmetical Restorations* called *Cryptarithmetic*. In *Arithmetical Restorations*, digits have been devoured by bookworms, leaving holes, whereas in *Cryptarithmetic*, they are replaced with symbols or letters. Given that digits consist of ten in total—0 in addition to those from 1 to 9—the symbols or letters serving as masks in *Cryptarithmetic* are necessarily limited to no more than ten.

“Which is more interesting—*Arithmetical Restorations* or *Cryptarithmetic*?” That varies from person to person, but I find both fascinating.

The key to solving *Arithmetical Restorations* typically lies in first identifying holes that must contain either 0 or 1.

In *Cryptarithmetic*, similar principles apply, but since identical letters or symbols appear in varied relationships, once you discern interconnections—like realizing “Ah! This letter must be double that one”—the remainder suddenly becomes solvable.

When solving Arithmetical Restorations and their kind, both the intellectual thrill and practical merit lie in advancing through pure deductive reasoning; a trial-and-error approach—muttering “Is it 1? Maybe 2? If not, then 3?” while mechanically substituting digits—holds no enjoyment.

As stated in the preface, intermediate and advanced *Arithmetical Restorations* are each as precious as gems and worthy of cherishing; I earnestly hope you will solve them with deliberate care.
Solving ten or even twenty problems a day will not only exhaust your mind but also, I believe, prevent you from grasping the true essence of Arithmetical Restorations.
Therefore, in designing this book’s “Arithmetical Restorations Competition,” I resolved to intermingle easy and challenging problems, grouping approximately four per venue.

Now, of the First through Thirtieth Venues, approximately the first third were relatively straightforward; beyond that, increasingly complex and abstruse problems were introduced—precisely why the joy that welled up when deduction succeeded was all the greater.When it came to the Twentieth Venue and beyond,I believed they would bring Arithmetical Restorations enthusiasts irresistible joy and excitement.

2. Simple Arithmetical Restorations and Their Solving Methods

First, I will begin by explaining the solving methods for simple Arithmetical Restorations.

**[Example 1]** There was an addition problem like the following on the next page.
The square holes were what we called the parts eaten by insects.
In this problem, there were four such instances.
You had to uncover this through deduction.

1□92

29□1

971□

+ 2□17

——————

17592

You started by considering the rightmost column’s □.
This was the units place.
From this calculation, it became clear that adding 2, 1, □, and 7 resulted in a units place of 2.
Excluding □, adding the known 2, 1, and 7 gave 10.
However, the units digit of the total in the units place was not 0 but 2.
Therefore, it was deduced that □ was not 0 but 2.
Thus you inserted that 2, resulting in the configuration shown in the upper left.

1□92

29□1

9712

+ 2□17

——————

17592

Next was the tens place. One had to remember that 1 had been carried over from the units place. There remained one hole. Setting that aside, adding the known digits—9, 1, 1—along with the 1 carried over from below resulted in 12. Yet the tens digit of this total showed 9 rather than 2. This discrepancy arose from excluding the hole in the tens column; subtracting 2 from 9 yielded 7, which became the determined digit for □. Reorganizing the figures accordingly produced the configuration visible at lower right.

1□92

2971

9712

+ 2□17

——————

17592

Next came the hundreds place. This approach wouldn’t work as it had before—there were two holes this time. Adding the known 9 and 7 along with the 1 carried over from the tens place gave us 17. Since this column’s total showed 5, the sum of the two missing digits might be either 8 or 18. We found ourselves at an impasse.

Yet there was no cause for alarm. When we examined the thousands place above, we noticed no holes in that column. Adding 1, 2, 9, and 2 produced 14, but the recorded total below stood at 17. Through this discrepancy, it became clear that the number carried over from the hundreds to thousands place was indeed 3—the result of subtracting 14 from 17.

In that case, the sum of the two holes in the hundreds place had to be 18 instead of 8.
If it wasn't 18, no carry-over of 3 would occur.
This settled it provisionally.
Any digits would theoretically work if their total came to 18.
But wait—that couldn't actually mean any digits.
When two single digits sum to 18, they must both inevitably be 9.
Since no digit in this column could exceed 9 under positional constraints—they absolutely had to both become 9.
Thus did we arrive at the solution shown above.

1992

2971

9712

+ 2917

——————

17592

Even within addition operations themselves, there are cases where “Cryptarithmetic”-style elements come into play. The next example problem illustrates this.

**Example 2**: An addition problem where the digit represented by the letter N appears in five locations.
Of course, every N represents the same digit.

2N8

2N2

88N

+ N2N

—————

2164

When looking at this arrangement, it might have seemed unclear where to begin, but after observing it for a while, one would discover the key.
It lay in how both the ones place and tens place contained four digits summing to 8, 2, and two Ns each.
Moreover, examining the totals below showed 4 in the ones column and 6 in the tens column.
This was it—the key!

What did this signify?
The number carried over from the ones place to the tens place had to be 2—obtained by subtracting 4 from 6.

How can we arrive at 2?
Adding 8 and 2 gives 10, so this first produces a carry-over of 1.
Since there are two additional Ns in this sum, their total must generate another carry-over of 1.
This means the sum of the two Ns becomes 14.
Therefore, N must be 7.

If you suspect trickery, go ahead—substitute 7 into all five Ns and verify the calculation yourself.
Through this method, we solved the puzzle without ever needing to address the N in the hundreds place.
It carries a faint thrill akin to cleaving through a knot with one swift stroke.

Let’s try a subtraction problem.
It’s an easy one, but….

**Example 3**: This was something even a first-year National School student could solve.
We began from the ones place and worked our way up.

□84562

–8□1□3

———————

6□7□7□

Since three could not be subtracted from two, ten was borrowed from the tens place and three was subtracted from twelve to get nine.

Next was the tens place.
The 6 had already been reduced to 5.
Since subtracting □ resulted in 7 through this operation，the □ was determined to be 8 via reverse calculation.
At this stage，10 had been borrowed from the hundreds place.

Next comes the hundreds place. The minuend’s digit 5 has become 4 after lending 1 to its rightward neighbor earlier.
Subtracting 1 from this yields 3 – thus determining □ to be 3.
Now we advance to thousands place.
Here’s where simplicity reigns!
□ must unequivocally equal 7.
Do note that 10 was borrowed from above during this step.

In the ten-thousands place, the 8 in the minuend had lent 1 to the lower place, making it effectively 7.
Subtracting 8 then resulted in 9.
The □ was determined to be 9.

The □ in the hundred-thousands place was 7.
This was because it had lent 1 downward, and since the result was 6, it had to undoubtedly be 7.
This completed it.
The result was as shown above.

784562

–87183

———————

697379

Next is the case of multiplication.
While multiplication doesn’t fundamentally differ in approach, I would like to present one particular strategy for your consideration.

**Example 4**: When looking at this problem, one finds missing-digit holes in eight places.
With so many gaps laid bare, one couldn't help feeling anxious—could this truly be unraveled through deductive reasoning alone?

97□

×      □8

——————

□□□0

9□□

——————

175□0

However, if one stares at this calculation for a while, a significant clue will eventually come to light.
That’s precisely where the fascination lies.

In other words, the first clue was that the hole in the tens place of the multiplier had to be 1. This became evident when observing that the fourth line from the top showed a three-digit number beginning with 9 in its hundreds place. Yet upon examining the multiplicand, it too proved to be a three-digit number with 9 occupying its hundreds place. From this comparison emerged an irrefutable conclusion—the tens place of the multiplier could only be 1. Consequently, both the first and fourth lines from the top now represented identical numbers. Through this reasoning, we successfully filled two of the eight missing digits as demonstrated above.

97□

× 18

——————

□□□0

97□

———————

175□0

The second clue emerged from multiplying the 8 in the multiplier's ones place by the □ in the multiplicand's ones place - their product's ones digit had to be 0, as evidenced below the 8.
This meant □ couldn't be arbitrary.
Through elimination, □ had to resolve to either 0 or 5.
Any other value would violate the third line's required rightmost 0.

With possibilities narrowed down, we advanced to decisive verification between these two candidates.

The □ was not 0.
This was because when taking the third line as 7,760 and adding it to the fourth line's 970, instead of arriving at something like 175... as in the fifth line, we ended up with 174..., leaving the calculation short.
Therefore, the □ in the ones place of the multiplicand had to be determined as 5.

When we calculated with □ as 5, the figures correctly aligned as shown in the lower left of the right-hand page.

975

× 18

——————

7800

975

——————

17550

Next was division.
Division possessed all the allure one could ask for in arithmetical restorations.
Superior arithmetical restorations generally took the form of division problems.
What followed next was an easy division-based arithmetical restoration.

**Example Five**: In this problem, both digits of the divisor were missing (represented by holes), the tens place of the quotient was also missing, and six holes remained within the calculation itself.

□3

——————

□□)949

□□

————

□□9

□□9

————

0

At first glance—despite careful examination—the crucial divisor remained entirely unknown, and even the quotient was only half determined; it seemed utterly impossible to make any progress.

Yet when one calmly settled their mind and steadily examined this calculation sheet, effective clues gradually began revealing themselves.
First came noticing that the hole in the divisor’s ones place was 3.
The reason unfolded thus:
When multiplying this divisor by 3—the quotient’s ones digit—the ones place of that product became 9 (as seen at the right end of the fifth line from the top).
In other words: To produce 9 in the ones place through multiplying 3 by an unknown number (the divisor’s ones digit), that number could only be 3.
This followed from 3×3=9.
Thus we filled one hole as shown above.

□3

——————

□3)949

□□

—————
□□9
□□9
—————
0

Next, you noticed that the tens digit of the quotient had to be either 2 or 1—both less than 3.
This was because multiplying the divisor by the ones digit of the quotient—shown in the fifth line from the top—resulted in the three-digit number □□9.
However, when multiplying by the tens digit of the quotient—seen in the third line from the top—the calculation produced only a two-digit number: □□.
In other words, the tens digit of the quotient was restricted to either 2 or 1.
This considerably narrowed the scope of inquiry.

Meanwhile, you noticed that the □ in the tens place of the divisor had to be 4 or greater. This was because when you multiplied the divisor’s □3 by the quotient’s ones digit 3 at the fifth line from the top, the ones place became 9. Next, multiplying the tens place of the divisor by the ones digit of the quotient (3) resulted in a two-digit number occupying the hundreds and tens places—this occurred because if the divisor’s tens digit were 3 or less, multiplying it by 3 would not yield a two-digit number (for instance, 3×3=9 or 2×3=6—both single digits). It had to absolutely be 34 (which gives 12) or higher. Therefore, you realized that the hole in the tens place of the divisor must be 4 or a number greater than 4.

Therefore, this time, you returned once again to examining the tens place of the quotient—that is, to the third line from the top.
As previously stated, the tens place of the quotient had to be either 1 or 2.
However, if this were 2, it would undermine the relationship you had just deduced—that "the tens digit of the divisor must be 4 or greater."
Assuming it to be the minimum value of 4, multiplying 2 (the tens digit of the quotient) by 43 gave 86. Subtracting this from 94 in the dividend left a remainder of 8—a single-digit number.
This meant that instead of taking the form □□9 in the fourth line from the top, it became □9, causing a mismatch in digit alignment.
Therefore, while the tens place of the quotient had been considered to be either 1 or 2, it was now definitively determined to be 1.
Now, with this, it was mostly solved.
If you filled in the solved portions into the holes and wrote them out, it appeared as shown above.

13

——————

□3)949

□3

————

□19

□19

————

0

As demonstrated here, once the tens digit of the quotient was determined to be 1, the right end of the third line naturally became 3. Then—since subtracting 3 from 4 gives 1—and given that the sixth line was 0—the digit below that 1 likewise had to be 1.

In the fifth line—where a calculation producing 1 in the tens place involved a mystery number that was a multiple of 3 (i.e., the tens digit of the divisor)—what could it be? The sequence now shifted to this problem. This became immediately clear: it was 7. Because 37 multiplied by 21 equals 777. Thus, you realized that the tens digit of the divisor was 7—meaning all crucial parts had now been solved.

For verification’s sake, testing whether 73 × 13 worked correctly revealed perfect alignment with what was shown above.
Long live deductive reasoning!
And there you had it.

13

＿＿＿＿

73)949

73

——————

219

219

————

0

Having covered methods for solving elementary problems to this extent, I would next introduce two or three approaches to more challenging ones.

3. Advanced Arithmetical Restorations and How to Solve Them

Even though they were advanced, there was no fundamental difference in the approach to solving them.
The challenging aspect—and consequently the aspect of great interest—lay in determining where the clues were hidden and from which point one should begin solving to achieve the most efficient progress.
Superior gem-like problems referred to those where the key’s hiding place lay in an utterly unexpected location as described here, and once you struck upon it and solved that part, the rest cascaded down and unraveled effortlessly.

As you all set out to solve these, whether you’re the great detective Akechi Kogoro, Sherlock Holmes, Philo Vance, or Ellery Queen, I urge you to exert your utmost intellectual prowess.

I then proceeded with explaining the example problems.

【Example Six】 This is a "cryptarithmetic puzzle."
Various letters are arranged here, but this does not mean all twenty-six letters of the alphabet are used; there are only ten types in total.
Specifically, they are the ten letters: C, M, L, Q, T, P, A, I, N, S.
Our task now is to act as detectives—using deductive reasoning—to determine which of these ten masked letters corresponds to each digit from 0 through 9.

QCN

______

CML)QTPAI

QIS

—————

QPA

CML

————

CCCI

CSNS

————

MI

At first glance, they could make neither head nor tail of it, feeling as bewildered as if they had crash-landed on Mars, utterly stranded with no foothold in sight.
However, if they observed this methodically—in true great detective fashion, as I always advised—they would eventually discover the splendid key to unlocking the secret door, finding themselves unable to suppress a triumphant smile.

Now, let us turn our attention to the detective work of this cryptarithmetic puzzle.

The great detective first set their sights on the CML in the fifth line from the top.
Wasn't this exactly identical to the divisor CML?
A monumental discovery! A monumental discovery!

Therefore, the digit C in the tens place of the answer that led to this fifth-step calculation had to be 1.
"Indeed," they murmured.
They proceeded to replace every instance of C in the calculation with 1, reorganizing it as shown above.

Q1N

______

1ML)QTPAI

QIS

—————

QPA

1ML

————

111I

1SNS

————

MI

Now, where should they uncover the second key?
"Ah, here we are," the detective realized.
The S at the far right of the third line emerged as the critical element.
The digit positioned above this S was P.
Curiously, the digit below this S also proved to be P.

Subtract S from P, and the result is P.
P − S = P.
This seemed like an odd relationship.
No—it wasn’t strange at all.
Such an equation could only hold true when S was 0.
That settled it.
The second key lay in this S being zero.

Accordingly, they replaced every S in the calculation sheet with 0 and revised it anew.
The puzzle now came into sharper focus.

No—it was too soon to rest easy.
He couldn’t know what obstacles lay ahead.

Q1N

______

1ML)QTPAI

QI0

—————

QPA

1ML

————

111I

10N0

――――

MI

At last, he set about discovering the third key.
Now, where was it located?
This time, it was quite formidable.
Well now—he hadn't noticed this.
This must be it—the third key! he thought.

They now came to the "S is 0" in the third line that had been previously determined—the insertion of 0 there resulted from multiplying L (the units digit of the divisor) by Q (the hundreds digit of the answer).
In other words, multiplying *L* and *Q* had resulted in the units digit of their product being 0.
In such cases, either *L* or *Q* must be 5, with the other being even.
(Who was it—the one who had said cases where *L* or *Q* was 0 should also be acceptable?…
But 0 had already been determined.
(He had just deduced that S was 0)—Now, this alone wasn’t conclusive, but there was another point requiring focus.
That digit at the far right of the seventh line was also 0.
This 0 had resulted from multiplying the divisor’s *L* by the units digit *N* of the answer.
Therefore, by the same logic as before, either *L* or *N* must be 5 and even.

Thus it was conclusively determined that *L* must be 5. The reasoning unfolded as follows: since earlier conditions required either *L* or *Q* to be 5 or even, and new constraints demanded the same of *L* or *N*, only *L*=5 could satisfy both requirements simultaneously ("Ah—did I hear an objection? Who suggested *L* might be even instead?"). That approach proves untenable—were *L* even, *Q* would necessarily become 5 per initial conditions. But this would disastrously conflict with subsequent requirements when examining *L* and *N* ("Can't you see how that would create irreconcilable contradictions?"). Through this rigorous elimination, *L* stood confirmed as 5 while both *Q* and *N* emerged as even numbers. These values were promptly inscribed into their respective positions.

Q1N

______

1M5)QTPAI

QI0

—————

QPA

1M5

————

111I

10N0

————

MI

If at this point your heart raced with the thought, “Arithmetical restorations are fascinating!”, that was proof the “demon” of arithmetical restorations had already possessed you.
Once this “demon” took hold—I’m afraid (?)—you would never again escape being a lifelong fan of *arithmetical restorations*.
I wasn’t trying to scare you, but that was simply how things stood.

Let’s cut out the digressions and proceed to the next step.
Where was the fourth key located?
In the fourth line’s Q, there was a 1 below it.
There was also a 1 below that.
Thus subtracting 1 from Q had resulted in 1.
This led to the surmise that Q was either 2—the sum of 1 and 1—or that Q had borrowed 1 from the lower digit and might actually be 3.
In other words, Q had to be either 2 or 3.
However—hold on—Q needed to be 2.
3 was not permitted.
Because it had just been determined earlier that both Q and N were even numbers.

21N

______

1M5)2TPAI

2I0

—————

2PA

1M5

————

111I

10N0

――――

MI

Since the fourth key was relatively straightforward, if you relax thinking the rest will be easy, a grave mistake will occur.

There, examine the middle of the sixth and seventh rows.
Subtracting 0N from 11 leaves M as the remainder.
Rewriting this relationship results in N + M = 11.
Since N had already been determined to be even, M must be an odd number.

However, since 1 and 5 had already been determined as odd numbers, M had to be either 3, 7, or 9.

Yet another restriction applied to M.
In the third segment, when multiplying the divisor 1M5 by 2 to obtain 2I0, M needed to be 4 or smaller.
This was because if M were 5 or higher, the calculation would produce a two-digit number, changing the third segment's leftmost digit from 2 to 3.
Combining this new condition—that M must be 4 or less—with the earlier conclusion that M could only be 3, 7, or 9 left no choice but M=3.
This became the fifth key.

Since the relationship N + M = 11 had been revealed earlier, substituting M with 3 naturally led us to conclude that N must be 8.

Thus, we determined that the divisor was 135 and the quotient was 218.
With this, the rest could be noisily pieced together.
Thus, when the calculation became as shown on the next page, this problem was splendidly solved.
It was the famous detective’s victory.

Now came the time to pop open some champagne!
If there was no champagne, they were to make do with iron pipe beer!

218

______

135)29467

270

—————

246

135

————

1117

1080

————

37

C=1

M=3

L=5

Q=2

T=9

P=4

A=6

I=7

N=8

S=0

As an example of an extremely difficult problem, I presented an example problem where only one digit remained while all others were holes.
Since many people likely did not believe such a problem could be solved through deduction alone, I wished to demonstrate precisely how deduction could systematically unravel it.

Example Problem 7: At a glance alone made clear this was an unmistakably difficult puzzle.
The sole known element was the quotient's thousands digit being seven.
Should someone somehow fill these voids with answers, skeptics might suspect either wild guesswork or prior knowledge of the solution.
Yet contrary to such assumptions through deduction's inexorable advance does this conundrum yield.
True—it demands efforts worthy of detective fiction's most arduous cases—but precisely this bone-wearying rigor renders arithmetical restorations irresistibly delectable to devotees.
Let us now commence our engagement.

□7□□□

＿＿＿＿＿＿＿＿＿

□□□）□□□□□□□□

□□□□

———————

□□□

□□□

——————

□□□□

□□□

——————

□□□□

□□□□

————

0

The first step was to look for 0.
This was straightforward.
The tens digit of the answer was 0.
This was because upon closely examining the eighth step，it descended two digits at once from above.
Therefore，it became irrefutably clear that the tens digit of the quotient was 0.
Therefore，0 was written in the □.

Next，it became clear that both the ten-thousands place and ones place of the quotient were digits greater than 7.
In other words，they were likely 8 or 9.
This was because if it had been 7，calculating as per fifth-step logic would have yielded a three-digit number，yet both third-step calculations—pertaining to ten-thousands place deductions—and ninth-step calculations—for ones place determinations—produced four-digit numbers.
Thus，both positions were confirmed to hold values of either 8 or 9，exceeding seven by necessity.

Next, focus on the leftmost part of the sixth step.
This must be 1.
In the seventh step where subtraction occurs after bringing down one digit, since there are no digits beneath it remains clear that this must undoubtedly be 1.

We now determine that the hundreds digit of the divisor in the second step's leftmost position must be 1.
This follows because multiplying 7 (the quotient's thousands digit) by this three-digit divisor produces a three-digit result in the fifth step - meaning the divisor's hundreds digit can only be 1.
Were it 2 or higher, this multiplication would yield four digits or more - an impossibility.

The leftmost digits of the third step and ninth step had to both be 1.
For once we had determined that hundreds digit of divisor was now 1, leftmost digit of this four-digit number could not become anything other than 1 no matter what number you multiplied by.

Thus leftmost digit of eighth step also had to be 1.
This was because eighth step was identical to ninth step.

Well, at this point, let us organize the original calculation and present it below.
Note that ▲ denotes that it is either 8 or 9.

▲7□0▲

＿＿＿＿＿＿＿＿＿

1□□）□□□□□□□□

1□□□

———————

□□□

□□□

——————

1□□□

□□□

——————
1□□□

1□□□

————
0

Thus, we had finally solved seven or eight holes, but thirty-two remained.
A profound sense of the long road ahead settled over us.

Now then，we mustered our spirits and proceeded to tackle the next step．
The tens digit of the divisor in the second step was two or less．
In other words，it had to be two，one，or zero．
This was because—as seen from examining how multiplying seven (the thousands digit of the quotient) by this three-digit divisor produced another three-digit number—the hundreds digit could only be one．

However, upon closer examination, if it were 0, that would be illogical, narrowing it down to either 2 or 1.
This was because if it were 0, the divisor would become 10□—and whether you multiplied this by 8 or by 9 (the digits presumed to occupy the ten-thousands place of the quotient)—both operations would yield three-digit numbers that could never match the four-digit figures shown in the third and ninth steps.

Next, we determined that the hundreds digit of the quotient in the first step was 8. This was because—as we had already deduced—the divisor was either 12□ or 11□; however, if we assumed this hundreds digit to be 7 and multiplied it by that divisor, we would get either 84 or 77. Subtracting this from the four-digit number 1□□□ in the sixth step left a remainder that inconveniently descended two digits—as seen in the leftmost 1 of the eighth step. This absolutely could not hold unless it was 8.

Once we determined that the hundreds digit of the quotient was 8, both its ten-thousands digit and ones digit had to be 9. The reason lay in how even 8 produced a three-digit number when multiplied by our narrowed divisor candidates. Yet since both ten-thousands place and ones place operations resulted in four-digit numbers—as evidenced by their respective steps—this could only mean they were necessarily 9.

With this, all digits of the quotient were now determined.
Thus, □7□□□ was determined to be 97809.

Moreover, the leftmost digit of the seventh step was 9.
This was because assuming it to be 8 would leave some digit beneath it; however, since nothing actually remained there, it had to be 9.

When we incorporated all these deductions into the calculation, it appeared as shown below.

97809

＿＿＿＿＿＿＿＿＿

1□□）□□□□□□□□

↑ 1□□□

2 ———————

Again □□□

ha □□□

1 ——————

1□□□

9□□

——————

1□□□

1□□□

————

0

Now, just one final push.

The tens digit of the divisor has already been narrowed down to either 2 or 1.
Solving this problem proves relatively straightforward.
Given the deductions made thus far, it appears far more likely to be 2 than 1.
Let us therefore tentatively assume it to be 2.

Thus, the divisor took the form of 12□.

If that held true, the ones digit of the divisor had to be 4 or smaller.
This was because in the seventh step, 124 × 8 = 992, while any value of 5 or higher—such as 125 × 8 = 1000—would produce a four-digit number.
However, since the seventh step showed a three-digit number, the ones digit of the divisor needed to be 4 or below.

When we assumed this digit to be 4, the divisor became 124.
Performing the calculation with this divisor and the previously determined quotient of 97809 yielded the result shown on the right.

97809

＿＿＿＿＿＿＿＿＿

124)□□□□□□□□

1116

―――――――――

□□□

868

――――――

1□□□

992

――――――

1□□□

1116

――――

0

When proceeding backward from this point, all missing digits were successfully filled without inconsistencies, resulting in the solution shown below.
This marked the answer.
Phew! That required considerable effort.

97809

＿＿＿＿＿＿＿＿＿

124)12128316

1116

———————

968

868

――――――

1003

992

——————

1116

1116

————

0

The solution method was by no means limited to this single approach; there might well be other, better methods available, and I would like you to give careful thought to this matter.

At this point, I decided to conclude the explanations of the example problems.

4. About the “Arithmetical Restorations” Competition

At last, I will now commence the “Arithmetical Restorations Competition.”
There are venues from the First to the Thirtieth.
Each venue presents approximately four problems.
I want you to tackle them thoroughly and derive the answers through the power of deduction.

The blank pages were left open for the convenience of performing detailed calculations or redoing them.

I advise against advancing through more than one venue per day.
If you rush through them, your head will ache, and I fear you won’t fully savor the gem-like essence of these arithmetical restorations.

Of the four problems in each venue, the first two or three were relatively easy, but those appearing later became considerably more difficult.

Moreover, while the initial venues were relatively manageable, as one progressed through them, they gradually grew more challenging.
From around the Twentieth Venue onward, I imagined they would demand such strenuous effort that steam might rise from one's head.
Yet in return, one could fully savor the charm of arithmetical restorations that richly rewarded every ounce of exertion.

Moreover, these answers were intentionally omitted.
If you flipped to the answers and reacted with remarks like “Ah! So this □ is a 9?” then the exquisite charm of arithmetical restorations would be lost.
If you found yourself stuck in your approach, you were advised to review the previous example problems, master the knack of solving arithmetical restorations, and then take on the problem once more.

If you passed the First Venue, you would be granted the title of First-tier Selectee.
Thus, as you pressed onward and passed the Thirtieth Venue, you naturally attained the highest honor as a Thirtieth-tier Selectee—though I regretted having no medals at hand to award, much as I would have liked to present them.

Even if you became a Thirtieth-tier Selectee, this book of arithmetical restorations did not thereby transform into something dull.
If you started anew and revisited the First Venue once more, you would likely discover fresh enjoyment.
Arithmetical Restorations were not so shallow that once you solved them once or twice, they became devoid of interest thereafter.
For this reason, to allow reworking problems two, three, four times over, I wished you to refrain from writing in this book with ink and instead use a soft pencil for your entries.

“With that, I hereby declare the venues open!”
“Come now, come now! The venue for the world’s rarest ‘Arithmetical Restorations Competition’ begins right here.”
“Those with confidence and interest, please proceed through this gate.”
“Please, go ahead, go ahead…”

“Arithmetical Restorations Competition” First Venue

“Now then, this is the First Venue.”
This marked the starting point of the “Arithmetical Restorations Competition.”
Addition, subtraction, multiplication, and division—they were all there.

(4) In problems where five holes form a number divided by 11—with its quotient displayed beneath—please note there remains an undivided remainder.
(5) Though this venue's most challenging problem may daunt you at first glance，persistent scrutiny will reveal its starting point.
“To those finding themselves at an impasse—might I extend a sliver of insight?”
“Consider this：what multiplier of eight yields four in its units place? Given its three-digit span... Ah，but divulging more would strip away all delight—my sincerest apologies.”

(1)★

□937

8□

109

+ 4□75

―――――――

74□5

(2)★

8□74□

\- 6□23

―――――――

□11□6

(3)★

□□□□1□

×             7

―――――――――

□999998

(4)★

11)□□□□□

￣￣￣￣￣￣

7996 Remainder 9

(5)★

□□

＿＿＿＿＿

28)1□□4

□□

―――――

□□4

□□4

――――

0

“Arithmetical Restorations Competition” Second Venue

We welcome your honored presence as a First-tier Selectee.
I earnestly hope you will brilliantly clear this venue too and attain the distinction of Second-tier Selectee.

Five problems await you in this venue as well.

(1) is simple.
(2) may require some head-scratching.
(3) poses no trouble either.
(4) appears daunting initially but reveals unexpected simplicity.
Direct your focus toward particular digits here and there on the right side.
(5) presents an intriguing challenge.
Pay heed to the three Cs and then the B... Ah—perhaps I should have kept that to myself...

(1)★

526
- □7□
――――――
347

(2)★

AB2DEF
×             2
――――――――
2DEFAB

(3)★

9)□□□□□
￣￣￣￣￣￣
4140 Remainder 8

(4)★

823

×    1□

――――――

□□□5

823

------

□□□□5

(5)★

DC

＿＿＿＿＿

ABC) DEDC

DAB

―――――

CDC

CDC

――――

0

“Arithmetical Restorations” Competition: Third Venue

This was the third venue.
Unlike before, things were about to get a bit more challenging.

For the missing digits in (1), we had to fill in numbers 8 or below.
(2) remained a breeze.
(3) promised tougher going!
What numbers must A and C be when their product fell in the thirties?
Since 7 and 3 had already been claimed, they stood excluded.
Zero too lay beyond consideration.
Nor could they dwell among the smaller numbers of three or below.
Our possibilities were narrowing perceptibly.
There came another factor to weigh.
At the final step, multiplying 3 by A yielded C in the units place.
Such specific pairings proved scarce indeed.
...By this point, you'd surely grasped it.
(4) held no terrors.
(5) brimmed with interest.

(1)★

2344

□□□□

+ 5720

――――――――

13252

(2)★

□7□6□

×        7

―――――――――

3□29□6

(3)★

A)3B7AC

￣￣￣￣￣￣

CB73

(4)★

AB□A

×      AB

――――――――

□□□□

□□□9

――――――――

□□□B□

(5)★

1□2

＿＿＿＿＿＿

□□)□□□64

□□

―――――

246

□□1

―――――

□□4

□□4

―――――

0

Arithmetical Restorations Competition: Fourth Venue

Congratulations on having achieved third-tier Selectee status in the previous venue.
However, it would appear the previous venue proved somewhat taxing.

For this venue—to compensate for that—I had lined up relatively easy yet engaging problems.
In (5), there are two A’s in the answer section where...

That became your admonishment—“You’re rambling too much.”
“Very well,” I conceded at last. “I shall keep silent.”

(1)★

674□

2□7

63

+  2□□8

―――――――

□□1□3

(2)★

7)□□□□□

￣￣￣￣￣￣

6618 remainder 1

(3)★

CB8A6

×         A

―――――――――

21CB8A

(4)★

2N8

2N2

88N

+  N2N

―――――――

2164

(5)★

A7A

______

□□)1□□□□

1□□

―――――

□□

□□

―――――

□□□

□□□

―――――

0

**Arithmetical Restorations Competition: Fifth Venue**

“Welcome! This is the Fifth Venue.”
“How are you finding it?”
“It’s grown rather engaging now, hasn’t it?”
“We’ve prepared five problems for this venue as well.”
Problems (1), (2), and (3) should prove solvable even while nodding off.

“And (4)?”
“I wonder if this calculation’s wrong—you multiplied a three-digit number but only got two digits!” someone exclaimed. Oh dear, how pitiful.

“(5) required focusing on the 4 in the third row…”
Ah—perhaps that had been an unnecessary hint.

(1)★

□□7□4□

×             7

―――――――――

1□2□8□1

(2)★

□84562

-    8□1□3

―――――――――

6□7□7□

(3)★

□2□

＿＿＿＿＿

8)3□24

(4)★

WTFR

×      STX

―――――――――

XRSY

SWQY

―――――――――

SYRQSY

(5)★

A□1

＿＿＿＿＿＿

1A)□56□□

□4□

――――――

□□□

□□□

―――――

1□

1□

―――

0

“Arithmetical Restorations Competition: Sixth Venue”

Since you have already attained the rank of Fifth-tier Selectee, I shall express respect for your proven skill and henceforth refrain from discreetly letting hints peek out from under my sleeve in this and subsequent venues.

“Now then, please proceed.”

(1)★

□□57□

×        9

―――――――――

3□1□02

(2)★

5483□

– □□2□□

――――――――

9639

(3)★

6)□□□□□
￣￣￣￣￣￣
9) □□□□  Remainder 1
￣￣￣￣￣￣
470 Remainder 1
(4)★
3□□□
×        9□
―――――――――
2□□□1
30□17
―――――――――
33□□□1
(5)★
7F

＿＿＿＿＿＿

ADC)BAEFC

BDG8

――――――

DEEC

1EEC

――――――

0

“Arithmetical Restorations Competition: Seventh Venue”

Yes, here we are at the Seventh Venue.
The problems are gradually becoming more interesting.

“Are you saying that neither (1), (2), nor (3) are particularly interesting?”
“Is it that they’re too simple to be engaging?”
“That’s right—after all, you’ve already attained Sixth-tier Selectee status.”
“However, I believe that warming up is necessary in all things.”

(4) and (5) were truly engaging.
“I am humbled by your praise.”

(1)★

2157

□□□□

+ 1625

―――――――

10342

(2)★

□□□

×  □7

―――――――

3□□4

5□□□

―――――――

□□□□□

(3)★

A7

＿＿＿＿

3)1A1

(4)★

ABCD

×  DCB

――――――――

BDKG

CHEB

DFCK

――――――――

LCCEGG

(5)★

□74

＿＿＿＿＿＿

31□)□□6□□

63□

――――――

□□8□

□□□□

――――――

□□□□

□□□□

――――――

□18

“Arithmetical Restorations Competition: Eighth Venue”

The Eighth Venue is here.
As per convention, five problems are lined up here.
Please proceed at your leisure.

“Huh? Are you asking me to speak like I did before? That’s rather troublesome.”
“The problems in this venue aren’t that difficult, I assure you.”

“(4), you say? This one’s simple enough.”
“Once you determine what B represents in the sixth row, the tens place of the divisor becomes clear, and then... well, mumble mumble.”

“(5), is it? So you’re genuinely requesting guidance here.”
“You claim there are no clues—with both divisor and quotient being complete blanks?”
“Oh, this one’s straightforward too.”
“To find which two factors produce the number in the fifth row... A touch of factorization knowledge would... Best stay quiet before earning a reprimand.”

(1)★

3456□8

□324□

+ □6□123

―――――――――

959045

(2)★

4□□□4

− □947□

――――――――

26205

(3)★

7□4□
× 53
――――――――
23□□7
38□□□
―――――――
410□□7

(4)★

STP

＿＿＿＿＿

BT)KDNG

PK

―――――

FN

TN

――――

BGG

NB

――――

N

(5)★

□□□

＿＿＿＿＿＿

□□□)□□□□□

□□□

――――――

□□□□

1851

――――――

□□□□

□5□□

――――――

0

“Arithmetical Restorations” Competition: Ninth Venue

I ended up being scolded after all.
Apparently because my unnecessary chatter diminished your interest...
In that case, covering your ears to my prattle should suffice to achieve our purpose... No—I shall refrain from complaints.

We had now arrived at the Ninth Venue.
Let us tackle problem (5) and fully savor the flavor of Arithmetical Restorations.

(1)★

45□7□

23□1

□2931

+ □246

―――――――――

104196

(2)★

57

＿＿＿＿

3) A7A

(3)★

1ABC57

× 3

―――――――――

ABC571

(4)★

□3

＿＿＿＿

□□)949

□□

————

□□9

□□9

————

0

(5)★

BQM

＿＿＿＿＿＿＿

XVS)XQARAM

XMAB

―――――――

CQA

BBA

―――――

VAM

VAM

――――

0

“Arithmetical Restorations Competition: Tenth Venue”

This marked the Tenth Venue.

This meant you had reached exactly one-third of all thirty venues.
Pass this venue, and you would become a Tenth-tier Selectee.

With full confidence, I trusted that you would pass this venue with ease.
But complacency had to be avoided!
Beyond this lay mountains, valleys, and rapids—if you doubted me, you were to proceed to the next venue.
May your eyes not grow dizzy there.

(1)★

1□92

29□1

971□

+ 2□17

——————

17592

(2)★

□)56□96

￣￣￣￣￣￣

81□8

(3)★

97□

×  □8

———————

□□□0

9□□

———————

175□0

(4)★

3N

＿＿＿＿

N)259

(5)★

□□□

＿＿＿＿＿＿＿

□□□)□□□□□1

□□89

———————

75□

□□□

—————

□81

□81

————

0

**“Arithmetical Restorations Competition: Eleventh Venue”**

**This is the Eleventh Venue.**
Please enjoy at your leisure.
No—I did not intend that as sarcasm.

Mr. Boon’s *Arithmetical Restorations Collection* contains a considerable number of problems and gathers many masterpieces, and I hold it in the highest regard.
Mr. Boon gave these Arithmetical Restorations to his students and made them study arithmetic with great enjoyment.
However, the students, through this opportunity, became thoroughly enamored with Arithmetical Restorations and eventually reached the point of devising problems together with their teacher.
The problems created by such students are displayed at the venues' exhibits.

(1)★

□)81□8

—————

□0□

(2)★

3□AB

×  □B

————————

□6□□A

□□□□

————————

□□□□9

(3)★

56N

N65

6N5

5N6

+ 65N

—————

3240

(4)★

□□□

＿＿＿＿＿＿

□4□)9□□□5

□□□

——————

2□□□

□□□□

——————

2□□

□□□

————

0

(5)★

GQRDQN

×     SLADE

———————————————

GGGTAQN

DERLTT

EALDDTT

ASRELN

LDTNNT

———————————————

ARLDQSQAQN

“Arithmetical Restorations Competition: Venue 12”

This was Venue 12.
Please step inside.

Generally speaking, it was none other than Mr. H.E. Dudeney—the Englishman who had researched and compiled these intriguing problems since antiquity—who first introduced arithmetical restorations and other mathematical puzzles to the world through his publications at an early date.
Of all mathematical puzzles ever devised up to now, those absent from Mr. Dudeney’s works remain so exceedingly rare that one could scarcely name them.

(1)★

7)3□5□6

￣￣￣￣￣￣

□93□

(2)★

□52□

×   3□

————————

□0□8

4□□2

————————

4□7□8

(3)★

A3B

– 3BA

——————

189

(4)★

EAK

＿＿＿＿＿

AE)ETFG

EK

—————

AF

AE

————

KG

KG

———

0

(5)★

2□□□

＿＿＿＿＿＿＿＿

1□5□)□□□□□□6

□7□□

————————

□□□□

5□□□

—————————

□□□□

8□□□

——————

□□□□□

□□□5□

——————

0

“Arithmetical Restorations Competition: Thirteenth Venue”

From this venue onward, the number of problems was reduced to four.
Although reduced by one from previous venues, their quality became significantly more challenging—I trust they will provide ample substance for your deductive faculties.

“They’re way too tough!”
“Ah, is that so? Is that so.”

“But since you have already attained the rank of Twelfth-tier Selectee, I thought it would be a pity not to provide challenges of this caliber—hence, as you can see…”

(1)★

□□7□□

□10□9

□006

+ □9

———————

2□□□□1

(2)★

285

× □□

—————————

1□2□

□□□

————————

□9□□

(3)★

□7□

＿＿＿＿＿＿

□□)17424

□□

——————

□52

□□□

—————

□□4

□□4

————

0

(4)★

ＰＬＰＲ

＿＿＿＿＿＿＿

QSV)TMBLRD

SBT

———————

RSL

QSV

——————

VPR

SBT

―――――

LDDD

LBLD

―――――

LBM

“Arithmetical Restorations Competition: Fourteenth Venue”

The Fourteenth Venue is here.
Now we’re getting to the real challenge!

(2) was an unprecedented fraction problem.
When encountering something unfamiliar, one might feel momentarily flustered.
That very moment of fluster makes for quite a satisfying sensation.
Wouldn’t you agree?

(3) proved equally fascinating, while the cryptarithmetic in (4) bore clear marks of meticulous craftsmanship—do observe how the numerals 3 and 4 glimmer like the Sphinx’s piercing gaze…

(1)★

4□□

×  □□

———————

19□□

3□0□

———————

□□0□□

(2)★

□□□□

5/□□□ = 9

□□□

Note: —
Insert digits from 1 through 9 (note that 5 and 9 have already been used).

(3)★

□□□
＿＿＿＿＿＿
□□□)□□□□5

742
——————
□□□□

1113
—————
□□□□

□□□□

————

0

(4)★

PXEP

＿＿＿＿＿＿＿＿

AZH)AZJAJJM

AYG3

——————

AMXJ

AMMZ

———————

4XJ

MMJ

—————

AYGM

AYGM

—————

0

Arithmetical Restorations Competition: Fifteenth Venue

This is the Fifteenth Venue.
It marks exactly the halfway point of this long journey.

(1) serves as a warm-up.
(2) won’t exactly have you cutting through a Gordian knot—though it unravels neatly in one go.
(3) provides some light entertainment for you.
(4) proves quite taxing indeed.
It takes at least three hours; two hours would be considered swift work by most standards. Some solvers report spending half a day on it—meticulous ones even two full days!
(After solving it, kindly consult page 116 of this volume—pardon my interruption.)

(1)★

□7□

×  □5

———————

□□9□

1□56

———————

16□□□

(2)★

9)NNN

￣￣￣￣

N7

(3)★

BTS

＿＿＿＿＿＿

PQA)TSBTS

CQB

——————

ASQT

ADQT

—————

AQQS

AQQS

————

0

(4)★

□7□□□

＿＿＿＿＿＿＿＿＿

□□□)□□□□□□□□

□□□□

———————

□□□

□□□

——————

□□□□

□□□

——————

□□□□

□□□□

————

0

“Arithmetical Restorations Competition: Sixteenth Venue”

“The Sixteenth Venue awaits here.”
“Come now—pray enter without delay.”
“Esteemed 15th-tier Selectees.”

We glossed over problems (1) and (2), resolved first to tackle the divisor’s tens place in (3), and marveled at how (4) multiplied symmetrical counterparts while pointedly rejecting symmetry in its solution...

The chatterbug stirred once more.

“This is quite a predicament. Shall we step back?”

(1)★

□0□

× □6

———————

5□2□

□□□6

———————

4□58□

(2)★

3A)1B3A

—————

B1

(3)★

□□□

——————

A□A)□□A□A

□□6

——————

□□□

□□8

—————

□□□

□□□

————

0

(4)★

ABCBA

× ABCBA

————————————

ABCBA

BXZXB

CZ G Z C

ＢＸＺＸＢ

ABCBA

————————————

ＡＮＡＰＪＬＪＸＡ

“Arithmetical Restorations Competition” Seventeenth Venue

This was the Seventeenth Venue.

At the previous venue, you were told, “Once you become a Sixteenth-tier Selectee, the problems just don’t feel challenging enough, do they?”

That was too bad for you.
The previous venue had been designed as a restorative break after crossing the halfway point, but since you remained so spirited, very well—so be it.
Next came four problems I had prepared; please savor them thoroughly.
However, as there was no guarantee the clock’s hands wouldn’t creep into tomorrow’s markings, I suggested setting an alarm before embarking.

(1)★

□□Ｅ７

- NNN

———————

EEE

(2)★

□□□□

——————

142)123□□□□

□□□□

————————

□□□

□□□

—————

100□

□□□

—————

□□□

□□□

———

0

(3)★

NGDI

—————————

TSD) WNITPA

TSD

—————————

THTP

ＴＷＮＮ

——————

PPNA

NADI

—————

TPT

(4)★

□□□

＿＿＿＿＿＿＿

□□□)□□□□84

□□□

———————

□□□

6□□

————

□3□

□□□

———

0

Arithmetical Restorations Competition: Eighteenth Venue

“From what I can observe, your eyelids appear rather swollen.”
“Ah, no—that slipped out.”

In our country, the most devoted individual to the study of "mathematical puzzles" was Mr. Fujimura Kousaburou, who resided in Osaka. The deity of puzzles had long been devoted to this endeavor, with an overwhelming number of related works—at least three or four volumes to my recollection—and furthermore, various game designs applying these mathematical puzzles had been devised and made commercially available. Particularly, *3D Shogi* stood as the masterpiece of masterpieces, and it was said to have caused Kimura Meijin to groan mightily.

(1)★

9)□□□□□□

—————————

6) 3925□ Remainder 4

—————————

□□□□ Remainder 3

(2)

□□□ 2
————— = —
□□□□ 4

Note: —

You must fill in the nine digits from 1 to 9 (note that 2 and 4 have already been used).

(3)★

IOR

＿＿＿＿＿＿＿

ACR)IDGDRR

ACR

—————————

COOR

SEOO

—————

UDSR

OOIR

————

RID

(4)★

□□7□

_________________

□□□□)□□9□□□□□

□□□□9

—————————

9□□□□

□□9□□

—————————

□□□□9

□□□□9

——————————

□9□□□

□□□□□

—————————

0

“Arithmetical Restorations” Competition: 19th Venue

Why is it that mathematical puzzles thrive only in Britain? It is quite an intriguing matter.
The five Western reference books I listed in the preface all belong to British publications.

In America, one seldom encounters such dreary material.
To be fair, it isn’t that America has none at all—there are indeed some—but they lack the academic gravitas seen in Britain.
In America, magic tricks and experiments seem to hold far greater appeal among boys than those puzzles do.

Ah, right!
This is the 19th Venue!

(1)★

8)□□□□□

￣￣￣￣￣￣

7) □□□□ Remainder 7

￣￣￣￣￣￣

772 Remainder 4

(2)★

2 1

1―×4―=□

□ 5

(3)★

AFBEC

________________

ABC)ATTTAAT

ABC

————————

GFT

GCH

——————————

CAA

BDF

—————————

FEA

FAE

————————

CFT

CFT

———

0

(4)

□□□□

________________

□2□)□□2□9□□

□□□□

————————

□□□

□□□

—————

□□□□

□□□0

————————

□7□

□□□

———

0

“Arithmetical Restorations Competition: Twentieth Venue”

There were old Japanese books that contained mathematical puzzles, and there was also a Chinese book on mathematical puzzles that I had acquired long ago in Yokohama’s Nankinmachi; I treasured both, but in the recent upheaval, they vanished without a trace.
It is most regrettable.

Ah, here we are at the Twentieth Venue.
If you pass this venue, you will have completed two-thirds of the entire journey; if you were a golfer, this would be the point where you finally advance to single digits.
Now, now—it would be better for you to puff out your chest after you have solved this venue’s four problems.

(1) ★

□A□

________________

□A□)□□□□□

□□□

——————————

□□BB

□□BA

—————————

□A□□

□□□□

————

0

(2) ★

□□□□□

___________________

□□□)□□□□□□□□

□□□

—————————

□□□

4□□

—————

□□□

3□□

————

□□□□

□4□□

————

0

(3) ★

NINIO

______________________

MARTIN)HARRINGTON

HMGIAN

———————————

HAOIGT

HMGIAN

——————————

MMITLON

LOGATI

—————————

MMGGTN

(4) ★

ADHA

___________________

ACC)ATEHBC

ACC

—————————

GBH

ECB

—————————

TEB

TCA

————

BAC

ACC

———

SW

“Arithmetical Restorations Competition: 21st Venue”

“It should have been visible by now, but it remains entirely unseen—what could be the matter?”
“No, it’s just that the figures of our honorable 20th-tier Selectees still haven’t quite...”
“Could it be that the bridges were washed away by a typhoon? Or perhaps they encountered an atomic bomb en route…”

Mr. Barnard, whom I introduced in the preface, was a Master of Science who served as an assistant teacher at Abbotsholm School.

“Oh! They’ve finally come into view!”
“What in the world happened to you all?”
“I was worried about you all.”
“What? Where is this place?”
“Oh my, are you all so out of breath?”
“This is the 21st Venue, but…”

(1) ★

□□7□□

× □7□

———————————

□0□7□

□□0□□

７□□０□

——————————

□□□□□□□

(2) ★

AM6

× MN

——————————

5FN

2DM

——————————

3F2N

(3) ★

□□□

＿＿＿＿＿＿＿

523)□□□667

□□□□

—————————

□□□6

□□□□

—————————

□□□7

□□□□

――――

193

(4) ★

DCAM

＿＿＿＿＿＿＿＿

AEQ)MBDNUQN

MDUE

————————

BQU

AEQ

——————————

QBDQ

QDCY

——————————

CNQN

CNQN

————————

0

“Arithmetical Restorations Competition: 22nd Venue”

This was indeed the 22nd Venue.

“You might say it’s gotten difficult.”
“That’s only natural.”
When we reach the 20th venue and above, we must ask you to strive much harder.
If you remain too carefree, you’ll turn into a fossil midway through. But if you get too worked up, your eyes will turn triangular, and it’ll take real effort to recover later.

However, the more you grapple with challenging problems, the more you come to feel the boundless allure of arithmetical restorations.
Wouldn’t you agree, 21st-tier Selectees?

(1) ★

BAAB

× ABC

—————————

B□□□A

□□□□Ｃ

□□□□□

—————————

□□□□□□□

(2) ★

2□□□

96―――― = 100

□□□

Note: ――
Fit the nine digits from 1 to 9 into [the puzzle] (9, 6, and 2 are already used).

(3) ★

□7□

＿＿＿＿＿＿

□7)□□□7□

□□

————————

7□7

□□□

————————

5□□

5□□

——————————

0

(4)★

RHUCU

＿＿＿＿＿＿＿＿

HSH)HMHECUT

HSH

————————

MSE

USU

——————————

CCC

ISI

―――――

RCRU

RIRI

—————————

CAT

ISI

―――

RMC

“Arithmetical Restorations Competition: 23rd Venue”

Speaking of Mr. Dudeney mentioned earlier, I shall list here the titles of his four renowned works.

The first was *Canterbury Puzzles and Other Curious Problems*.
Next came *Amusements in Mathematics*.
This book had become extraordinarily famous as the foundational source for mathematical puzzles featured in Japanese magazines and similar publications.
Its original title was *Amusements in Mathematics* (Rest assured—this isn’t a cryptarithmetic conundrum;
no need to moisten your pencil tip).
The third work was *The World's Best Word Puzzles*.
Finally, the fourth—*The Latest Puzzles and How to Solve Them*—appeared in this book’s preface as previously noted.

This marks the entrance to the 23rd Venue.

(1)★

ABCDE

× FBC

—————————

AEDFFB

GGGEFEA

ABCDE

—————————

ＦＨＤＥＥＴＪＢ

(2)

□ □

(□□+□―)=(□□―)

□ □

Note: —

On the left side of =, insert the odd numbers 1,3,5,7,9; on the right side of =, insert the even numbers 2,4,6,8.

(3)★

EMA

＿＿＿＿＿

MA)UEMA

MA

—————————

TM

ＡＳ

———

EMA

EMA

———

0

(4)★

□□□□

＿＿＿＿＿＿＿＿＿

3□□4)□□□□□□□□

□□□96

—————————

□□□□

3□□□

——————

□□□□□

□5□2

——————————

□□□□□

□□□1□

—————————

0

“Arithmetical Restorations Competition” 24th Venue

This is the 24th Venue.

I mentioned briefly earlier about Mr. Boon—who must not be forgotten in Arithmetical Restorations—but this teacher was the head of mathematics at Dulwich University.

The publisher was Mills & Boon Ltd., but the relationship between this Boon and Mr. Boon remained unclear.

Mr. Boon’s classic *Puzzle Papers in Arithmetic* was truly a work to be cherished, but the fact that the “solution method” explanation for the example problem at its beginning contained an error was a touch of charm (the book, page 9).

(1)★

AB7

\- 7BA

——————————

1AB

(2)★

□□□□

×  □□□

——————————

9□□□

19□4□

26□□2

——————————

□□□5□□□

(3)

□□□

＿＿＿＿＿＿＿

628)□□□270

□□□□

—————————

□□□7

□□□□

—————————

□□□0

□□□

——————————

482

(4)★

TCCCAZX

______________________

MSR)MSRTRABRC

MSR

——————————————

TRAB

TZLC

——————————

STBR

TBZR

——————————

SMCC

SSBZ

————

L

“Arithmetical Restorations” Competition: 25th Venue

This way, this way!
At last, the 25th Venue had arrived.
Were you to pass this, only five venues would remain.
Ah—I ought not to have uttered such trivialities.

The Selectees must now be devoting themselves wholeheartedly to these Arithmetical Restorations.
By this stage—to such an extent that I no longer needed to solemnly proclaim praises for Arithmetical Restorations—each Selectee would surely be utterly absorbed: licking pencil tips or scrubbing their heads vigorously with eraser ends.

If you pass this, you shall thereby become a 25th-tier Selectee.
This is the 25th tier!

(1)★

ABC

– CBA

——————————

3BA

(2)

□□ □□ 1

———— = ——— = ——

□□□ □□ 3

Note: ——

Insert the nine digits from 1 to 9 into the □s.

(3)★

QAR

___________________

QTAN)QPTZDN

QTAN

—————————

DQDD

TNPR

——————————

XDPN

XDPN

————

0

(4)★

□5□

________________

□□□□)□55□□5□

□□5□□

————————————

□□□□□

□□□□□

——————————

□□□□

□□□□

——————————

0

“Arithmetical Restorations Competition: 26th Venue”

The 26th Venue stands before you.

At long last, those infernal problems make their grand entrance.
You must fortify your resolve with utmost determination.

Mr. Ackermann, whose name appeared in the works listed in the preface, was a Bachelor of Engineering graduate from the University of London.
The numerous puzzle problems featured in that literature were compilations of those broadcast by Britain’s 2LO station; while arithmetical restorations and cryptarithmetic amounted to a mere four problems, the other hundred or so spanned mathematics, science, and engineering—all quite intriguing—and Mr. Ackermann’s explanations were remarkably straightforward.
I could not help but feel envious of his country’s broadcasting stations and citizens for airing such programs.

(1)★

□7

___________

7□)□775

2□□

————

□□5

□□5

———

0

(2)★

1 1

4――×□―― = 9

□ 7

(3)

□□□

________________

□□□)□80□□□

□□□

—————————

□□□□

8□□□

—————————

□□□

□□□

———

0

(4)★

ＣＷＰＳＱＮＧＭ

× ＭＧＮＱＳＰＷＣ

―――――――――――――――――

ＢＷＢＢＢＢＢＢＣ

ＰＢＮＳＢＤＧＱＷ

ＳＢＳＣＧＳＣＧＰ

ＱＢＢＧＷＭＰＤＳ

NCDPMWGBQ

GCGCPGCPN

ＭＣＳＮＤＢＰＱＧ

ＣＷＰＳＱＮＧＭ

―――――――――――――――――

ＭＧＤＳＰＤＢＣＤＤＤＧＢＣＳＣ

“Arithmetical Restorations” Competition: 27th Venue

As I had mentioned at the outset, there existed various approaches to solving Arithmetical Restorations.
For distinguished Selectees of your caliber, investigating every conceivable solving technique would likely prove most engrossing.

Alongside the Sherlock Holmesian deductive methods we have examined until now, there also stood an approach that extracted systems of multivariate equations from arithmetic notations before tackling them through vigorous algebraic manipulation.
This constituted what we might term an algebraic methodology.

While this algebraic path indeed surpassed Holmesian deduction in precision and efficiency when strictly measured, amateur enthusiasts' interest still gravitated more profoundly toward the great detective's method.
We now arrive at our Twenty-Seventh Venue.

While this method excels in accuracy and speed compared to Sherlock Holmes' approach, I believe amateur enthusiasts will still find the great detective's method more compelling.

This marks the 27th Venue.

(1)★

AB
× BA
—————————
□□1
□1□
—————————
□□□□
(2)★
□□□□

94―――― = 100

□□□

Note: ――
Fit in the nine digits from 1 to 9 (9 and 4 are already used).

(3)★

B□A

＿＿＿＿＿＿＿

Ａ□Ｂ）□□□Ｃ□□

□□□B

—————————

□□BC

□□□□

—————————

□□□Ａ

□□C□

――――

0

(4)★

A7□0A

＿＿＿＿＿＿＿＿＿

□□□）1□□□□□□□

1□□□

―――――――――

□□□

□□□

—————————

100□

9□□

——————————

□□□□

□□□□

――――

0

**Arithmetical Restorations Competition: 28th Venue**

The number of venues had gradually dwindled until only three remained.
I ask that you apply your utmost effort.

Esteemed 27th-tier Selectees.

Once you completely solved all problems, what remained afterward would leave you too desolate to persevere.
Where then could one seek intellectual rapture akin to arithmetical restorations? I surmise there are not few Selectees lamenting this—
I venture to perceive it thus.

Those finding themselves in such straits, I entreat you to begin anew from the start.
Thereupon, contrary to expectation, you shall discover no flavor of reheated fare; rather, these arithmetical restoration problems will astound you with undiminished freshness.

(1)★

AYM8P

×  ARZ

————————————

EGM6A

YPMRK

AYMKP

————————————

ＴＲＴＭＡＲＡ

(2)★

B09

＿＿＿＿＿＿

AB)□0□□3

3□□

―――――

A□3

A□3

———

0

(3)★

ＲＳＲ

＿＿＿＿＿＿

PR)MTVVR

MVR

——————————

KKV

KMD

――――

MVR

MVR

———

0

(4)★

Repeating Decimal

┌───┴───┐

・ ・

□□.□□□□□□□□□□

＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿

□□□□□□）□□□□□□□

□□□□□□

——————————

□□□□□□□

□□□□□□

———————————

□□□□□□ □

□□□□□□ □

———————————

□□□□□ □□

□□□□ □□

———————————

□□□□ □□□

□□□□ □□□

———————————

□□□ □□□□

□□□ □□□□

———————————

□□ □□□□□

□□ □□□□□

———————————

□ □□□□□□

□ □□□□□□

———————————————

□□□□□□

□□□□□□

—————————

□□□□□□

“Arithmetical Restorations Competition: Twenty-Ninth Venue”

“Are you saying you suffered terribly at the previous venue…?”
“My condolences.”
“I scarcely know how to address this.”

“Do you mean to claim you’d underestimated Arithmetical Restorations until now?”
“My, did that problem really leave you so chastened…?”

“Are you inquiring whether I’ve attempted that one?”
“Well, precisely so.”
“Truthfully, I’ve been working on it for some time now but still haven’t reached a solution.”
“It appears my secret is out—that I remain a mere 27th-tier Selectee and not yet a true master of Arithmetical Restorations.”

Now then, this is the Twenty-Ninth Venue.

(1)★

3□

＿＿＿＿＿

□□)2□□9

23□

――――

5□9

5□9

―――

0

(2)

6 1

□――×5―― = □□

7 □

(3)★

XLQAB

× AKLB

———————————

FHALB

LHFBK

XCBCDK

BBQKC

—————————

FFLBQCXB

(4)★

□□7□□

＿＿＿＿＿＿＿＿＿＿＿

□□□□７□）□□７□□□□□□□

□□□□□□

———————————

□□□□□7□

□□□□□□□

——————————

□7□□□□

□7□□□□

――――――――

□□□□□□□

□□□□7□□

————————

□□□□□□

□□□□□□

――――――

0

"Arithmetical Restorations Competition: Thirtieth Venue"

At last, we arrive at the final venue.
Now I ask you to summon every last drop of your reasoning power.
If you can pass here，you will be recognized as a highest-tier selectee in arithmetical restorations.

It’s hard to bid farewell.
Is that so?
What if you were to repeat this book from the beginning...
Ah—do you instead declare your wish to confront fresh problems, new intellectual trials?

Unfortunately，all the materials I have at hand have already been presented，and nothing remains thereafter．
Ah，that’s right．
To those desiring similar stimulation，I recommend perusing *The Black Chamber* by that renowned American cryptanalyst Major Yardley．

Now then, please take your place at the final tee.

(1)★

2N1

4N6

+ 162

——————

829

(2)

□ □

□(□□− + □□−) = 100

□ □

Note: —

The left side of the equals sign must be filled with the nine distinct digits from 1 to 9.

(3)★

3Y□□

＿＿＿＿＿＿

□X)6□X7Y

□Y

——————

XX

7□

————

□□7

□□□

————

□Y

□Y

――

0

(4)★

□□□□.□□□□

＿＿＿＿＿＿＿＿＿＿＿＿

□□□)□□□□□□

□□□

――――――――

□□□

□□□

—————

□□□

□□□

―――――

□□ □

□□ □

―――――――

□□□□

□□□□

―――――

0