Digital Puzzles

The digital puzzles play with the nine figures 1 to 9, and sometimes the nought as well. Dudeney gave them a class to themselves because, as he put it, very little was generally known of the laws behind them beyond the old trick of casting out nines. That trick is the thread through this chapter. Add the digits of a number, add the digits of the result, and so on until one figure is left: Dudeney called this the digital root. It is the remainder the number leaves on division by 9 (with 9 standing for a remainder of nought), because 10 leaves remainder 1 on division by 9, so every power of 10 does too, and a number differs from the sum of its digits by a multiple of 9. His own first example: 15,763,530,163,289 has digit sum 59, then 14, then 5, and since a square always has digital root 1, 4, 7 or 9, the number is not a square.

Unless a puzzle says otherwise, “the nine digits” means 1 to 9, each used exactly once, and a number never begins with a nought. Where fractions or decimals are allowed, the puzzle says so.

No. 76. The Barrel of Beer

No. 77. Digits and Squares

No. 78. Odd and Even Digits

No. 79. The Lockers Puzzle

No. 80. The Three Groups

No. 81. The Nine Counters

No. 82. The Ten Counters

No. 83. Digital Multiplication

No. 84. The Pierrot’s Puzzle

No. 85. The Cab Numbers

No. 86. Queer Multiplication

No. 87. The Number-Checks Puzzle

No. 88. Digital Division

No. 89. Adding the Digits

No. 90. The Century Puzzle

No. 91. More Mixed Fractions

No. 92. Digital Square Numbers

No. 93. The Mystic Eleven

No. 94. The Digital Century

No. 95. The Four Sevens

No. 96. The Dice Numbers

The Barrel of Beer

The wine was sold in the proportion one to two, so the wine sold came to three equal parts and its total is a multiple of 3.

The six barrels hold 119 gallons, which leaves remainder 2 on division by 3. The beer barrel must therefore leave remainder 2 as well. The remainders of 15, 31, 19, 20, 16 and 18 are 0, 1, 1, 2, 1 and 0, and only the 20-gallon barrel qualifies.

That fixes the beer, and the rest must be checked, because a multiple of 3 is necessary but not enough. The wine totals 99 gallons, so the first customer had 33 and the second 66. Among 15, 31, 19, 16 and 18 the only barrels making 33 are 15 and 18, and the other three make 66. The arrangement is unique.

Dudeney’s answer is the same, and it rests on reading the story as saying that all the wine was sold. If the dealer could keep some wine back as well, the puzzle falls apart: there are then seven ways to make the sale, and the beer could be any barrel except the 18. For instance, with beer in the 15-gallon barrel he might sell 18 gallons to one man and 20 and 16 to the other, keeping 31 and 19.

Answer the 20-gallon barrel; 15 + 18 to one buyer, 31 + 19 + 16 to the other

Digits and Squares

Call the top row $n$. The three rows add up to $n + 2n + 3n = 6n$, and since the nine digits add up to 45, the digital roots say that $6n$ leaves the same remainder on division by 9 as 45 does, namely 0. So 9 divides $6n$, which means 3 divides $n$. The bottom row $3n$ must be at most 987, the largest three-figure number with different digits, so $n \le 329$, and $n \ge 123$. That leaves 69 multiples of 3 to try, and they can be run through in a few minutes with the further check that $n$, $2n$ and $3n$ share no digit and contain no nought. Exactly four survive: $$\begin{gather*} 192,\ 384,\ 576; \qquad 219,\ 438,\ 657;\\ 273,\ 546,\ 819; \qquad 327,\ 654,\ 981. \end{gather*}$$ Dudeney’s four are these four, and there are no others.

Answer top rows 219, 273 and 327 (besides 192)

Odd and Even Digits

The crux is that whole numbers cannot do it, and nor can decimals.

Any number has the same remainder on division by 9 as its digit sum. If the odd digits are written as whole numbers, the total therefore leaves the same remainder as $1+3+5+7+9 = 25$, which is 7; the even digits give the remainder of 20, which is 2. The totals can never be equal.

Terminating decimals do not escape. Suppose some decimals on the two sides gave equal totals. Multiply both sides by a power of 10 large enough to clear every decimal point. Each term, such as 7.9, becomes a whole number with the same digits followed by noughts, such as 790, so its digit sum is unchanged, and the argument above applies to the enlarged totals. They would leave remainders 7 and 2 on division by 9, so they are not equal, and neither were the originals.

So a proper fraction is needed, and with one we have $$79 + 5 + \tfrac{1}{3} = 84 + \tfrac{2}{6} = 84\tfrac{1}{3}.$$ (The odd side can also be written $75 + 9 + \tfrac13$.) This is Dudeney’s answer, and it is better than he claimed. He called it the simplest solution; in fact, allowing whole numbers and proper fractions but no decimals, $84\tfrac13$ is the only total that works at all.

The puzzle’s wording forbids only recurring decimals, so terminating ones seem to be allowed, and mixing them with fractions opens five more totals. The smallest is $$.79 + \tfrac{3}{15} = .24 + \tfrac{6}{8} = \tfrac{99}{100},$$ and the others are $1.17$, $1\tfrac{2}{3}$, $4\tfrac{1}{34}$ (for instance $3.5 + \tfrac{9}{17} = 4 + \tfrac{2}{68}$) and $8\tfrac{11}{15}$.

Answer $79 + 5 + \frac13 = 84 + \frac26$, the only total without decimals

The Lockers Puzzle

The crux is a rule Dudeney gives in his answer:

the digit sum of the total is fixed by the figure left out and the number of carries. Write $s(x)$ for the digit sum of $x$. Adding column by column, a column that reaches 10 or more keeps its units digit and carries 1, which loses exactly 9 from the digit total. So if the sum $x + y = z$ has $c$ carries, $s(x) + s(y) = s(z) + 9c$. The nine figures used add to $45 - o$, where $o$ is the one left out, so $$2\,s(z) = 45 - o - 9c.$$

The smallest total. The hundreds figures of $x$ and $y$ are different and not nought, so $z$ is at least 300, and a total in the 300s needs hundreds figures 1 and 2 with nothing carried into the hundreds. The tens figure of $z$ cannot be 0, since two different tens figures cannot add to 0, and a total of 10 would carry into the hundreds. It cannot be 1, 2 or 3, which are already in use, and it cannot be 4: the tens figures would have to add to 4, or 3 with a carry, using figures other than 1, 2, 3 and 4, which is impossible. With tens figure 5, the tens figures of $x$ and $y$ add to 5 or 4 from $\{0, 4, 6, 7, 8, 9\}$, so they are 0 and 4 with a carry from the units. The units then come from 6, 7, 8 and 9 and add to $10 + u$, where $u$ is the units of $z$. The pairs give 13, 14, 15, 15, 16 and 17, so $u$ is 3, 4, 5, 6 or 7, and 3, 4 and 5 are already in use. The least is $u = 6$ from $7 + 9$, and $$107 + 249 = 356.$$ The largest total. A total in the 990s repeats a figure, so the best hope is $98u$, with digit sum $17 + u$. The rule gives $o = 11 - 2u - 9c$. Two carries would make $o$ negative. With one carry, $o = 2 - 2u$, so $u$ is at most 1. With no carries, $u$ is 2, 3, 4 or 5 ($u = 1$ would leave out the 9, which is in use) and the tens figures of $x$ and $y$ must add to exactly 8. Checking each case: for $u = 5$ (leaving out 1) the free figures are 0, 2, 3, 4, 6, 7, the units must be 2 and 3, and no two of 0, 4, 6, 7 make 8; for $u = 4$ (leaving out 3) no two free figures make 4; for $u = 3$ (leaving out 5) the units are 1 and 2 and no two of 0, 4, 6, 7 make 8; for $u = 2$ (leaving out 7) no two free figures make 2. So 981 is the largest, and it is made as $$235 + 746 = 981 \quad\text{or}\quad 324 + 657 = 981.$$ The middle cupboard. The totals 356 and 981 use six figures, so the middle total is made of three of 0, 2, 4 and 7. A search of every such sum shows that only three of the candidates can be made: $134 + 586 = 720$, $134 + 568 = 702$ and $138 + 269 = 407$, each in essentially one way.

In each sum, any column may have its top two figures exchanged, giving 8 layouts of every basic sum. Cupboard A then has 8 layouts, B has $3 \times 8 = 24$ and C has $2 \times 8 = 16$, so there are $8 \times 24 \times 16 = 3072$ ways to number the lockers, exactly as Dudeney counted. His answer agrees throughout.

Answer A: $107+249=356$; B: $134+586=720$ (or 702, or 407); C: $235+746=981$; 3072 layouts

The Three Groups

Casting out nines cuts the field sharply. If $a \times b = c$ uses the nine digits, the digit sums of $a$, $b$ and $c$ add to 45, so the remainders $r$, $s$ and $t$ of $a$, $b$, $c$ on division by 9 satisfy $r + s + t \equiv 0$ and $rs \equiv t$. Hence $rs + r + s \equiv 0$, that is, $$(r+1)(s+1) \equiv 1 \pmod 9.$$ The only pairs of remainders whose product is 1 modulo 9 are $1 \times 1$, $2 \times 5$, $4 \times 7$ and $8 \times 8$, so $(r, s)$ must be one of $(0,0)$, $(1,4)$, $(3,6)$, $(7,7)$ or these reversed. For instance, $12 \times 483$ has remainders 3 and 6, and $4 \times 1738$ has 4 and 1. Only six of the eighty-one pairs of remainders survive, and the rest is patient trial. It yields exactly Dudeney’s nine: $$\begin{array}{lll} 12 \times 483 = 5796 & 42 \times 138 = 5796 & 18 \times 297 = 5346 \\ 27 \times 198 = 5346 & 39 \times 186 = 7254 & 48 \times 159 = 7632 \\ 28 \times 157 = 4396 & 4 \times 1738 = 6952 & 4 \times 1963 = 7852 \end{array}$$ Seven have the two-three-four shape and two the one-four-four shape.

Answer nine solutions, as listed; Dudeney’s list is complete

The Nine Counters

Here Dudeney slipped. His best product, $174 \times 32 = 96 \times 58 = 5568$, is only the third largest. The largest is $$532 \times 14 = 76 \times 98 = 7448,$$ and $584 \times 12 = 73 \times 96 = 7008$ also beats his.

The methodical route is to work from the two-by-two product, since it is squeezed hardest. Four different figures make two two-figure numbers whose product is at most $96 \times 87 = 8352$. So list the products of two two-figure numbers with four different digits from the largest down, and for each ask whether the five figures left over can be arranged as a three-figure number times a two-figure number with the same product. The first product to survive is $76 \times 98$, whose leftover figures 1, 2, 3, 4 and 5 give $532 \times 14$. The surprise is that the winning three-figure number, 532, is built from the five smallest figures: the large figures do more good in the two-by-two sum, where both factors are squeezed into two places.

Altogether there are eleven pairs of equal products, from Dudeney’s example 3634 up to 7448.

Answer $532 \times 14 = 76 \times 98 = 7448$ (Dudeney’s 5568 is not the largest)

The Ten Counters

The crux is that each sum must use exactly five counters.

A number of $m$ figures lies between $10^{m-1}$ and $10^m$, so a product of an $m$-figure and an $n$-figure number is at least $10^{m+n-2}$ and less than $10^{m+n}$. If one sum used four counters its product would be below $10^4$, while a sum of six counters gives at least $10^4$; four against six is impossible, and three against seven or worse is more so. Each side is therefore a one-figure number times a four-figure number, or a two-figure number times a three-figure number.

For the smallest product Dudeney’s idea is the natural one: take the smallest multipliers, 1 and 2, and look for two four-figure numbers from the other eight counters, one double the other, with the smaller as low as possible. Candidates such as 3045 and 3405 fail, and the first that works is $2 \times 3485 = 6970 = 1 \times 6970$. The same product also arises as $2 \times 3485 = 10 \times 697$, so the smallest product has two arrangements, and Dudeney gave one of them.

For the largest, his answer $915 \times 64 = 732 \times 80 = 58{,}560$ is right, and again there is a twin: $915 \times 64 = 8 \times 7320$. That nothing larger can be made, and nothing smaller than 6970, is the verdict of a search through every pair of equal products, not of an argument.

Answer smallest $2 \times 3485 = 1 \times 6970 = 6970$; largest $915 \times 64 = 732 \times 80 = 58{,}560$

Digital Multiplication

Here too Dudeney slipped, at the top end. The same counting as in the previous puzzle shows that one sum uses four figures and the other five: three figures give a product below 1000 and six give at least $10^4$, so the split cannot be more lopsided. Four figures give a product below $10^4$, so the common product has four figures, and its digit sum is at most 36.

Dudeney’s smallest is right: $23 \times 174 = 58 \times 69 = 4002$, with digit sum 6, and it is the only arrangement that low. His largest, $9 \times 654 = 18 \times 327 = 5886$ with digit sum 27, can be beaten: $$7 \times 984 = 56 \times 123 = 6888,$$ whose digits add up to 30. That is the true maximum, reached in this one way only. There is no clever route to either extreme; Dudeney himself said the answer comes only by trial, and the trial here is a full search.

Answer smallest $23 \times 174 = 58 \times 69 = 4002$ (sum 6); largest $7 \times 984 = 56 \times 123 = 6888$ (sum 30, not Dudeney’s 27)

The Pierrot’s Puzzle

The crux is casting out nines again. If $x \times y$ is written with the same digits as $x$ and $y$ together, then the product and the sum $x + y$ have the same digit sum, so $xy \equiv x + y \pmod 9$. Rearranged, this is $$(x - 1)(y - 1) \equiv 1 \pmod 9,$$ and as in the Three Groups the only products of remainders equal to 1 modulo 9 are $1 \times 1$, $2 \times 5$, $4 \times 7$ and $8 \times 8$. So the digital roots of the two factors must be 2 and 2, 3 and 6, 5 and 8, or 9 and 9. For instance, 15 and 93 have roots 6 and 3. Dudeney found the same four classes, and they cut the trials to a small fraction; the rest is a short, patient search. Using the digits 1 to 9 there are exactly six cases: $$\begin{array}{lll} 15 \times 93 = 1395 & 21 \times 87 = 1827 & 27 \times 81 = 2187 \\ 35 \times 41 = 1435 & 9 \times 351 = 3159 & 8 \times 473 = 3784 \end{array}$$ These are Dudeney’s six, though his text prints the second as 1287, a slip for 1827.

The puzzle says only “any four digits”, and nothing in it bars the nought. Allow it and six more cases appear, $21 \times 60 = 1260$, $30 \times 51 = 1530$, $6 \times 201 = 1206$, $6 \times 210 = 1260$, $3 \times 501 = 1503$ and $3 \times 510 = 1530$. All six are the two three-digit cases with a nought slipped in, which is presumably why Dudeney did not count them, but his words do not exclude them. The three-digit claim holds on either reading.

Answer the six listed (Dudeney’s six, with $21 \times 87 = 1827$); twelve if the nought is allowed

The Cab Numbers

The digital roots say at once that both cab numbers are multiples of 3. The two numbers together use digits adding to 45, so $x + y$ is a multiple of 9, and so is the product $xy$, whose digits also add to 45. Writing $y \equiv -x \pmod 9$ gives $xy \equiv -x^2 \pmod 9$, so 9 divides $x^2$, which forces 3 to divide $x$, and then $y$ too. That thins the field, but the largest product is still a matter of search. Dudeney’s answer, $$96 \times 8{,}745{,}231 = 839{,}542{,}176,$$ which he offered with an “I think”, is correct: nothing larger exists. The runner-up is $96 \times 8{,}723{,}451 = 837{,}451{,}296$. Both factors of the winner are indeed multiples of 3, with digit sums 15 and 30.

Answer $96 \times 8{,}745{,}231 = 839{,}542{,}176$, as Dudeney thought

Queer Multiplication

Casting out nines, usually the quickest way to kill a digital puzzle, gives no help here. The eight digits other than 6 add to 39, so the eight-figure number leaves remainder 3 on division by 9, and six times it leaves remainder 0, exactly what a product using all nine digits needs. Nothing stands in the way, and the puzzle is honest trial. Dudeney’s answer, $$6 \times 32{,}547{,}891 = 195{,}287{,}346,$$ is correct, but “far from easy” undersells how many there are: 87 arrangements work, from $6 \times 21{,}578{,}943 = 129{,}473{,}658$ up to $6 \times 98{,}745{,}231 = 592{,}471{,}386$. The trial can be shortened by working from the right, since the last digit of the number fixes the last digit of the product, and the carries then fix each further digit in turn.

Answer $6 \times 32{,}547{,}891 = 195{,}287{,}346$ (one of 87)

The Number-Checks Puzzle

Counting figures settles the shape first. If the factors have $m$ and $n$ figures, the product has $m + n$ or $m + n - 1$ figures, and all three together use ten. The second case would need $2(m + n) - 1 = 10$, which is impossible, so the product has five figures and the two factors share the other five, split one and four or two and three. On the ring, then, the product is a run of five neighbouring discs, which can start at any of ten places, and the remaining run of five is cut in one of four places: forty cases, of which thirty remain once runs starting with the nought are set aside.

Casting out nines thins these further. The ten digits add to 45, so, exactly as in the Three Groups, the remainders $r$ and $s$ of the factors on division by 9 must satisfy $(r + 1)(s + 1) \equiv 1 \pmod 9$. Running through the thirty cases with this check leaves just five multiplications to do, and only one of them works: $$715 \times 46 = 32{,}890,$$ with the groups 7, 1, 5 then 4, 6 then 3, 2, 8, 9, 0 in order round the ring. Here 715 and 46 leave remainders 4 and 1, and $5 \times 2 = 10 \equiv 1$, as required. This is Dudeney’s answer and it is the only one, even if the first and second groups may be any two of the three runs. Reading the ring the other way round gives no solution at all.

Answer $715 \times 46 = 32{,}890$, the only answer

Digital Division

Write the division as $N = kD$. Multiplying a number by $k$ from 2 to 9 adds at most one figure, so if $D$ has $m$ figures then $N$ has $m$ or $m + 1$, and the nine digits force $m = 4$: the divisor has four figures and the dividend five.

Casting out nines then explains a pattern Dudeney noticed. The digits of $N$ and $D$ add to 45, so $N + D = (k + 1)D$ is a multiple of 9. For $k = 8$ this says nothing, since $9D$ always is. For $k = 2$ and $k = 5$ it says $3D$ or $6D$ is a multiple of 9, so $D$ is a multiple of 3. For $k = 3, 4, 6, 7$ and 9 the factor $k + 1$ is 4, 5, 7, 8 or 10, none of them a multiple of 3, so $D$ itself must be a multiple of 9. This is Dudeney’s classification of the digital roots: 9 and 9 for the five cases just named, 3 and 6 in some order for halves and fifths, and anything at all for eighths, which is why eighths have by far the most solutions.

Within these limits the search is short, and it confirms each of Dudeney’s eight smallest pairs: $$\begin{array}{ll} 13458 = 2 \times 6729 & 17469 = 3 \times 5823 \\ 15768 = 4 \times 3942 & 13485 = 5 \times 2697 \\ 17658 = 6 \times 2943 & 16758 = 7 \times 2394 \\ 25496 = 8 \times 3187 & 57429 = 9 \times 6381 \end{array}$$ The numbers of solutions for $k = 2$ to 9 are 12, 2, 4, 12, 3, 7, 46 and 3.

Answer Dudeney’s eight pairs are all correct and all smallest

Adding the Digits

Everything turns on the digit 1, which the small amounts all want. The smallest amount has £1, and then no other part may use a 1. That rules out 10s. to 19s., 1d., 10d. and 11d., and the fractions ¼ and ½, so the fraction is ¾ and the shillings $s$ and pence $d$ are single digits other than 1, 3 and 4. The digit sum is $1 + 3 + 4 + s + d = 8 + s + d$, and its two figures must avoid 0, 1, 3, 4, $s$ and $d$.

Now try the shillings in increasing order. With $s = 2$ the sum is $10 + d$, somewhere from 15 to 19, and always contains a 1. With $s = 5$ the possible sums are 15, 19, 20, 21 and 22; with $s = 6$ they are 16, 19, 21, 22 and 23; with $s = 7$ they are 17, 20, 21, 23 and 24. Each contains a 0, a 1, a 3, a 4 or a repeated figure. With $s = 8$ the pence 2, 5, 6 and 7 give 18, 21, 22 and 23, which fail, but $d = 9$ gives 25, which is clear. So the smallest amount is $$\text{£1 8s. 9¾d.},$$ whose digits 1, 8, 9, 3, 4 add to 25, exactly Dudeney’s answer.

His claim for the largest amount also holds. The digit sum of at least four different digits is at least 10, so it takes two figures, and the fraction takes two more. With the shillings and pence taking at least one each, the pounds have at most three figures, and when they do, all nine digits are in use and the shillings and pence are single figures. The digit sum $10a + b$ then equals $45 - a - b$, so $11a + 2b = 45$, whose only solution in digits is $a = 3$, $b = 6$. The best possible pounds, 987, leave 1, 2, 4 and 5 for the shillings, pence and a fraction without a 3, and the largest choice is 5s. 4½d.

Answer £1 8s. 9¾d., digit sum 25; Dudeney’s largest, £987 5s. 4½d., is confirmed

The Century Puzzle

Write the mixed number as $a + b/c$.

The fraction must be a whole number, $100 - a$, so $b = (100 - a)c$, and $a$ has one or two figures. Casting out nines gives a sharp filter. The digits of $a$, $b$ and $c$ add to 45, so $a + b + c \equiv 0 \pmod 9$, and substituting for $b$, with $101 \equiv 2$, $$a + (2 - a)\,c \equiv 0 \pmod 9 .$$ If $a$ leaves remainder 2 on division by 3, then $2 - a$ is a multiple of 3, and reading the condition modulo 3 would make $a$ a multiple of 3, a contradiction; so such whole parts are out at once. For every other $a$ the factor $2 - a$ is prime to 9 and the condition fixes the remainder of $c$. The search that remains is modest. It gives exactly eleven ways, as Dudeney said: $$\begin{array}{llll} 96 + \frac{2148}{537} & 96 + \frac{1752}{438} & 96 + \frac{1428}{357} & 94 + \frac{1578}{263} \\[4pt] 91 + \frac{7524}{836} & 91 + \frac{5823}{647} & 91 + \frac{5742}{638} & 82 + \frac{3546}{197} \\[4pt] 81 + \frac{7524}{396} & 81 + \frac{5643}{297} & 3 + \frac{69258}{714} & \end{array}$$ The odd one out is $3 + \frac{69258}{714}$, where the fraction is 97. It passes the filter: with $a = 3$ the condition reads $3 - c \equiv 0$, and 714 does leave remainder 3.

Answer $3 + \frac{69258}{714}$; eleven ways in all, as Dudeney said

More Mixed Fractions

The same filter applies to any target $N$: with $N = a + b/c$ and $b = (N - a)c$, the digits give $a + (N - a + 1)c \equiv 0 \pmod 9$. It prunes the search, but it does not settle which targets are impossible; for that there is no substitute for trial, as Dudeney said. The trial shows that 15 and 18 cannot be done, and that 16, 20 and 27 can each be done in exactly one way, so Dudeney’s remark that he had found only one arrangement for these three is the whole truth. His answers for the other numbers are all correct: $$\begin{array}{lll} 13 = 9 + \frac{5472}{1368} & 14 = 9 + \frac{6435}{1287} & 16 = 12 + \frac{3576}{894} \\[4pt] 20 = 6 + \frac{13258}{947} & 27 = 15 + \frac{9432}{786} & 36 = 24 + \frac{9756}{813} \\[4pt] 40 = 27 + \frac{5148}{396} & 69 = 65 + \frac{1892}{473} & 72 = 59 + \frac{3614}{278} \\[4pt] 94 = 75 + \frac{3648}{192} & & \end{array}$$ The numbers of ways for 13, 14, 36, 40, 69, 72 and 94 are 2, 2, 3, 7, 3, 7 and 8. Dudeney got round 15 and 18 with a fraction stacked on a fraction, and rightly said it was fairer to keep to the plain form, under which they are impossible.

Answer 15 and 18 are impossible; 16, 20 and 27 have one way each; the rest as listed

Digital Square Numbers

A square using 1 to 9 once has nine figures, between 123,456,789 and 987,654,321, so its root lies between 11,112 and 31,426.

Its digits add to 45, so it is a multiple of 9, and therefore its root is a multiple of 3. That leaves under seven thousand roots to square, a long afternoon by hand in 1917 and an instant by machine. There are thirty such squares, and Dudeney’s two extremes are right: $$11{,}826^2 = 139{,}854{,}276, \qquad 30{,}384^2 = 923{,}187{,}456.$$

Answer smallest $11{,}826^2 = 139{,}854{,}276$; largest $30{,}384^2 = 923{,}187{,}456$

The Mystic Eleven

The crux is the test for 11. Since $10 \equiv -1 \pmod{11}$, a number is divisible by 11 exactly when the digits in the odd places (first, third, and so on) and those in the even places have sums differing by a multiple of 11. Dudeney makes the same point: the difference need only be a multiple of 11, not nought.

The largest. The best conceivable start is 98765, so try it. The last four places, sixth to ninth, take four of 0, 1, 2, 3, 4. The odd places so far hold $9 + 7 + 5 = 21$ and the even places $8 + 6 = 14$, so the seventh and ninth figures must beat the sixth and eighth by 4 (or fall short by 7, which four small digits cannot manage, as the sixth and eighth would need 3 and 4 and the others would have to add to nought). A surplus of 4 comes from 3 and 4 against 1 and 2 (leaving out 0), or from 2 and 3 against 0 and 1 (leaving out 4). The first gives the larger sixth figure, and the largest arrangement is $$987{,}652{,}413,$$ with odd places $9 + 7 + 5 + 4 + 3 = 28$ and even places $8 + 6 + 2 + 1 = 17$.

The smallest. The best start is 10234. The odd places hold $1 + 2 + 4 = 7$ and the even places $0 + 3 = 3$, and the last four figures come from 5 to 9, one of them left out. The sixth and eighth must now beat the seventh and ninth by exactly 4 (the other way round, the seventh and ninth would have to win by 7, which needs a pair adding to 18). If the pair sums are $S$ and $S + 4$, the four figures add to $2S + 4$, which is $35$ less the digit left out, so the omitted digit is odd. Leaving out 7 would need $S = 12$ from 5, 6, 8, 9, which fails; leaving out 5 gives 6, 7 against 8, 9; leaving out 9 gives 5, 6 against 7, 8, and this has the smaller sixth figure. The smallest arrangement is $$102{,}347{,}586.$$ Both are Dudeney’s answers.

Answer largest 987,652,413; smallest 102,347,586

The Digital Century

The crux is that long numbers do the work, so few signs are needed.

With one sign the expression is two numbers joined by a single operation, and none of the eight ways of cutting 123456789 in two gives 100 by adding, subtracting, multiplying or dividing; a leading minus does not help. With two signs, three numbers and two operations, a search of every cut and every pair of operations still finds nothing that reaches 100, and brackets are no use with a single operation, so at least three signs are needed.

Three signs do suffice, and the answer Dudeney gave as his best, $$123 - 45 - 67 + 89 = 100,$$ uses three signs and four strokes. He said only that he did not think it would ever be beaten. It cannot be: among all expressions with three signs, brackets included, it is the only one that makes 100, so it is best on both counts. (The one three-sign form the search below does not list, a minus in front of a bracketed pair, $-(A - B)$, would need a tail of the figures to exceed the head by exactly 100, which never happens.)

Answer $123 - 45 - 67 + 89 = 100$: three signs, four strokes, and the only three-sign answer

The Four Sevens

The trick is the decimal point, which turns a 7 into a tenth of itself: $7 \div .7 = 10$.

Two of those multiplied together give Dudeney’s answer, $$\frac{7}{.7} \times \frac{7}{.7} = 10 \times 10 = 100,$$ and, as he says, any figure would serve in place of 7.

The decimal point allows more than he mentions. Allowing the sevens to be run together into 77 and .77 as well, a search through every arrangement of four sevens with the four operations and brackets finds eighteen expressions equal to 100. Most are Dudeney’s answer regrouped, such as $7 \times 7 \div (.7 \times .7)$. Three are different in kind: $$\frac{77}{.77} = 100, \qquad \frac{77 - 7}{.7} = 100, \qquad \frac{7}{.77 - .7} = 100 .$$ The first is perhaps the neatest of all, and like Dudeney’s it works with any figure in place of the 7.

Answer $\tfrac{7}{.7} \times \tfrac{7}{.7}$, or $\tfrac{77}{.77}$

The Dice Numbers

The crux is symmetry of position. Take any four different figures $a$, $b$, $c$, $d$. They can be arranged in 24 orders, and in those 24 numbers each figure stands in each of the four places exactly 6 times. So the units column adds up to $6(a+b+c+d)$, and so do the tens, hundreds and thousands, and the 24 numbers together make $$6(a+b+c+d)(1000 + 100 + 10 + 1) = 6666\,(a+b+c+d).$$ The dice can show the seven figures 1, 2, 3, 4, 5, 6 and 9, and since a 6 and a 9 on different dice are different figures, any four of the seven may be used together: $\binom74 = 35$ sets. Each figure belongs to $\binom63 = 20$ of them, so the figure-sums of all 35 sets add up to $20 \times (1+2+3+4+5+6+9) = 600$. The grand total is $6666 \times 600 = 3{,}999{,}600$, as Dudeney says, and adding all 840 numbers one by one confirms it.

Dudeney went on to the nine figures 1 to 9 in general. His table of how many sets of four different figures have each sum (one set for 10 and 11, two for 12, three for 13, and so on up to 11 for 18, 19, 21 and 22 and 12 for 20) is correct, and so is his total of 16,798,320 for all numbers of four different non-zero figures. The same argument gives that at once: each figure lies in $\binom83 = 56$ sets, the sets’ figures add up to $56 \times 45 = 2520$, and $6666 \times 2520 = 16{,}798{,}320$. His formula for $n$ figures, $5(10^n - 1)\,8!/(9-n)!$, follows the same way, and agrees with direct addition for $n$ up to 6.

Answer 3,999,600