Digits, Bases and Representation

Problems

Solutions

Solution: IOQM 2024, Q3

Look at the pattern: $5^2 = 25$, $5^3 = 125$, $5^4 = 625$, all ending in $25$. One line of algebra shows why. If a number ends in $25$, write it as $100m + 25$; multiplying by $5$ gives $500m + 125 = 100(5m+1) + 25$, which again ends in $25$.

Since $5^2$ ends in $25$, so does every higher power, and in particular $5^{2024}$. The last two digits are $25$.

Answer 25

Solution: IOQM 2026, Q4

If the numbers have tens digits $a_1, a_2, a_3$ and units digits $b_1, b_2, b_3$, their sum is $$10(a_1 + a_2 + a_3) + (b_1 + b_2 + b_3).$$ Suppose some units digit $b$ were larger than some tens digit $a$. Swapping the two raises the sum by $10b + a - (10a + b) = 9(b - a) > 0$, so that choice was not the largest. In the best choice every tens digit beats every units digit: the tens digits are $9, 8, 7$ and the units digits are $6, 5, 4$.

That gives $$N = 10 \cdot 24 + 15 = 255,$$ for instance as $96 + 85 + 74$, and the sum of the digits of $N$ is $2 + 5 + 5 = 12$.

Answer 12

Solution: PRMO 2014, Q2

Apply the rule twice: $$2014 \mapsto 2^3 + 0^3 + 1^3 + 4^3 = 73, \qquad 73 \mapsto 7^3 + 3^3 = 343 + 27 = 370.$$ So the third term is $370$. Now apply it once more: $$370 \mapsto 3^3 + 7^3 + 0^3 = 27 + 343 + 0 = 370.$$ The sequence has reached a fixed point and stays there. Since the $2014$th term is far beyond the third, $$x_{2014} = 370.$$

The three-digit numbers equal to the sum of the cubes of their digits are exactly $153$, $370$, $371$ and $407$, and $1$ is a fixed point too. A sequence of this kind always falls into one of those or into a short cycle. It is worth knowing that the descent is quick: any four-digit number maps to at most $4 \times 9^3 = 2916$, so the sequence is trapped in a small range almost immediately.

Answer 370

Solution: PRMO 2015 Part A, Q6

The number $100$ contributes nothing, since its only non-zero digit is odd, so the sum is really over $1$ through $99$. It is convenient to pad these to two digits and include $00$ as well, writing every number from $00$ to $99$. The extra entry $00$ contributes nothing either, and now the list is completely regular: the units digit runs through $0,1,\ldots,9$ exactly ten times, and so does the tens digit.

The even digits are $0, 2, 4, 6, 8$, whose sum is $20$. Each digit position therefore contributes $10 \times 20 = 200$ to the total, and there are two positions, so $$E(1) + E(2) + \cdots + E(100) = 200 + 200 = 400.$$

The move that made this easy was refusing to walk through the numbers one at a time. Whenever a question sums a digit-based function over a full block like $1$ to $100$, padding to a fixed width and counting one column at a time is almost always the shorter road.

Answer 400

Solution: PRMO 2015 Part A, Q7

Write $N = 10a + b$ with $a$ from $1$ to $9$ and $b$ from $0$ to $9$. The reversed number is $10b + a$, and $$N + (10b + a) = 11a + 11b = 11(a+b).$$ For this to be a perfect square, the prime $11$ must appear to an even power, so $11$ must divide $a + b$. Since $a + b$ is at most $9 + 9 = 18$ and is positive, the only possibility is $a + b = 11$, which indeed gives $11 \times 11 = 121 = 11^2$.

It remains to count the two-digit numbers whose digits sum to $11$. Choosing $a$ determines $b = 11 - a$, and we need $b$ to be a digit, so $11 - a \le 9$, that is $a \ge 2$. Every $a$ from $2$ to $9$ works, giving $29, 38, 47, 56, 65, 74, 83, 92$, which is $8$ numbers.

Answer 8

Solution: PRMO 2018, Q3

Write the number out by place value: $$\overline{abccba} = 100001a + 10010b + 1100c.$$ Reduce each coefficient modulo $7$. Since $7 \times 14285 = 99995$, we get $100001 \equiv 6$; since $7 \times 1430 = 10010$ exactly, the middle coefficient vanishes; and since $7 \times 157 = 1099$, we get $1100 \equiv 1$. So $$\overline{abccba} \equiv 6a + c \pmod 7.$$

Divisibility by $7$ therefore means $c \equiv -6a \equiv a \pmod 7$, and $b$ is free to be any of the five odd digits.

Count the pairs $(a,c)$ with $a$ from $1$ to $9$, $c$ from $0$ to $9$ and $c \equiv a \pmod 7$. For a given $a$ the admissible $c$ are those in $\{0,\ldots,9\}$ congruent to $a$, and there are two of them when $a \equiv 0$, $1$ or $2 \pmod 7$ and one otherwise: $$\begin{array}{c|ccccccccc} a & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \#c & 2 & 2 & 1 & 1 & 1 & 1 & 2 & 2 & 2 \end{array}$$ That is $14$ pairs. Multiplying by the $5$ choices of the odd digit $b$, $$14 \times 5 = 70.$$

Answer 70

Solution: PRMO 2018, Q4

In base $b$ the two factors are $166_b = b^2 + 6b + 6$ and $56_b = 5b + 6$, and the product is $8590_b = 8b^3 + 5b^2 + 9b$. Multiplying out the left-hand side, $$(b^2 + 6b + 6)(5b + 6) = 5b^3 + 36b^2 + 66b + 36.$$ Setting the two equal and collecting terms, $$3b^3 - 31b^2 - 57b - 36 = 0.$$

Only three bases need testing. The digits appearing are $1$, $6$, $5$, $8$, $9$ and $0$, so $b \ge 10$; and rearranging the cubic as $b\left(3b^2 - 31b - 57\right) = 36$ shows that a whole number $b$ must divide $36$. The divisors of $36$ that are at least $10$ are $12$, $18$ and $36$.

Try $b = 12$: $3 \times 1728 - 31 \times 144 - 57 \times 12 - 36 = 5184 - 4464 - 684 - 36 = 0$, so $12$ is a root. Dividing it out, $$3b^3 - 31b^2 - 57b - 36 = (b - 12)(3b^2 + 5b + 3),$$ and the quadratic factor has discriminant $25 - 36 < 0$, so it has no real root at all. The only base is $b = 12$, which is at least $10$ and larger than every digit used, so the sum of all possible values is $12$.

Answer 12

Solution: PRMO 2019, Q6

We need digits $a, b, c$ from $1$ to $9$ with $a^2 + b^2 = c^2$. There is no need to know anything about Pythagorean triples: the only squares in play are $$1,\ 4,\ 9,\ 16,\ 25,\ 36,\ 49,\ 64,\ 81,$$ and ask which is the sum of two others. The following table tests $a<b<c$; checking $a^2<c^2/2$ makes each unordered pair appear once: $$\begin{array}{c|l} c& c^2-a^2\quad(1\le a<c/\sqrt2)\\\hline 2&3\\ 3&8,5\\ 4&15,12\\ 5&24,21,16\\ 6&35,32,27,20\\ 7&48,45,40,33\\ 8&63,60,55,48,39\\ 9&80,77,72,65,56,45 \end{array}$$ The only square in the right-hand column is $16=4^2$, in the row $c=5$, with $a=3$. Equal positive legs cannot work, since $2a^2=c^2$ would have an odd exponent of $2$ on the left and an even one on the right. Thus the only possibilities are $(a,b,c)=(3,4,5)$ and $(4,3,5)$. So $\overline{abc}$ is $345$ or $435$. Factorising, $$345 = 3 \times 5 \times 23, \qquad 435 = 3 \times 5 \times 29.$$ The largest prime factor appearing is $29$.

Answer 29

Solution: IOQM 2020, Q2

The whole difficulty of a problem like this is remembering what a string of digits actually means. Writing $503$ in base $b$ is shorthand for $5b^2 + 0 \cdot b + 3$, and writing $305$ in base $b+2$ is shorthand for $3(b+2)^2 + 0 \cdot (b+2) + 5$. Both expressions describe the same number $N$, so we may set them against each other: $$5b^2 + 3 = 3(b+2)^2 + 5 = 3b^2 + 12b + 17.$$

Collecting terms leaves $2b^2 - 12b - 14 = 0$, and dividing through by $2$ gives $b^2 - 6b - 7 = 0$, which factors cleanly as $(b-7)(b+1) = 0$. Only one of these roots can be a base, since a base must be positive and must also exceed every digit written in it, and the digit $5$ appears. That settles $b = 7$.

From there the arithmetic is short. We have $N = 5 \cdot 49 + 3 = 248$, and as a final check it is worth confirming that $248$ really is $305$ in base nine, which it is, since $3 \cdot 81 + 5 = 248$. The product of the digits of $N$ is therefore $2 \cdot 4 \cdot 8 = 64$.

Answer 64

Solution: IOQM 2024, Q8

The printed word “integer” leaves the domain implicit. We use the intended reading that $n$ is positive. If digit sums of negative integers were defined by ignoring their minus signs, there would be no smallest $n$: every positive solution would give a negative one via $n\mapsto-n-1$, and the positive constructions below continue without bound.

The carry can be seen in $2399+1=2400$: two nines disappear and the preceding digit increases by one, so the digit sum changes by $1-2\cdot9$. Now suppose $n$ ends in exactly $k$ nines. Adding $1$ turns those $k$ nines into zeros and increases the preceding digit by $1$, so $$S(n+1) = S(n) - 9k + 1.$$ Both digit sums are to be multiples of $5$, so $9k - 1 \equiv 0 \pmod 5$, that is $4k \equiv 1 \pmod 5$, which gives $k \equiv 4 \pmod 5$. In particular $k = 0$ is impossible, and the smallest allowed value is $k = 4$.

So $n$ ends in four nines: write $n = 10^4m + 9999$ with $m \ge 0$ and the last digit of $m$ not equal to $9$. Then $$S(n) = S(m) + 36,$$ and requiring $5 \mid S(n)$ gives $S(m) \equiv 4 \pmod 5$. This already rules out $m = 0$, which would give $n = 9999$ with digit sum $36$, not a multiple of $5$. We want the smallest such $n$, hence the smallest such $m$, and the digit sums of $1, 2, 3, 4$ are $1, 2, 3, 4$, so $m = 4$ is the first that works.

Therefore $n = 49999$, and a check confirms it: $S(49999) = 40$ and $S(50000) = 5$, both divisible by $5$. The first two digits are $4$ and $9$.

Answer 49

Solution: IOQM 2025 Part SEP, Q7

Once $a$ and $b$ are chosen, $c = a+b$ is forced, and the only requirement is that $c$ really is a digit, that is $a + b \le 9$. Also $a \ge 1$ because the number has three digits.

For each $a$ from $1$ to $9$ the digit $b$ may be anything from $0$ to $9-a$, which is $10-a$ choices. Summing, $$9 + 8 + 7 + \cdots + 1 = 45.$$

Answer 45

Solution: IOQM 2026, Q9

A number with $k$ digits has digit sum at most $9k$. Since $2026 = 9 \cdot 225 + 1$, a number with $225$ digits reaches at most $2025$, so $N$ has at least $226$ digits, and a $226$-digit number is smaller than any longer one.

Among the $226$-digit numbers with digit sum $2026$, the smallest is the one with the smallest leading digit. The other $225$ digits add up to at most $2025$, so the leading digit is at least $2026 - 2025 = 1$, and it equals $1$ only when the other digits are all $9$. So $$N = 1\underbrace{99\ldots9}_{225}, \qquad N + 1 = 2\underbrace{00\ldots0}_{225},$$ and the leading digit of $N + 1$ is $2$.

Answer 2

Solution: IOQM 2026, Q14

Write $A = 100a + 10b + c$, so that $B = 100c + 10b + a$. Then $$A - B = 99a - 99c = 9 \cdot 11 \cdot (a - c).$$ The digits are distinct and non-zero, so $|a - c|$ is one of $1, 2, \ldots, 8$, whose prime factors are among $2$, $3$, $5$ and $7$. Every $|A - B|$ is therefore $11$ times a number whose prime factors are at most $7$, and the largest prime factor is $11$, for instance $|123 - 321| = 198 = 2 \cdot 9 \cdot 11$.

Answer 11

Solution: PRMO 2013, Q6

To make a number small you want few digits, so you want every digit as large as possible. With digit sum $2013 = 9 \times 223 + 6$, that means $223$ nines and a leading $6$: $$N = 6\underbrace{99\cdots9}_{223} = 7 \times 10^{223} - 1.$$ Nothing with $223$ digits or fewer can reach a digit sum of $2013$, since $9 \times 223 = 2007$, and with $224$ digits the leading digit is at least $2013-9\cdot223=6$. Equality forces all the remaining digits to be $9$, so this really is the smallest.

Now compute, using the closed form: $$5N + 2013 = 35 \times 10^{223} - 5 + 2013 = 35 \times 10^{223} + 2008.$$ The first term is $35$ followed by $223$ zeros, and adding $2008$ only fills in the last four of those zeros, with no carrying at all. So the digits are $3$, $5$, then a long run of zeros, then $2$, $0$, $0$, $8$, and $$S(5N + 2013) = 3 + 5 + 2 + 0 + 0 + 8 = 18.$$

Answer 18

Solution: PRMO 2013, Q20

A bit is a binary digit, $0$ or $1$. For example, $001011$ represents $8+2+1=11$ and has three ones; leading zeros let every number below $64$ use six positions. A natural number below $64$ is therefore a six-bit string, and we want those with exactly three ones, of which there are $\binom63 = 20$.

Rather than list them, count each bit’s contribution. A given position is a one in exactly those strings whose other two ones are chosen from the remaining five positions, that is $\binom52 = 10$ of them. So each position contributes its place value ten times, and the total is $$10 \times (1 + 2 + 4 + 8 + 16 + 32) = 10 \times 63 = 630.$$

The same argument gives the sum for any number of ones: with $j$ ones among $m$ bits the total is $\binom{m-1}{j-1}\left(2^m - 1\right)$.

Answer 630

Solution: PRMO 2015 Part A, Q20

Let the digits be $d+3, d+2, d+1, d$ read from the left, where $d \ge 0$ and $d + 3 \le 9$, so $d$ runs from $0$ to $6$ and there are seven such numbers. Their values are $$n = 1000(d+3) + 100(d+2) + 10(d+1) + d = 1111d + 3210.$$

Both constants behave well modulo $37$. Since $37 \times 30 = 1110$, we have $1111 \equiv 1$, and since $37 \times 86 = 3182$, we have $3210 \equiv 28$. Therefore $$n \equiv d + 28 \pmod{37}.$$ As $d$ runs from $0$ to $6$ the right-hand side runs from $28$ to $34$, all of which are already between $0$ and $36$, so they are the remainders themselves and they are seven distinct values. Their sum is $$28 + 29 + 30 + 31 + 32 + 33 + 34 = \frac{7(28+34)}{2} = 217.$$

The reason $37$ appears rather than some other modulus is that $37$ divides $111$, and therefore divides $1110$, which is what makes $1111$ leave the convenient remainder $1$. A problem that pairs repdigit-like numbers with the modulus $37$ is nearly always leaning on that fact.

Answer 217

Solution: PRMO 2017, Q1

A number is divisible by $3$ exactly when its digit sum is, so the two conditions together say that the digit sum is divisible by both $7$ and $3$, that is by $21$.

A number below $1000$ has at most three digits, so its digit sum is at most $27$. The only positive multiple of $21$ available is $21$ itself, and the digit sum is positive because the number is.

So we must count the numbers below $1000$ whose digits sum to $21$. Write every such number with three digits, padding with leading zeros, which changes nothing: a number with fewer digits has digit sum at most $18$, so none is lost. We need $$a + b + c = 21, \qquad 0 \le a, b, c \le 9.$$ Replace each digit by its complement, $a' = 9 - a$ and so on. Then $a' + b' + c' = 27 - 21 = 6$ with each $a', b', c' \ge 0$, and the upper bound of $9$ is automatic since the total is only $6$. Counting those is the standard stars and bars: lay out six stars in a row and choose two of the eight positions in the resulting line of stars and bars to be the bars, which cuts the six into three groups. So the number of solutions is $$\binom{6 + 2}{2} = \binom{8}{2} = 28.$$

Answer 28

Solution: PRMO 2018, Q20

For a one-digit example, the product of the digits of $7$ is $7$; for $72$ it is $7\cdot2=14<72$. The leading digit has its full place value in the number, while the digit product only multiplies it by the smaller remaining digits. This suggests bounding the product by the number itself.

Write $P(n)$ for the product of the digits. Two facts bound it. It is at least $0$, and it is at most $n$: for a one-digit number the two are equal, and for a number with more digits the leading digit alone contributes at least $10^{k}$ times itself while the product of the rest is at most $9^k$, so $P(n) < n$.

Applying both to $P(n) = n^2 - 15n - 27$: $$n^2 - 15n - 27 \ge 0 \qquad \text{and} \qquad n^2 - 16n - 27 \le 0.$$ Each quadratic changes sign once in the range of interest, so integer tests settle both without a square root. The first is $-11$ at $n = 16$ and $7$ at $n = 17$, so $n \ge 17$; the second is $-10$ at $n = 17$ and $9$ at $n = 18$, so $n \le 17$. So $n = 17$ is the only candidate, and it works: $P(17) = 7$ and $17^2 - 15 \cdot 17 - 27 = 289 - 255 - 27 = 7$.

The sum of all such $n$ is therefore $17$.

Answer 17

Solution: IOQM 2020, Q8

Problems that describe a number through its digits are usually solved by asking what the digits are permitted to be, rather than by any clever algebra. Here the five digits are $k$, $k+1$, $k+2$, $3k$ and $k+3$, and each of them must be a single digit lying between $0$ and $9$, while the leading digit $k$ must also be non-zero because the number has five digits.

The binding restriction is the fourth digit. Requiring $3k \leq 9$ forces $k \leq 3$, and combined with $k \geq 1$ this leaves only three candidates, which we may write out in full: $$k=1: \; 12334, \qquad k=2: \; 23465, \qquad k=3: \; 34596.$$

Testing each is now a matter of locating it between neighbouring squares. Since $111^2 = 12321$ and $112^2 = 12544$, the first candidate is not a square. Since $153^2 = 23409$ and $154^2 = 23716$, neither is the second. The third is a square exactly, because $186^2 = 34596$.

So $m = 186$, and the sum of its digits is $1 + 8 + 6 = 15$.

Answer 15

Solution: IOQM 2021 Part A, Q4

First rule out a zero among the three digits, since a zero cannot lead and the counting below would not be symmetric. If the digits are $0$, $p$, $q$, then the leading place must carry $p$ or $q$, and there are $2 \cdot 3^5 = 486$ numbers. Each of $p$ and $q$ heads $3^5 = 243$ of them, while in each of the five lower places every one of the three digits appears $2 \cdot 3^4 = 162$ times. The total is therefore $$\bigl(243 \cdot 100000 + 162 \cdot 11111\bigr)(p+q) = 26099982\,(p+q).$$ Since $p$ and $q$ are distinct non-zero digits, $p + q \leq 17$, and the total is at most $26099982 \cdot 17 = 443699694$. That falls short of $593999406$, so none of the digits is zero.

With all three digits non-zero, there are $3^6 = 729$ six-digit numbers built from $a$, $b$ and $c$. Fix one place, say the leading one. Of those $729$, exactly a third have $a$ there, a third have $b$ and a third have $c$, so each digit appears $3^5 = 243$ times in that place, and by symmetry the same is true of every place.

Summing over all six places, each of which is worth its place value, gives $$243 \cdot (a+b+c) \cdot (100000 + 10000 + \cdots + 1) = 243 \cdot 111111 \cdot (a+b+c),$$ which is $26999973\,(a+b+c)$. Setting this equal to $593999406$ and dividing, $a + b + c = 22$. Notice how little the huge number actually told us: only the digit sum, and nothing about which digit is which.

That leaves an optimisation. We want the largest remainder of the three-digit number $\overline{abc}$ on division by $100$, which is just the two-digit ending $\overline{bc}$, so we want $10b + c$ as large as possible subject to $a, b, c$ being distinct digits summing to $22$ with $a \neq 0$. Taking $b = 9$ and $c = 8$ gives $a = 5$, and all three are distinct, so $\overline{bc} = 98$ is attainable. Nothing larger is possible, since $b = 9, c = 9$ would repeat a digit.

Answer 98

Solution: IOQM 2024, Q21

If $\lfloor n/9 \rfloor$ is the three-digit repeated-digit number $\overline{ddd}$, then $$9 \cdot \overline{ddd} \le n \le 9 \cdot \overline{ddd} + 8,$$ so $n \le 9 \cdot 999 + 8 = 9000 - 1$.

The second condition says $\lfloor (n-172)/4 \rfloor$ is a rearrangement of the digits $2, 0, 2, 4$, and every such four-digit number is at least $2024$. Hence $$n - 172 \ge 4 \cdot 2024, \qquad n \ge 8268.$$

Combining, $8268 \le n \le 8999$, and the only repeated-digit multiple in range is $\overline{ddd} = 999$, giving $$8991 \le n \le 8999.$$ For those values $n - 172$ runs from $8819$ to $8827$, so $\lfloor (n-172)/4 \rfloor$ is $2204$, $2205$ or $2206$. Only $2204$ uses the digits $2, 0, 2, 4$.

Finally, $\lfloor (n-172)/4 \rfloor = 2204$ means $8816 \le n - 172 \le 8819$, that is $8988 \le n \le 8991$. Intersecting with the earlier range leaves the single value $$n = 8991,$$ and the remainder on division by $100$ is $91$.

Answer 91

Solution: IOQM 2025 Part SEP, Q28

In the first three brackets, a term $x^5$ must take $x$ from the first and $x^4$ from the third, because $5=4+1$; the middle bracket contributes its constant $3$. The same binary choice controls the much larger exponent in the question.

Each bracket contributes either its power of $x$ or its constant. The powers on offer are $x^{1}, x^{2}, x^{4}, \ldots, x^{1024}$, that is $x^{2^k}$ for $k = 0, 1, \ldots, 10$, and the constant in the bracket with $x^{2^k}$ is $2k+1$.

Choosing a set $S$ of brackets to supply their power of $x$ produces the exponent $\sum_{k \in S} 2^k$, and by the uniqueness of binary representation there is exactly one $S$ giving $2025$. Writing it out, $$2025 = 1024 + 512 + 256 + 128 + 64 + 32 + 8 + 1,$$ so $S = \{0, 3, 5, 6, 7, 8, 9, 10\}$ and the brackets left over are those with $k = 1, 2, 4$.

Their constants are $3$, $5$ and $9$, so $$N = 3 \cdot 5 \cdot 9 = 135,$$ and the remainder on division by $100$ is $35$.

Answer 35

Solution: IOQM 2025 Part SEP, Q28

By place value, $$\overline{ababab} = \overline{ab} \cdot 10101, \qquad 10101 = 3 \cdot 7 \cdot 13 \cdot 37.$$ So the equation reads $$5(2m+1)(2m+3)(2m+5) = 10101\,t, \qquad t = \overline{ab}.$$ Since $5$ divides the left side but not $10101$, the two-digit number $t$ is a multiple of $5$.

Now bound $m$ exactly. At $m = 12$ the left side is $5 \cdot 25 \cdot 27 \cdot 29 = 97875$, still five digits, and at $m = 13$ it is $5 \cdot 27 \cdot 29 \cdot 31 = 121365$. At the other end $m = 27$ gives $5 \cdot 55 \cdot 57 \cdot 59 = 924825$, while $m = 28$ gives $1025715$, which has seven digits. The left side increases with $m$, so $$13 \leq m \leq 27, \qquad \text{and hence} \qquad 27 \leq 2m+1 < 2m+5 \leq 59.$$

Rather than test fifteen values, use the factor $37$: it must divide one of the three consecutive odd numbers, and the only multiple of $37$ between $27$ and $59$ is $37$ itself. So one of $2m+1, 2m+3, 2m+5$ equals $37$, and the three possibilities are $$33 \cdot 35 \cdot 37, \qquad 35 \cdot 37 \cdot 39, \qquad 37 \cdot 39 \cdot 41.$$ Multiplying by $5$ gives $213675$, $252525$ and $295815$, and only the middle one has the required form, $\overline{ababab}$ with $a = 2$ and $b = 5$.

That case is $2m+1 = 35$, so $m = 17$, and $$m + a + b = 17 + 2 + 5 = 24.$$

Answer 24

Solution: IOQM 2025 Part SEP, Q7

Expanding by place value, $$\overline{abcab} = (10a+b)\cdot 1000 + 100c + (10a+b) = 1001M + 100c,$$ where $M = \overline{ab}$.

Two conditions follow. Since $M$ divides $1001M$, divisibility of the number by $M$ says exactly $$M \mid 100c.$$ And since $13 \mid 1001$, the number is congruent to $100c \equiv 9c \pmod{13}$, so it fails to be divisible by $13$ exactly when $13 \nmid 9c$, that is when $c \ne 0$.

We want to maximise the digit sum $2(a+b) + c$, where $a + b$ is the digit sum of $M$. For each value of $c$ from $1$ to $9$, take the two-digit divisor of $100c$ whose own digit sum is largest. The candidate divisors and their best digit sums can be checked row by row: $$\begin{array}{c|l|c|c} c&\text{two-digit divisors of }100c&\text{best }M&2(a+b)+c\\\hline 1&10,20,25,50&25&15 \\ 2&10,20,25,40,50&25&16 \\ 3&10,12,15,20,25,30,50,60,75&75&27 \\ 4&10,16,20,25,40,50,80&80&20 \\ 5&10,20,25,50&25&19 \\ 6&10,12,15,20,24,25,30,40,50,60,75&75&30 \\ 7&10,14,20,25,28,35,50,70&28&27 \\ 8&10,16,20,25,32,40,50,80&80&24 \\ 9&10,12,15,18,20,25,30,36,45,50,60,75,90&75&33 \\ \end{array}$$ Every value of $c$ is there, so nothing is left to check. The winner is $M = 75$ with $c = 9$, giving the number $75975$.

Both conditions check out: $75975 = 75 \cdot 1013$, and $75975 = 13 \cdot 5844 + 3$. Its digit sum is $7+5+9+7+5 = 33$.

Answer 33

Solution: IOQM 2026, Q20

Say $M$ has $k + 1$ digits, and write $M = 3 \cdot 10^k + x$, where $x$ is the number formed by the other $k$ digits. Moving the $3$ to the end gives $N = 10x + 3$, and $M = 4N$ reads $$3 \cdot 10^k + x = 40x + 12, \qquad 39x = 3 \cdot 10^k - 12, \qquad 13x = 10^k - 4.$$

So $10^k$ must leave remainder $4$ on division by $13$. Each remainder is $10$ times the one before, reduced: $$10^1 \to 10, \quad 10^2 \to 9, \quad 10^3 \to 12, \quad 10^4 \to 3, \quad 10^5 \to 4.$$ The first success is $k = 5$, with $x = 99996/13 = 7692$, so $$M = 307692, \qquad N = 76923, \qquad 4 \times 76923 = 307692.$$

The five digits after the $3$ are $0, 7, 6, 9, 2$, so the rearranged six-digit string is $076923$, representing the number $N=76923$. The condition $N = M/4$ holds all the same, and there is no way round it: $x = (10^k - 4)/13$ is less than $10^{k-1}$, so the second digit of $M$ is $0$ for every $k$ that works. The sum of the digits of $M$ is $3 + 0 + 7 + 6 + 9 + 2 = 27$.

Answer 27

Solution: PRMO 2018, Q19

The first few partial sums are $6$, $72$, $738$, $7404$ and $74070$. They suggest a repeating block near the front, so a closed form is useful for exposing the digits of the hundred-term sum.

A closed form

Each term is $\underbrace{66\cdots6}_{k} = \tfrac23(10^k - 1)$, so $$N = \frac23\sum_{k=1}^{100}\left(10^k - 1\right) = \frac23\left(\frac{10^{101} - 10}{9} - 100\right),$$ which tidies up to $N = \tfrac{2}{27}\left(10^{101} - 910\right)$.

The digits

Let $A$ be the number written as thirty-three copies of the block $740$ followed by a final $7$, so $A$ has $100$ digits. Summing the geometric series inside it, $$A = 7400 \cdot \frac{10^{99} - 1}{999} + 7.$$ Multiplying by $27$ and using $199800 = 200 \times 999$, $$27A = 200\left(10^{99}-1\right) + 189 = 2 \cdot 10^{101} - 11.$$ Therefore $$27N = 2 \cdot 10^{101} - 1820 = \left(27A + 11\right) - 1820 = 27A - 1809,$$ and since $1809 = 27 \times 67$, $$N = A - 67.$$

The subtraction touches only the last four digits. $A$ ends in $7407$, and $7407 - 67 = 7340$, so $$N = \underbrace{740\,740\,\cdots\,740}_{32 \text{ blocks}}\,7340,$$ a hundred digits in all.

Counting the sevens

Each of the $32$ blocks contributes one $7$, and the tail $7340$ contributes one more: $$32 + 1 = 33.$$

Answer 33

Solution: PRMO 2018, Q30

Since the coefficients are non-negative integers and $P(1) = \sum a_i = 4$, every coefficient is at most $4$. That is precisely the condition for the digits of a base-five numeral, and $$P(5) = \sum_i a_i 5^i = 136$$ says that the coefficients are the base-five digits of $136$.

Expand $136$ in base five: $136 = 125 + 11 = 1 \cdot 5^3 + 0 \cdot 5^2 + 2 \cdot 5 + 1$, so $$136 = 1021_{\,5}, \qquad P(x) = x^3 + 2x + 1.$$ As a check, the digits do sum to $1 + 0 + 2 + 1 = 4 = P(1)$, as they must, and base-five expansions are unique, so this is the only such polynomial.

Hence $$P(3) = 27 + 6 + 1 = 34.$$

Answer 34

Solution: IOQM 2023, Q19

The digits of $2357$ have product $210$ and sum $17$. Prefixing a $1$ gives $12357$, with the same product and sum $18$. Ones therefore add length without changing the product, which is the resource to exploit.

Since $P(n) \ne 0$, no digit is $0$. Since $P(n)$ is squarefree, no prime may occur twice in the product of the digits. That rules out the digits $4 = 2^2$, $8 = 2^3$ and $9 = 3^2$ outright, and it also means that among the remaining digits $2, 3, 5, 6, 7$ we may use each prime once only. In particular $$P(n) \le 2 \cdot 3 \cdot 5 \cdot 7 = 210,$$ with equality when the non-unit digits are $2, 3, 5, 7$ (or $6, 5, 7$, since $6 = 2 \cdot 3$).

Digits equal to $1$ change nothing in the product and add $1$ each to the sum, so they are the way to make a number long. Suppose the non-unit digits have product $p$ and sum $t$, and there are $k$ ones. Then $$P(n) = p, \qquad S(n) = k + t,$$ and the requirement is that $k+t$ is a proper divisor of $p$. The number of digits is $k + (\text{number of non-unit digits})$, so for a fixed set of non-unit digits we want $k$ as large as possible, that is $k + t$ as large as possible, and the largest proper divisor available is $p$ divided by its smallest prime factor.

Take $p = 210$ with non-unit digits $2, 3, 5, 7$, so $t = 17$ and there are four of them. The largest proper divisor of $210$ is $105$, so $k = 105 - 17 = 88$ and the number has $$88 + 4 = 92$$ digits. The number consisting of $88$ ones followed by $2, 3, 5, 7$ has digit sum $88 + 17 = 105$, digit product $210$, and $105$ is indeed a proper divisor of $210$.

Every other choice is shorter. At $p = 210$ the only other grouping of the digits is $6, 5, 7$, which raises $t$ to $18$ and drops to three non-unit digits, giving $87 + 3 = 90$.

Every smaller $p$ falls far short, and a single bound covers all of them at once. The admissible products are the divisors of $210$, so $p < 210$ means $p \le 105$. Write $D = S(n)$, a proper divisor of $p$, so that $D \le p/2 \le 52$. If there are $r$ non-unit digits with sum $t$, then $t \ge 2r$, since each such digit is at least $2$, and the number of digits is $$(D - t) + r \;\le\; D - 2r + r \;=\; D - r \;\le\; 51.$$ That is nowhere near $92$, so the maximum is $92$.

Answer 92

Solution: IOQM 2023, Q26

Let $f(N)$ be the number of ways to pay $N$ Bens using notes of value $1, 2, 4, 8, \ldots$ with at most two of each. Look at how many one-Ben notes are used. That number has the same parity as $N$ and is at most $2$, so it is forced up to one choice:

If $N$ is odd, exactly one one-Ben note is used, and the rest, $N - 1$, is paid in notes of value at least $2$; halving every note turns this into a payment of $\tfrac{N-1}2$ under the same rules. If $N$ is even, either no one-Ben note is used, giving a payment of $\tfrac N2$ after halving, or two are used, giving $\tfrac{N}{2} - 1$. Hence $$f(2k+1) = f(k), \qquad f(2k) = f(k) + f(k-1),$$ with $f(0) = 1$ and $f(1) = 1$.

A check against the problem’s own examples: $f(2) = f(1) + f(0) = 2$ and $f(5) = f(2) = 2$, both as stated.

Now unwind $f(100)$, keeping only what is needed: $$\begin{aligned} f(100) &= f(50) + f(49), \\ f(50) &= f(25) + f(24), \qquad f(49) = f(24), \\ f(25) &= f(12), \qquad f(24) = f(12) + f(11), \\ f(12) &= f(6) + f(5), \qquad f(11) = f(5), \\ f(6) &= f(3) + f(2) = 1 + 2 = 3, \qquad f(5) = f(2) = 2. \end{aligned}$$ Working back up: $f(12) = 5$, $f(11) = 2$, $f(25) = 5$, $f(24) = 7$, $f(50) = 12$, $f(49) = 7$, and finally $$f(100) = 12 + 7 = 19.$$

Answer 19

Solution: IOQM 2025 Part SEP, Q28

Adding the two conditions compares $ac+bd$ with $a+b+c+d$. Subtracting the matching linear terms makes $ac-a-c$ and $bd-b-d$, each one short of a product of non-negative integers. Complete those products by adding $1$ to each: $$(a-1)(c-1) = ac - (a+c) + 1 = ac - bd + 1,$$ $$(b-1)(d-1) = bd - (b+d) + 1 = bd - ac + 1.$$ Adding, everything cancels except the constants: $$(a-1)(c-1) + (b-1)(d-1) = 2.$$

Every digit is at least $1$, so both products are non-negative integers, and there are exactly three ways to split $2$.

The split $1 + 1$

Then $(a-1)(c-1) = 1$ forces $a = c = 2$, and likewise $b = d = 2$. Both conditions hold, since $2 + 2 = 2 \cdot 2$. This is the single number $2222$.

The split $0 + 2$

Then one of $a, c$ is $1$, and $(b-1)(d-1) = 2$ makes $\{b,d\} = \{2,3\}$. So $bd = 6$, and $a + c = 6$ gives the other of $a, c$ as $5$. The remaining condition $b + d = ac$ reads $5 = 5$, so it holds. That is $\{a,c\} = \{1,5\}$ with $\{b,d\} = \{2,3\}$: two orders for each pair, hence $2 \cdot 2 = 4$ numbers, $$1253, \quad 1352, \quad 5213, \quad 5312.$$

The split $2 + 0$

The same argument with the roles reversed, which the conditions permit because swapping the pair $(a,c)$ with the pair $(b,d)$ leaves them unchanged. Here $\{a,c\} = \{2,3\}$ and $\{b,d\} = \{1,5\}$, giving $2135$, $2531$, $3125$ and $3521$.

The three splits are exhaustive, so the total is $$4 + 4 + 1 = 9.$$

Answer 9