2013 USAMO Problems/Problem 5

Problem

Given positive integers $m$ and $n$ , prove that there is a positive integer $c$ such that the numbers $cm$ and $cn$ have the same number of occurrences of each non-zero digit when written in base ten.

Solution 1

This solution is adopted from the official solution. Both the problem and the solution were suggested by Richard Stong.

Without Loss of Generality, suppose $m \geq n \geq 1$ . By prime factorization of $n$ , we can find a positive integer $c_1$ such that $c_1n=10^s n_1$ where $n_1$ is relatively prime to $10$ . If a positive $k$ is larger than $s$ , then $(10^k c_1 m - c_1 n)= 10^s t$ , where $t=10^{k-s} c_1m-n_1>0$ is always relatively prime to $10$ .

Choose a $k$ large enough so that $t$ is larger than $c_1m$ . We can find an integer $b\geq 1$ such that $10^b-1$ is divisible by $t$ , and also larger than $10c_1m$ . For example, let $b=\varphi(t)$ and use Euler's theorem. Now, let $c_2=(10^b-1)/t$ , and $c=c_1c_2$ . We claim that $c$ is the desired number.

Indeed, since both $c_1m$ and $n_1$ are less than $t$ , we see that the decimal expansion of both the fraction $(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)$ and $n_1/t=(c_2n_1)/(10^b-1)$ are repeated in $b$ -digit. And we also see that $10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)$ , therefore the two repeated $b$ -digit expansions are cyclic shift of one another.

This proves that $cm$ and $c_2n_1$ have the same number of occurrences of non-zero digits. Furthermore, $cn = c_2c_1n=10^s c_2n_1$ also have the same number of occurrences of non-zero digits with $c_2n_1$ .

Solution 2

This is a rephrasing of the above solution.

It is enough to solve the problem when $m,n$ are replaced by $km,kn$ for any positive integer $k$ . In particular, by taking $k=2^a5^b$ for appropriate values of $a,b$ , we may assume $n=10^sn_1$ where $n_1$ is relatively prime to 10.

Furthermore, adding or removing trailing zeros from $m$ and $n$ doesn't affect the claim, so we may further assume $\gcd(n,10)=1$ and that $m$ has a xillion trailing zeros (enough to make $m$ way bigger than $n$ , and also so that $m$ has at least one trailing zero).

For clarity of exposition, we will also multiply $m,n$ by a small number to make the units digit of $n$ be $1$ (though this is not necessary for the solution to work).

The point is that, for any positive integer $X$ , most nonzero digits appear the same number of times in $X$ and $X+999\dots999$ if there are enough $9$ s; In particular, if the units digits of $X$ is $1$ , then all nonzero digits appear the same number of times as long as there are at least as many $9$ s as digits in $X$ .

So we will pick $c$ to satisfy:

$mc=nc+999\dots999$ where the number of $9$ s is more than the number of digits of $nc$
The units digit of $nc$ is $1$ .

Because we made the units of $n$ to be $1$ , the second condition is equivalent to making the units digit of $c$ to be $1$ .

The first condition is equivalent to $(m-n)c=999\dots999$ . Because $m$ has at least one trailing 0, the units digit of $m-n$ is 9, so $\gcd(m-n,10)=1$ and there is some $c$ so that $(m-n)c=999\dots999$ , and the units digit of $c$ must be $1$ which agrees with the other condition.

Finally, as $m\gg n, (m-n)c\approx mc\gg nc$ so it is possible to make the number of $9$ s more than the number of digits in $nc$ .

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2013 USAMO Problems/Problem 5

Problem

Solution 1

Solution 2