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Difference between revisions of "2013 USAMO Problems/Problem 5"

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Given postive integers <math>m</math> and <math>n</math>, prove that there is a positive integer <math>c</math> such that the numbers <math>cm</math> and <math>cn</math> have the same number of occurrences of each non-zero digit when written in base ten.
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== Problem ==
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{{MAA Notice}}
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Given positive integers <math>m</math> and <math>n</math>, prove that there is a positive integer <math>c</math> such that the numbers <math>cm</math> and <math>cn</math> have the same number of occurrences of each non-zero digit when written in base ten.
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== Solution 1 ==
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This solution is adopted from the [http://web.archive.org/web/20130606075948/http://amc.maa.org/usamo/2013/2013USAMO_Day1_Day2_Final_S.pdf official solution]. Both the problem and the solution were suggested by Richard Stong.
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Without Loss of Generality, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</math>.
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Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\varphi(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.
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Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another.
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This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.
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== Solution 2 ==
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This is a rephrasing of the above solution.
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It is enough to solve the problem when <math>m,n</math> are replaced by <math>km,kn</math> for any positive integer <math>k</math>. In particular, by taking <math>k=2^a5^b</math> for appropriate values of <math>a,b</math>, we may assume <math>n=10^sn_1</math> where <math>n_1</math> is relatively prime to 10.
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Furthermore, adding or removing trailing zeros from <math>m</math> and <math>n</math> doesn't affect the claim, so we may further assume <math>\gcd(n,10)=1</math> and that <math>m</math> has a xillion trailing zeros (enough to make <math>m</math> way bigger than <math>n</math>, and also so that <math>m</math> has at least one trailing zero).
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For clarity of exposition, we will also multiply <math>m,n</math> by a small number to make the units digit of <math>n</math> be <math>1</math> (though this is not necessary for the solution to work).
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The point is that, for any positive integer <math>X</math>, most nonzero digits appear the same number of times in <math>X</math> and <math>X+999\dots999</math> if there are enough <math>9</math>s; In particular, if the units digits of <math>X</math> is <math>1</math>, then all nonzero digits appear the same number of times as long as there are at least as many <math>9</math>s as digits in <math>X</math>.
  
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==Solution==
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So we will pick <math>c</math> to satisfy:
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* <math>mc=nc+999\dots999</math> where the number of <math>9</math>s is more than the number of digits of <math>nc</math>
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* The units digit of <math>nc</math> is <math>1</math>.
  
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This solution is adopted from the [http://web.archive.org/web/20130606075948/http://amc.maa.org/usamo/2013/2013USAMO_Day1_Day2_Final_S.pdf official solution]. Both the problem and the solution were suggested by Richard Stong.
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Because we made the units of <math>n</math> to be <math>1</math>, the second condition is equivalent to making the units digit of <math>c</math> to be <math>1</math>.
  
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The first condition is equivalent to <math>(m-n)c=999\dots999</math>. Because <math>m</math> has at least one trailing 0, the units digit of <math>m-n</math> is 9, so <math>\gcd(m-n,10)=1</math> and there is some <math>c</math> so that <math>(m-n)c=999\dots999</math>, and the units digit of <math>c</math> must be <math>1</math> which agrees with the other condition.
  
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WLOG, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</math>. Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\varphi(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.
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Finally, as <math>m\gg n, (m-n)c\approx mc\gg nc</math> so it is possible to make the number of <math>9</math>s more than the number of digits in <math>nc</math>.
  
  
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Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another. This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.
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{{MAA Notice}}

Latest revision as of 23:31, 6 September 2023

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 $nc$.


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