1971 Canadian MO Problems/Problem 7

Revision as of 15:45, 10 June 2020 by Vietajumping (talk | contribs) (Solution)


Let $n$ be a five digit number (whose first digit is non-zero) and let $m$ be the four digit number formed from n by removing its middle digit. Determine all $n$ such that $n/m$ is an integer.


Let $n=10000a+1000b+100c+10d+e$ and $m=1000a+100b+10d+e$, where $a$, $b$, $c$, $d$, and $e$ are base-10 digits and $a\neq 0$. If $n/m$ is an integer, then $m|n$, or


This implies that


Clearly we have that $8(1000a+100b+10d+e)<9000a+900b+100c<10(1000a+100b+10d+e)$, as $a$ is positive. Therefore, this quotient must be equal to 9 (note that this does not mean n/m = 9), and


This simplifies to $90d+9e=100c$. The only way that this could happen is that $c=0$. Then $d=e=0$. Therefore the only values of $n$ such that $n/m$ is an integer are multiples of 1000. It is not hard to show that these are all acceptable values.

See Also

1971 Canadian MO (Problems)
Preceded by
Problem 6
1 2 3 4 5 6 7 8 Followed by
Problem 8
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