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Difference between revisions of "1971 Canadian MO Problems/Problem 7"

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<cmath>1000a+100b+10d+e|9000a+900b+100c.</cmath>
 
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Clearly we have that <math>8(1000a+100b+10d+e)<9000a+900b+100c<10(1000a+100b+10d+e)</math>, as <math>a</math> is positive. Therefore, this quotient must be equal to 9, and
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Clearly we have that <math>8(1000a+100b+10d+e)<9000a+900b+100c<10(1000a+100b+10d+e)</math>, as <math>a</math> is positive. Therefore, this quotient must be equal to 9 (note that this does not mean <math>n/m = 9</math>), and
  
 
<cmath>9000a+900b+90d+9e=9000a+900b+100c.</cmath>
 
<cmath>9000a+900b+90d+9e=9000a+900b+100c.</cmath>

Latest revision as of 15:47, 10 June 2020

Problem

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.

Solution

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

\[1000a+100b+10d+e|10000a+1000b+100c+10d+e.\]

This implies that

\[1000a+100b+10d+e|9000a+900b+100c.\]

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

\[9000a+900b+90d+9e=9000a+900b+100c.\]

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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