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

(Created page with "== Problem == Let <math>n</math> be a five digit number (whose first digit is non-zero) and let <math>m</math> be the four digit number formed from n by removing its middle digit...")
 
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<cmath>9000a+900b+90d+9e=9000a+900b+100c.</cmath>
 
<cmath>9000a+900b+90d+9e=9000a+900b+100c.</cmath>
  
This simplifies to <math>90d+9e=100c</math>. The only way that this could happen is that <math>c=0</math>. Then <math>d=e=0</math>. Therefore the only values of <math>n</math> such that <math>n/m</math> is an integer are multiples of 1000.
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This simplifies to <math>90d+9e=100c</math>. The only way that this could happen is that <math>c=0</math>. Then <math>d=e=0</math>. Therefore the only values of <math>n</math> such that <math>n/m</math> is an integer are multiples of 1000. It is not hard to show that these are all acceptable values.
  
 
== See Also ==
 
== See Also ==
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{{Old CanadaMO box|num-b=6|num-a=8|year=1971}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 15:23, 12 September 2012

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 $n/m$ must be equal to 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