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

Revision as of 15:44, 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, 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.

@@ Line 11: / Line 11: @@
 <cmath>1000a+100b+10d+e|9000a+900b+100c.</cmath>
-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
+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
 <cmath>9000a+900b+90d+9e=9000a+900b+100c.</cmath>

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

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

Revision as of 15:44, 10 June 2020

Problem

Solution

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