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. | + | 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 == | ||
+ | {{Old CanadaMO box|num-b=6|num-a=8|year=1971}} | ||
+ | [[Category:Intermediate Number Theory Problems]] |
Revision as of 14:23, 12 September 2012
Problem
Let be a five digit number (whose first digit is non-zero) and let be the four digit number formed from n by removing its middle digit. Determine all such that is an integer.
Solution
Let and , where , , , , and are base-10 digits and . If is an integer, then , or
This implies that
Clearly we have that , as is positive. therefore must be equal to 9, and
This simplifies to . The only way that this could happen is that . Then . Therefore the only values of such that 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 |