Third Millennium International Mathematical Olympiad 2009
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Find all natural numbers M for which |
Solution by Sadik Shahidain, Cranbury, NJ, USA.
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Copyright © 1996-2012 Alexander Bogomolny
Rewriting the left side as
| (7M - M2)M - (M2)M = 2009 |
Factoring the left
| (7M - 2M2)((7M - M2)M-1 + M(7M - M2)M-2(M2) + ... + M(7M - M2)(M2)M-2 + (M2)M-1) = 2009 |
Therefore 7M - 2M2 must be a factor of 2009, and the possibilities are
| 7M - 2M2 = {1 or 7 or 41 or 49 or 287 or 2009} |
and M < 4, where M is a natural number. 2 is the only viable option and substituting back into the original equation gives 2 as the answer.
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