2007 Cyprus MO/Lyceum/Problem 17: Difference between revisions

Latest revision as of 20:16, 6 May 2007

The last digit of the number $a=2^{2007}+3^{2007}+5^{2007}+7^{2007}$ is

$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 8$

$2^{2007\bmod4}\equiv2^3\equiv8\mod10$

$3^{2007\bmod4}\equiv3^3\equiv7\mod10$

$5^{2007\bmod4}\equiv5^3\equiv5\mod10$

$7^{2007\bmod4}\equiv7^3\equiv3\mod10$

$8+7+5+3\equiv3\mod10$

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 ==See also==
 {{CYMO box|year=2007|l=Lyceum|num-b=16|num-a=18}}
-[[Category:Introductory Combinatorics Problems]]