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

Revision as of 20:04, 6 May 2007

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

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

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

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

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

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

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

@@ Line 5: / Line 5: @@
 ==Solution==
-<math>2^{2007\bmod4}\equiv2^3\equiv8\bmod10</math>
+<math>2^{2007\bmod4}\equiv2^3\equiv8\mod10</math>
-<math>3^{2007\bmod4}\equiv3^3\equiv7\bmod10</math>
+<math>3^{2007\bmod4}\equiv3^3\equiv7\mod10</math>
-<math>5^{2007\bmod4}\equiv5^3\equiv5\bmod10</math>
+<math>5^{2007\bmod4}\equiv5^3\equiv5\mod10</math>
-<math>7^{2007\bmod4}\equiv7^3\equiv3\bmod10</math>
+<math>7^{2007\bmod4}\equiv7^3\equiv3\mod10</math>
-<math>8+7+5+3\equiv3\bmod10</math>
+<math>8+7+5+3\equiv3\mod10</math>
 ==See also==