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90! and its last two nonzero digits

Source: AMC 12A 2010, Problem 23

February 10, 2010
modular arithmeticalgebrapolynomialAMCAMC 10AMC 10 Anumber theory

Problem Statement

The number obtained from the last two nonzero digits of 90! 90! is equal to n n. What is n n? <spanclass=′latex−bold′>(A)</span> 12<spanclass=′latex−bold′>(B)</span> 32<spanclass=′latex−bold′>(C)</span> 48<spanclass=′latex−bold′>(D)</span> 52<spanclass=′latex−bold′>(E)</span> 68 <span class='latex-bold'>(A)</span>\ 12 \qquad <span class='latex-bold'>(B)</span>\ 32 \qquad <span class='latex-bold'>(C)</span>\ 48 \qquad <span class='latex-bold'>(D)</span>\ 52 \qquad <span class='latex-bold'>(E)</span>\ 68
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