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sum of 99! th power of first 24 odd prime numbers

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April 23, 2009
modular arithmeticnumber theoryprime numbers

Problem Statement

Let p1<p2<…<p24 p_{1} <p_{2} <\ldots <p_{24} be the prime numbers on the interval [3,100] \left[3,100\right]. Find the smallest value of a≥0 a\ge 0 such that \sum _{i\equal{}1}^{24}p_{i}^{99!} \equiv a\, \, \left(mod\, 100\right).
<spanclass=′latex−bold′>(A)</span> 24<spanclass=′latex−bold′>(B)</span> 25<spanclass=′latex−bold′>(C)</span> 48<spanclass=′latex−bold′>(D)</span> 50<spanclass=′latex−bold′>(E)</span> 99<span class='latex-bold'>(A)</span>\ 24 \qquad<span class='latex-bold'>(B)</span>\ 25 \qquad<span class='latex-bold'>(C)</span>\ 48 \qquad<span class='latex-bold'>(D)</span>\ 50 \qquad<span class='latex-bold'>(E)</span>\ 99
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