Let p1<p2<…<p24 be the prime numbers on the interval [3,100]. Find the smallest value of a≥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 modular arithmeticnumber theoryprime numbers