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Find the exponent of 3 in 100!A. 47B. 48C. 49D. 46 |
Answer» Let `E_(p)(n)` denote the exponent of p in n. Then, `E_(p)(n!)=[(n)/(p)]+[(n)/(p^(2))]+....+[(n)/(p^(s))]` where s is the largest positive integer such that `p^(s)lenltp^(s+1)` Here, n=100, p=3 `:.3^(4)lt100lt3^(5)` `:.s=4` So, `E_(3)(100!)=[(100)/(3)]+[(100)/(3^(2))]+[(100)/(3^(3))]+[(100)/(3^(4))]` `=33+11+3+1=48` Hence, the exponent of 3 in 100! is 48. |
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