InterviewSolution
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InterviewSolution
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Prove that 33! Is divisible by `2^(19)` and what is the largest integer n such that 33! Is divisible by `2^(n)`? |
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Answer» In terms of prime factors, 33! Can be written as `2^(a)*3^(b)*7^(d)` . . . Now, `E_(2)(33!)=[(33)/(2)]+[(33)/(2^(2))]+[(33)/(2^(3))]=[(33)/(2^(4))]+[(33)/(2^(5))]` + . . . `=16+8+4+2+1+0` . . =31 Hence, the exponent of 2 in 33!. now , 33! is divisible by `2^(31)` which is also divisible by `2^(19)` `therefore`Largest value of n in 31. |
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