InterviewSolution
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InterviewSolution
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Find the number of zeros at the end of 100!. |
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Answer» In terms of prime factors, 100! Can be written as `2^(a)*3^(b)*5^(c)*7^(d)` . . . Now, `E_(1)(100!)=[(100)/(2)]+[100/2^(2)]+[(100)/(2^(3))]+[(100)/(2^(4))]+[(100)/(2^(5))]+[(100)/(2^(6))]` `=50+25+12+6+3+1=97` and `E_(5)(100!)=[(100)/(5)]+[(100)/(5^(2))]` `=20+4=24` `therefore100!=2^(97)*3^(b)*5^(24)*7^(d)=2^(73)*3^(b)*(2xx5)^(24)*7^(d)` . . . `=2^(73)*3^(b)*(10)^(24)*7^(d)` Hence, number of zeros at the end of 100! is 24. or exponent of 10 in 100!=min(97,24)=24. |
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