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10 Jan 2017, 03:58
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
67% (01:34) correct
33%
(01:42)
wrong
based on 107
sessions
History
Date
Time
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If x = a!, which of the following values of positive integer a is the least possible value for which the last 8 digits of the integer x will all be zero?
A. 21
B. 29
C. 38
D. 40
E. 100
A. 21
B. 29
C. 38
D. 40
E. 100
10 Jan 2017, 04:48
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x = a! = (Some number)*10^8
10^8 = (5*2)^8 --> a! must contain 8 5's in it.
A. 21 --> 21/5 = 4
B. 29 --> 29/5 + 29/25 = 4 + 1 = 5
C. 38 --> 38/5 + 38/25 = 7 + 1 = 8
D. 40
E. 100
Answer: C
10^8 = (5*2)^8 --> a! must contain 8 5's in it.
A. 21 --> 21/5 = 4
B. 29 --> 29/5 + 29/25 = 4 + 1 = 5
C. 38 --> 38/5 + 38/25 = 7 + 1 = 8
D. 40
E. 100
Answer: C
Sirakri
10 Jan 2017, 05:07
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Bunuel
By trailing zeroes formula, w.k.t., No of Trailing Zeroes of n!=n/5+n/5^2+...+n/5^k, where 5^k<n
The question asks us for the value of a when a! has 8 trailing zeroes.
By trial and error:
TZ. 38!= 38/5+38/25=7+1=8
Hence C.
