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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:
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66% (01:36) correct
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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
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
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.
