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24 Apr 2021, 03:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
56% (02:29) correct
44%
(02:00)
wrong
based on 301
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If y is the highest power of 'x' that can divide 101! without leaving a remainder,then for which among the following values of x will y be the highest?
A)111
B)462
C)74
D)33
E)210
Source:Wizako
A)111
B)462
C)74
D)33
E)210
Source:Wizako
24 Apr 2021, 14:17
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Jihyo
We need to scan the answer and check their factors. We want small factors since big factors would give low y.
A: 111 = 3*37. 37 is a huge prime, probably not A.
B: 462 = 6*77 = 2*3*7*11. y is determined by the number of 11's in 101!, since 2, 3, and 7's appear way more often than 11's.
C: 74 = 2*37. 37 is too big. This choice eliminates A and C, because A and C give the same y.
D: 33= 3*11. This choice eliminates B and D, because B and D give the same y.
E: 210 = 7*30 = 2*3*5*7. This choice has the lowest factors, besides that, we can eliminate ABCD.
Ans: E
General Discussion
24 Apr 2021, 06:19
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E - you need to look at the largest prime factor for each of the options and then see whose largest prime factor is the lowest.
