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Jihyo
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If y is the highest power of 'x' that can divide 101! 24 Apr 2021, 03:45
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53% (02:32) correct 47% (01:58) wrong based on 258 sessions

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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
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E
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If y is the highest power of 'x' that can divide 101! 24 Apr 2021, 14:17
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Jihyo
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
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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
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If y is the highest power of 'x' that can divide 101! 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.
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If y is the highest power of 'x' that can divide 101! 24 Apr 2021, 20:43
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Hello math experts! I am having a hard time understanding the question.What does 'y is the highest power of 'x' that can divide 101!' means here?
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If y is the highest power of 'x' that can divide 101! 24 Apr 2021, 21:11
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Jihyo
Hello math experts! I am having a hard time understanding the question.What does 'y is the highest power of 'x' that can divide 101!' means here?
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y is your exponent. X is the factor. The smaller the factor, the larger the exponent you will need to divide into 101! The question is asking what Is the smallest factor (and hence the highest power of y) that divides into 101!

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If y is the highest power of 'x' that can divide 101! 01 Dec 2024, 00:06
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101! will have multiple zeros in last, hence, we need to check out of these which will be divisible by 10(pairs 2 and 5).
Only 210 satisfy that requirement.

Please correct me if i am wrong.
Jihyo
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
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If y is the highest power of 'x' that can divide 101! 01 Dec 2024, 06:12
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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?

Highest power of 2 in 101! = 50 + 25 + 12 + 6 + 3 + 1 = 97
Highest power of 3 in 101! = 33 + 11 + 3 + 1 = 48
Highest power of 5 in 101! = 20 + 4 = 24
Highest power of 7 in 101! = 14 + 2 = 16
Highest power of 11 in 101! = 9
Highest power of 37 in 101! = 1

A)111 = 3*37 : Highest power of 111 in 101! = min (48,1) = 1
B)462 = 2*3*7*11: Highest power of 462 in 101! = min (97,48,16,9) = 9
C)74 = 2*37: Highest power of 74 in 101! = min (97,1) = 1
D)33 = 3*11: Highest power of 33 in 101! = min (48,9) = 9
E)210 = 2*3*5*7: Highest power of 210 in 101! = min (97,48,24,16) =16

IMO E
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