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For any integer k greater than 1, the symbol k* denotes the product of

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For any integer k greater than 1, the symbol k* denotes the product of  [#permalink]

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New post 01 Nov 2019, 05:28
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Question Stats:

59% (02:16) correct 41% (02:16) wrong based on 66 sessions

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For any integer k greater than 1, the symbol k* denotes the product of all integers between 1 and k, inclusive. If k* is a multiple of 3,675, what is the least possible value of k?

(A) 12
(B) 14
(C) 15
(D) 21
(E) 25


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Re: For any integer k greater than 1, the symbol k* denotes the product of  [#permalink]

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New post 01 Nov 2019, 07:07
1
factor of 3675 = 3*5^2*7^2
now answer has to be one which is divisible by 3*5^2*7^2
A; 12! ; 12*11*10*9*8*7*6! is not completely divisibly by 3*5^2*7^2 as 7 gets left in dr.
B ; 14! ; is completely divisibly by 3*5^2*7^2
IMO B correct
Bunuel wrote:
For any integer k greater than 1, the symbol k* denotes the product of all integers between 1 and k, inclusive. If k* is a multiple of 3,675, what is the least possible value of k?

(A) 12
(B) 14
(C) 15
(D) 21
(E) 25


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Re: For any integer k greater than 1, the symbol k* denotes the product of  [#permalink]

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New post 02 Nov 2019, 07:03
Bunuel wrote:
For any integer k greater than 1, the symbol k* denotes the product of all integers between 1 and k, inclusive. If k* is a multiple of 3,675, what is the least possible value of k?

(A) 12
(B) 14
(C) 15
(D) 21
(E) 25


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3675 = \(3*5^2*7^2\)
--> The fact that 2 multiples of 7 are there implies that there should be at least multiples from 1 to 14
--> Least possible value of k = 14

IMO Option B
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For any integer k greater than 1, the symbol k* denotes the product of  [#permalink]

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New post 24 Nov 2019, 08:00
Bunuel wrote:
For any integer k greater than 1, the symbol k* denotes the product of all integers between 1 and k, inclusive. If k* is a multiple of 3,675, what is the least possible value of k?

(A) 12
(B) 14
(C) 15
(D) 21
(E) 25


The largest power of a prime in x! (factorial) is the sum of quotients of x divided by powers of that prime less than x:
ie. 3's in 7! 7/3=2, so there are 3^2 in 7!
ie. 3's in 16! 16/3+16/9=5+1=6, so there are 3^6 in 16!

\(3675=5^27^23\)
\(f(k*)=5^27^23x\)
\(k≥14: 14/5=2…(5^2)…14/7=2…(7^2)…14/3+14/3^2=5…(3^5)\)
\(least(k)=14\)

Ans (B)
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For any integer k greater than 1, the symbol k* denotes the product of   [#permalink] 24 Nov 2019, 08:00
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