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Kaushik786
Joined: 20 May 2018
Last visit: 13 Dec 2023
Posts: 71
Given Kudos: 71
Location: India
Schools: Booth '22 Anderson '22 CBS '22 Johnson '22
GMAT 1: 740 Q50 V41
09 Mar 2021, 02:33
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A
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:
73% (01:54) correct
27%
(02:05)
wrong
based on 196
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If P is the product of all the integers from 1 to 36 inclusive, what is the greatest integer i for which 12^i is a factor of P ?
A. 17
B. 18
C. 19
D. 20
E. 21
A. 17
B. 18
C. 19
D. 20
E. 21
09 Mar 2021, 02:55
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Kaushik786
P = 36!
We need to find the highest power of 12 in 36!
12 = 2² x 3
Highest power of 2 in 36! is:
36/2 = 18
18/2 = 9
9/2 = 4.5 => 4
4/2 = 2
2/2 = 1
So we have: 18+9+4+2+1 = 34
So, highest power of 2² = 34/2 = 17
Highest power of 3 in 36! is:
36/3 = 12
12/3 = 4
4/3 = 1.33 => 1
So we have: 12+4+1 = 17
So, highest power of 3 = 17
Hence, highest power of 12 in 36! is also 17
Answer A
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General Discussion
09 Mar 2021, 07:57
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Kaushik786
the formula is:
\(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}+ ... till n>k^2\)
Expressing 12 as \(2^2 * 3\) and taking individual cases:
Greatest integer of \(2^2\) in 36! = Greatest integer of 3 in 36! = 17 ------------ option A
