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kiran120680
Moderator - Masters Forum
Joined: 18 Feb 2019
Last visit: 27 Jul 2022
Posts: 706
Given Kudos: 276
Location: India
Schools: AGSM '20 EDHEC Sept 2019 Intake Great Lakes '21 IE
GMAT 1: 460 Q42 V13
GPA: 3.6
31 May 2019, 08:58
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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:
75%
(hard)
Question Stats:
60% (02:23) correct
40%
(02:33)
wrong
based on 638
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The function f(n) is defined as the product of all integers from 1 to n, inclusive, and the function g(n) is defined as the product of all odd integers from 1 to n, inclusive, where n is a positive integer. If p is a prime factor of {f(150)/g(150)} + 1, then which of the following must be true?
A. p < 10
B. 10 < p < 25
C. 25 < p < 50
D. 50 < p < 75
E. p > 75
A. p < 10
B. 10 < p < 25
C. 25 < p < 50
D. 50 < p < 75
E. p > 75
31 May 2019, 09:06
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f(150) is just 150!, and when we divide 150! by all of the odd integers up to 150, we end up just with the product of the even integers from 2 to 150 inclusive. So we want to know about the smallest prime factor of
(150)(148)(146)...(6)(4)(2) + 1
If we factor out one 2 from each term in that big product, we can rewrite it more succinctly:
(150)(148)(146)...(6)(4)(2) + 1 = (2)(75)(2)(74)(2)(73)...(2)(3)(2)(2)(2)(1) + 1 = 2^75 * 75! + 1
Now, 75! is divisible by every prime less than 75. So if you take any such prime, say 47, then 2^75 * 75! will be divisible by 47, and 2^75 * 75! + 1 will be exactly 1 greater than a multiple of 47. That's another way of saying "the remainder will be 1 when we divide 2^75 * 75! + 1 by 47". So when we divide this huge number by any prime less than 75, we always get a remainder of exactly 1. Its smallest prime divisor must therefore be greater than 75.
This is, incidentally, a close copy of the famous official GMATPrep question (famous because for years it was posted to GMAT forums on a near daily basis because it was so hard) here:
https://gmatclub.com/forum/for-every-po ... 12521.html
(150)(148)(146)...(6)(4)(2) + 1
If we factor out one 2 from each term in that big product, we can rewrite it more succinctly:
(150)(148)(146)...(6)(4)(2) + 1 = (2)(75)(2)(74)(2)(73)...(2)(3)(2)(2)(2)(1) + 1 = 2^75 * 75! + 1
Now, 75! is divisible by every prime less than 75. So if you take any such prime, say 47, then 2^75 * 75! will be divisible by 47, and 2^75 * 75! + 1 will be exactly 1 greater than a multiple of 47. That's another way of saying "the remainder will be 1 when we divide 2^75 * 75! + 1 by 47". So when we divide this huge number by any prime less than 75, we always get a remainder of exactly 1. Its smallest prime divisor must therefore be greater than 75.
This is, incidentally, a close copy of the famous official GMATPrep question (famous because for years it was posted to GMAT forums on a near daily basis because it was so hard) here:
https://gmatclub.com/forum/for-every-po ... 12521.html
General Discussion
10 Jan 2026, 03:04
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f(150) = The product of all integers from 1 to 150, inclusive = 1 * 2 * 3 * ... * 149 * 150
g(150) = The product of all odd integers from 1 to n, inclusive = 1 * 3 * 5 * ... * 149
=> \(\frac{f(150)}{g(150)}\) + 1 = \(\frac{ 1 * 2 * 3 * ... * 149 * 150 }{ 1 * 3 * 5 * ... * 149}\) + 1
Now all odd terms will cancel out and only the even terms will remain
= 2 * 4 * 6 * ... * 150 + 1
= \(2^75\) * (1 * 2 * ... * 75) + 1
That means when the number is divided by any number from 1 to 75 then it will give 1 as remainder
=> Number is not divisible by any number from 1 to 75
=> Any prime number which has to divide this number has to be > 75
=> p > 75
So, Answer will be E
Hope it helps!
Watch the following video to MASTER Functions and Custom Characters
g(150) = The product of all odd integers from 1 to n, inclusive = 1 * 3 * 5 * ... * 149
=> \(\frac{f(150)}{g(150)}\) + 1 = \(\frac{ 1 * 2 * 3 * ... * 149 * 150 }{ 1 * 3 * 5 * ... * 149}\) + 1
Now all odd terms will cancel out and only the even terms will remain
= 2 * 4 * 6 * ... * 150 + 1
= \(2^75\) * (1 * 2 * ... * 75) + 1
That means when the number is divided by any number from 1 to 75 then it will give 1 as remainder
=> Number is not divisible by any number from 1 to 75
=> Any prime number which has to divide this number has to be > 75
=> p > 75
So, Answer will be E
Hope it helps!
Watch the following video to MASTER Functions and Custom Characters
