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amyhussein
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This has been discussed repeatedly in this forum. Just do a search before posting.
Refer to this post >> https://www.beatthegmat.com/integers-t102459.html#444130
integers
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When posting questions, please use the spoiler function to hide the correct answer. This will allow others to attempt the question without seeing the final answer.amyhussein wrote:h(n) is the product of all positive integers from 2 to n all inclusive. If p is the smallest prime of h(100)+1. what is the value of p?
a) 0 to 20
b) 20 to 25
c) 30
d) 40
e) >40
Important Concept: If k is a positive integer that's greater than 1, and if k is a factor (divisor) of N, then k is not a divisor of N+1
For example, since 7 is a factor of 350, we know that 7 is not a factor of (350+1)
Similarly, since 8 is a factor of 312, we know that 8 is not a factor of 313
Now let's examine h(100)
h(100) = (2)(4)(6)(8)....(96)(98)(100)
Factor to get: h(100) = 2[(1)(2)(3)(4)....(48)(49)(50)]
Since 2 is in the product of h(100), we know that 2 is a factor of h(100), which means that 2 is not a factor of h(100) +1 (based on the above rule)
Similarly, since 3 is in the product of h(100), we know that 3 is a factor of h(100), which means that 3 is not a factor of h(100) +1 (based on the above rule)
Similarly, since 5 is in the product of h(100), we know that 5 is a factor of h(100), which means that 5 is not a factor of h(100) +1 (based on the above rule)
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Similarly, since 47 is in the product of h(100), we know that 47 is a factor of h(100), which means that 47 is not a factor of h(100) +1 (based on the above rule)
So, we can see that none of the primes from 2 to 47 can be factors of h(100) +1, which means the smallest prime factor of h(100)+ 1 must be greater than 47.
Answer = E
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srcc25anu
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i think the correct question states that H(n) is the product of all positive EVEN integers from 2 to n all inclusive.
assuming this change in question stem:
h(100) = {2*4*6*8* ..... 98*100} => 2^50 (1*2*3*4* ..... 48*49*50)
hence H(100) is divisible by all numbers from 1 - 50
H(100) + 1 will not be divisible by any numbers from 1 - 50
therefore the samllest prime number that divides H(100)+1 should be greater than 50 hence Answer is E
assuming this change in question stem:
h(100) = {2*4*6*8* ..... 98*100} => 2^50 (1*2*3*4* ..... 48*49*50)
hence H(100) is divisible by all numbers from 1 - 50
H(100) + 1 will not be divisible by any numbers from 1 - 50
therefore the samllest prime number that divides H(100)+1 should be greater than 50 hence Answer is E
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Brent@GMATPrepNow
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Good catch. The original question does, indeed, specify even integers only.srcc25anu wrote:i think the correct question states that H(n) is the product of all positive EVEN integers from 2 to n all inclusive.
assuming this change in question stem:
h(100) = {2*4*6*8* ..... 98*100} => 2^50 (1*2*3*4* ..... 48*49*50)
hence H(100) is divisible by all numbers from 1 - 50
H(100) + 1 will not be divisible by any numbers from 1 - 50
therefore the samllest prime number that divides H(100)+1 should be greater than 50 hence Answer is E
Cheers,
Brent
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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