For every positive even integer "n" , the function h(n)is defined to be the product of all the even integers from 2 to n , inclusive. If "p" is the smallest prime factor of h(100)+1 , then p is
a) Between 2 and 10
b) Between 10 and 20
c) Between 20 and 30
d) Between 30 and 40
e) Greater than 40
Folks , what could be the best approach to this problem ?
Arthemetic : Question on factors and prime numbers
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Important Concept: If integer k is greater than 1, and k is a factor (divisor) of N, then k is not a divisor of N+1For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, the p is
A: Between 2 & 10
B: Between 10 & 20
C: Between 20 & 30
D: Between 30 & 40
E: Greater than 40
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)
= (2x1)(2x2)(2x3)(2x4)....(2x48)(2x49)(2x50)
Factor out all of the 2's to get: h(100) = [2^50][(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
Cheers,
Brent
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Brent,
In your explanation above, can I stop at this step to determine the answer:
Factor out all of the 2's to get: h(100) = [2^50][(1)(2)(3)(4)....(48)(49)(50)]
The above step indicates that all number 2 through 50 are factors of h(100), which implies they cannot be factors of h(100)+1. And hence any number that could be a factor of h(100)+1 should be greater than 50 ?
In your explanation above, can I stop at this step to determine the answer:
Factor out all of the 2's to get: h(100) = [2^50][(1)(2)(3)(4)....(48)(49)(50)]
The above step indicates that all number 2 through 50 are factors of h(100), which implies they cannot be factors of h(100)+1. And hence any number that could be a factor of h(100)+1 should be greater than 50 ?
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
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Precisely!!gmat_chanakya wrote:Brent,
In your explanation above, can I stop at this step to determine the answer:
Factor out all of the 2's to get: h(100) = [2^50][(1)(2)(3)(4)....(48)(49)(50)]
The above step indicates that all number 2 through 50 are factors of h(100), which implies they cannot be factors of h(100)+1. And hence any number that could be a factor of h(100)+1 should be greater than 50 ?
Cheers,
Brent