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help me

by shruti93 » Sun Jul 28, 2013 4:16 am
Q1. IF N= 2^64 AND N^N= 2^K. WHAT IS K?
Q2, IF H(N)= PRODUCT OF ALL EVEN INTEGERS BETWEEN 2 AND N.THEN THE SMALLEST PRIME NUMBER WHICH DIVIDES H(100)+1 LIES BETWEEN ?
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by Brent@GMATPrepNow » Sun Jul 28, 2013 6:17 am
Here's the original question in its entirety:
Q2. For 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
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)
= (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

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by Brent@GMATPrepNow » Sun Jul 28, 2013 6:22 am
shruti93 wrote:Q1. IF N = 2^64 AND N^N= 2^K. WHAT IS K?
If N = 2^64, then . . .
N^N = (2^64)^(2^64)
= 2^(64 times 2^64) [apply the Power of a Power rule]
= 2^(2^6 times 2^64) [rewrite 64 as 2^6]
= 2^(2^70) [apply Product rule]

So, K = [spoiler]2^70[/spoiler]

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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