Functions Problem Solving

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Functions Problem Solving

by chaithu_bunny » Wed Apr 17, 2013 6:21 pm
For every positive 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
between 10 and 20
[C] between 20 and 30
[D] between 30 and 40
[E] greater than 40


Correct Answer is E

Kindly help me with the strategy to be followed to answer this question.

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by Anju@Gurome » Wed Apr 17, 2013 6:31 pm
chaithu_bunny wrote:For every positive 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:
h(100) = 2*4*6* ... *100
= (2*1)*(2*2)*(2*3)* ... *(2*50)
= 2^(50)*(1*2*3 ... *50)
Then h(100) + 1 = 2^(50)*(1*2*3 ... *50) + 1
Now, h(100) + 1 cannot have any prime factors less than 50, because dividing this value by any of these prime numbers will give a remainder of 1.

The correct answer is E.
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by chaithu_bunny » Wed Apr 17, 2013 6:53 pm
Anju@Gurome wrote: Then h(100) + 1 = 2^(50)*(1*2*3 ... *50) + 1
Now, h(100) + 1 cannot have any prime factors less than 50, because dividing this value by any of these prime numbers will give a remainder of 1.

The correct answer is E.
Hi Anju, I couldn't understand the above statement, i.e., Now, h(100) + 1 cannot have any prime factors less than 50, because dividing this value by any of these prime numbers will give a remainder of 1.

Could you explain in detail please?

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by GMATGuruNY » Wed Apr 17, 2013 6:58 pm
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
Since the difference between them is 1, h(100) and h(100)+1 are consecutive integers.
Consecutive integers are COPRIMES: they share no factors other than 1.

Let's examine why:

If x is a multiple of 2, the next largest multiple of 2 is x+2.
If x is a multiple of 3, the next largest multiple of 3 is x+3.

Using this logic, if we go from x to x+1, we get only to the next largest multiple of 1.
So 1 is the only factor common both to x and to x+1.
In other words, x and x+1 are COPRIMES.

Thus:
h(100) and h(100)+1 are COPRIMES. They share no factors other than 1.

h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100
Factoring out 2, we get:
h(100) = 2^50 (1 * 2 * 3 *... * 47 * 48 * 49 * 50)

Looking at the set of parentheses on the right, we can see that every prime number between 1 and 50 is a factor of h(100).
Since h(100) and h(100)+1 are coprimes, NONE of the prime numbers between 1 and 50 can be a factor of h(100)+1.

Thus, the smallest prime factor of h(100) + 1 must be greater than 50.

The correct answer is E.
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by chaithu_bunny » Wed Apr 17, 2013 7:03 pm
GMATGuruNY wrote:
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
Since the difference between them is 1, h(100) and h(100)+1 are consecutive integers.
Consecutive integers are COPRIMES: they share no factors other than 1.

Let's examine why:

If x is a multiple of 2, the next largest multiple of 2 is x+2.
If x is a multiple of 3, the next largest multiple of 3 is x+3.

Using this logic, if we go from x to x+1, we get only to the next largest multiple of 1.
So 1 is the only factor common both to x and to x+1.
In other words, x and x+1 are COPRIMES.

Thus:
h(100) and h(100)+1 are COPRIMES. They share no factors other than 1.

h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100
Factoring out 2, we get:
h(100) = 2^50 (1 * 2 * 3 *... * 47 * 48 * 49 * 50)

Looking at the set of parentheses on the right, we can see that every prime number between 1 and 50 is a factor of h(100).
Since h(100) and h(100)+1 are coprimes, NONE of the prime numbers between 1 and 50 can be a factor of h(100)+1.

Thus, the smallest prime factor of h(100) + 1 must be greater than 50.

The correct answer is E.
Thanks a lot GMATGuruNY. I've got it now...

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by Brent@GMATPrepNow » Wed Apr 17, 2013 7:42 pm
chaithu_bunny wrote:For every positive 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
between 10 and 20
[C] between 20 and 30
[D] between 30 and 40
[E] greater than 40


Correct Answer is E

Kindly help me with the strategy to be followed to answer this question.


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

Cheers,
Brent
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