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GMAT Prep test Question

by tanvis1120 » Wed Aug 20, 2014 4:25 pm
h(n) = 2x4x6x8x....x96x98xn (n is a positive even number)
h(100) = 2x4x6x8x....x96x98x100
Hence, h(100)+1 = (2x4x6x8x....x96x98x100)+1

As the final number would end with 1 in its units place (Something x 100 plus 1), it can have 3, 7 or 11 as prime factor. Accordingly I hit the option A.
But OA is E.
What is wrong here in my method ?

Thanks.
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by GMATGuruNY » Wed Aug 20, 2014 7:15 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 from every value above, we get:
h(100) = 2��(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 [email protected] » Wed Aug 20, 2014 8:38 pm
Hi tanvis1120,

This question shows up every so often in this Forum and it is definitely tougher than a typical GMAT question.

As a general rule, Quant questions are almost always based on a pattern of some kind (math formula, math rule, Number Property, etc.). If you can't immediately deduce a pattern, then you might have to "play around" a bit with the question to try to deduce what the pattern is. In the broad sense, it's critical thinking: here's a weird situation - what can I do to figure it out?

Based on the description of the function in the prompt, we can run some "TESTS" to try to figure things out....

The H(n) is the product of all the even integers from 2 to n, inclusive.

So....
H(4) = 2x4 = 8
The prime factors of 8 are (2)(2)(2)
If we do H(4) + 1 = 9, then the prime factors are (3)(3)
NOTICE how NONE of the prime factors of 8 are in 9? That's interesting....

H(6) = 2x4x6 = 48
The prime factors of 48 are (2)(2)(2)(2)(3)
If we do H(6) + 1 = 49, then the prime factors are (7)(7)
NOTICE how NONE of the prime factors of 48 are in 49? That's interesting....and probably a pattern, since it's happened TWICE NOW.

From here, I'd have to deduce that this pattern holds true. With H(100), I know that there are LOTS of primes that go in (the largest of which is 47, which can be "found" in 94). I have to assume that NONE of them will go into H(100) + 1. Thus, the smallest prime would have to be greater than 47. The question doesn't actually ask us for the exact answer though (the answer choices are "ranges").

The takeaway from all of this is that you shouldn't be afraid to "play" with a question a bit. In the end, you don't have to be a brilliant mathematician to answer this question, but you're also not allowed to just stare at it either.

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by GMATinsight » Wed Aug 20, 2014 8:38 pm
For every positive integer n, the function h(n) is defined as the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) - 1, then p is :

1)between 2 and 10
2)between 10 and 20
3)between 20 and 30
4)between 30 and 40
5)greater than 40
As per the definition of the question

h(100) = 2 x 4 x 6 x 8 x 10 x 12 x 14 ... and so on...98 x 100 (Total 50 terms)

=> h(100) = 2^50 (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10......and so on...x 48 x 49 x 50)

This means h(100) is multiple of all prime numbers between 1 and 50

therefore h(100)+1 will leave a remainder of 1 when divided by any prime number from 1 to 50

therefore, p, which is a factor of h(100)+1, will certainly be greater than a prime numbers greater than 50

Hence, "p" must be greater than 40 as per the following options

Answer: Option - E
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by Brent@GMATPrepNow » Wed Aug 20, 2014 8:53 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
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+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

Cheers,
Brent
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by Matt@VeritasPrep » Wed Aug 20, 2014 11:25 pm
tanvis1120 wrote:As the final number would end with 1 in its units place (Something x 100 plus 1), it can have 3, 7 or 11 as prime factor. What is wrong here in my method ?
I'm not sure anyone else answered this part of your question, so let me give it a shot.

If a number ends in 1, it could have literally ANY odd factor that is not divisible by 5. For instance, 221 = 13 * 17, which has neither 3, 7, nor 11 as prime factors. I like your idea here, but it's probably safer on questions that ask about units digits than on questions that ask about factors.

The other problem is that h(n) is divisible by 3, 7, and 11. (6, 14, and 22 are all factors of h(n), and they divide by 3, 7, and 11, respectively.) Since h(n) + 1 shares no prime factors with h(n), as Mitch explained, h(n) + 1 thus can't be divisible by 3, 7, or 11.
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