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Re: gmat prep question [#permalink]
02 Mar 2007, 16:43

Stormgal wrote:

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 2and 10 b. between 10 and 20 c. between 20 and 30 d. between 30 and 40 e. greater than 40

h(n) = 2*(n/2)!
h(100) = 2*(50)!

since h(100) is divisible by every number from 1 to 50, h(100) wont be divisible by any those numbers (i think), so E would be the answer.

someone please correct me if i am wrong about the assumption i made here..

Function & prime factor Qs [#permalink]
24 Mar 2007, 17:02

1) 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, 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

smallest prime factor [#permalink]
18 Apr 2007, 19:33

For every positive interger n, the function h(n) is def. 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 bet. 2 and 10
b. bet. 10-20
c. bet. 20-30
d. bet 30-40
e. greater than 40

I'm not quite sure how to approach this question and I'm not sure what they're asking. Any help is appreciated. OA will follow.
TIA

Re: smallest prime factor [#permalink]
18 Apr 2007, 20:03

ayl989 wrote:

For every positive interger n, the function h(n) is def. 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 bet. 2 and 10 b. bet. 10-20 c. bet. 20-30 d. bet 30-40 e. greater than 40

I'm not quite sure how to approach this question and I'm not sure what they're asking. Any help is appreciated. OA will follow. TIA

I am taking a shot. Others can chime in

H(100) = 2 x 4 x 6 x ... x 100
= 2(1 x 2 x 3 x ... x 50)

so H(100) + 1 = 2(1 x 2 x 3 x ... x 50) + 1

Now any divisor of the above expression must divide both terms. Any prime factor below 50 will always divide the first term completely but it cant divide 1. So I think that required prime factor should be > 50

Help with a GMAT question [#permalink]
21 Apr 2007, 10:24

Can someone help me with this question? I felt a tickle in my brain when I saw this

For every positive even integer n, 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

I know the answer is E, but I haven't a clue as to why or how. HELP!

Re: Help with a GMAT question [#permalink]
22 Apr 2007, 11:51

afiggy1 wrote:

Can someone help me with this question? I felt a tickle in my brain when I saw this :oops:

For every positive even integer n, 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

I know the answer is E, but I haven't a clue as to why or how. HELP! :cry:

I am using . as multiplication.
h(100) = 2 . 4 . 6 . ... . 98 . 100
or h(100) = 2(1 . 2. 3. 4. ... . 49.50)

Now h(100) + 1 = 2(1 . 2. 3. 4. ... . 49.50) + 1
now every number from 2 t0 50 will divide the first term but not the second term so the divisor has to be > 50
Answer E

I am not sure if this is the most elegant answer but it makes sense to me. Any comments?

Re: Help with a GMAT question [#permalink]
22 Apr 2007, 12:44

techjanson wrote:

afiggy1 wrote:

Can someone help me with this question? I felt a tickle in my brain when I saw this

For every positive even integer n, 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

I know the answer is E, but I haven't a clue as to why or how. HELP!

I am using . as multiplication. h(100) = 2 . 4 . 6 . ... . 98 . 100 or h(100) = 2(1 . 2. 3. 4. ... . 49.50)

Now h(100) + 1 = 2(1 . 2. 3. 4. ... . 49.50) + 1 now every number from 2 t0 50 will divide the first term but not the second term so the divisor has to be > 50 Answer E

I am not sure if this is the most elegant answer but it makes sense to me. Any comments?

Hey Tech -

Thanks for the explanation ... what puzzles me about this question is the part where it mentions p as being "the smallest prime factor"; intuitively, doesn't that sound like p would be one of the first 10 primes? Maybe my thinking is screwed up, but that's what I initially gravitate towards.

This was the third question of my practice GMAT that I downloaded from the site. It must be pretty easy if you know a certain rule, but I am stumped...

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
B) Between 10 and 20
C) Between 20 and 30
D) Between 30 and 40
E) Greater than 40

This was the third question of my practice GMAT that I downloaded from the site. It must be pretty easy if you know a certain rule, but I am stumped...

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 B) Between 10 and 20 C) Between 20 and 30 D) Between 30 and 40 E) Greater than 40

Thanks Goalsnr. Wilfred's explanation was a little confusing for me, but it got me on the right track and I get it now. I posted the solution in my own words to help out anyone else confused by Wilfred's...

h(100) is a multiple of all even numbers from 2 to 100, so it's also a multiple of all numbers from 1 to 50 (Dividing the even numbers by 2).

h(100) + 1 is a number next to h(100), so it can't have a factor from 2 to 50. So, the smallest prime factor should be greater than 50. Hence the answer is E.

Edit: I seem to have repeated the solution already provided.

GmatPrep 1: Real tough problem [functions] [#permalink]
01 Jun 2007, 07:24

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 100
b. between 10 and 20
c. between 20 and 30
d. between 30 and 40
e. greater than 40

Prime number problem [#permalink]
05 Jun 2007, 08:01

How do you solve this? It's from the gmat prep practice exam.

For every positve 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

gmatclubot

Prime number problem
[#permalink]
05 Jun 2007, 08:01