Please help on this one: how can I solve it fast?

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Anton8103: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Sun May 17, 2009 6:54 pm

by Anton8103 » Sat May 08, 2010 9:56 am

5) 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 between

a. 2 and 10
b. 10 and 20
c. 20 and 30
d. 30 and 40
e. > 40

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Source: Beat The GMAT — Problem Solving | Sat May 08, 2010 9:56 am

akhpad: Legendary Member; Posts: 809; Joined: Wed Mar 24, 2010 10:10 pmFollowed by: 4 members; Thanked: 50 times

by akhpad » Sat May 08, 2010 10:14 am

This problem has been discussed many times. Please search it on Beat the GMAT.

Answer >50

Hence E

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gmatmachoman: Legendary Member; Posts: 2326; Joined: Mon Jul 28, 2008 3:54 amFollowed by: 2 members; Thanked: 173 times; GMAT Score:710

by gmatmachoman » Sat May 08, 2010 10:17 am

akhp77 wrote:This problem has been discussed many times. Please search it on Beat the GMAT.

Answer >50

Hence E

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shashank.ism: Legendary Member; Posts: 1022; Joined: Mon Jul 20, 2009 11:49 pmFollowed by: 2 members; Location: Gandhinagar; Thanked: 41 times

by shashank.ism » Sat May 08, 2010 7:17 pm

Anton8103 wrote:5) 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 between

a. 2 and 10
b. 10 and 20
c. 20 and 30
d. 30 and 40
e. > 40

h(n) = 2x3x4................xn = n!/1=n!
so h(100) + 1 = 100! + 1
since 100! is divisible by all the numbers between 1 to 100 and further numbers too so whenever 100! + 1 would be divided by these numbers they will leave a remainder as 1.

so smallest prime factor would be surely [spoiler]greater than 100 and hence >40
ans E[/spoiler]

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kstv: Legendary Member; Posts: 610; Joined: Fri Jan 15, 2010 12:33 amFollowed by: 2 members; Thanked: 47 times

by kstv » Sat May 08, 2010 9:14 pm

try

https://www.manhattangmat.com/forums/for ... t1152.html

@ shashank
h(100) = 2*4*6.......100
= 2(1*2*3......50)
h(100) + 1 = 2(1*2....50)+1

p = 53?

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