Math question - please help

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Math question - please help [#permalink]

24 Oct 2007, 12:45

Please review and appreciate your feedback with the detail description of the answer. Thanks in advance!!

For every positive even integer n, the funciton 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

1, between 2 and 10,

2, between 10 and 20

3, between 20 and 30

4, between 30 and 40

5, greater than 40

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24 Oct 2007, 12:48

A. 2 is the smallest even prime in the function, so 2 + 1 would be 3.

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Re: Math question - please help [#permalink]

24 Oct 2007, 12:53

pinal2 wrote:

For every positive even integer n, the funciton 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

1, between 2 and 10,

2, between 10 and 20

3, between 20 and 30

4, between 30 and 40

5, greater than 40

The answer is (E)

h(100) = 2 x 4 x 6 x 8 x ... x 100
= (2 x 1) x (2 x 2) x (2 x 3) ... x (2 x 50)
= (2^50) x 50!

Now what can be the smallest prime that can divide (2^50) x 50! + 1?
Let's call (2^50) x 50! "First Term"

Let start with number 2: 2 can divide First Term but it cannot divide 1. Out
number 3: 3 can divide First Term but it cannot divide 1. Out
...
number 50: 50 can divide First Term but it cannot divide 1.

So the smallest prime number of h(100) + 1 is clearly greater than 50.

The answer is E.

Last edited by devilmirror on 24 Oct 2007, 12:59, edited 1 time in total.

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24 Oct 2007, 12:56

This question was discussed many many times in this forum:

In order to better understand this problem , I suggest to plug in some values.

h(n) where n is positive and even (e.g 2,4,6) and h(n) is the prouduct of all even numbers from 2 to n.

n=10

h(10) = 10*8*6*4*2

so after we understand what this problem wants , we can now solve it !

h(100)+1 = (100*98*96*94....6*4*2) + 1 can be simplified to (2^50)*50!+1

what is the smallest prime factor ?

For any factorial + 1 the smallest factor (apart from 1) is greater then any of the members of the factorial

2! + 1 = 3: smallest factor is 3
3! + 1 = 7: smallest factor is 7
4! + 1 = 25: smallest factor is 5
5! + 1 = 121: smallest factor is 11
6! + 1 = 721: smallest factor is 103

50! + 1 = a number greater then 50

hope this will help.

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24 Oct 2007, 15:04

It helped a lot!! Thanks guys!!

[#permalink] 24 Oct 2007, 15:04

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Math question - please help

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Math question - please help

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