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 Post subject: How to calculate the sum of n integersPosted: Tue Mar 23, 2010 7:18 pm

Joined: Tue Mar 23, 2010 7:14 pm
Posts: 1
Hi. I have these questions I can't do. Please help me to answer them.

1) For every positive even integer n, the function h(n) is 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?

The correct answer is 'greater than 40', but there's no explanation

2) If n is the product of the integers from1 to 20 inclusive, what is the greatest integer k for which 2^k is a factor of n?

A. 10
B. 12
C. 15
D. 18
E. 20

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 Post subject: Re: How to calculate the sum of n integersPosted: Wed Mar 24, 2010 6:06 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
h(100) = 2*4*6* ... 98*100
h(100) =2*1 * 2*2 * 2*3 * .... * 2*49 * 2*50
h(100) = 2^50 * 1*2*3*4 ... 49*50

so all numbers from 2 to 50 are factors of h(n) ... since they are factors of h(n), they will not be factors of a number one larger than h(n) ( h(n) + 1). So, the smallest prime factor must be larger than 50.

n = 1*2*3 * .... 19*20

we need to find how many "2's" we can get from the factorization of n

2 = 2
4 = 2*2
6 = 3*2
8 = 2*2*2
10 = 5*2
12 = 3*2*2
14 = 7*2
16 = 2*2*2*2
18 = 3*3*2
20 = 5*2*2

So, there are a total of 18, so 2^18 is a factor of n

-- Veritas Help

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