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For every positive even integer n, the function h(n) is [#permalink]

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27 Aug 2005, 04:55

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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 10 (B) between 10 and 20 (C) between 20 and 30 (D) between 30 and 40 (E) greater than 40

For every positive even integer n, the function f(h) 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

Thanks, Yaron

First thing I think there are some mistakes in the question I think the bolded parts above should be "function f(n)" and f(100) + 1

Solution

f(100) + 1 = 2(1x2x3...50) + 1

2^50x 50! +1

So the number should be something like xyz....0000001

So the prime factor should be much higher than 40...infact over 1000 probably...

For every positive 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 0 and 20
B. Between 10 and 20
C. Between 20 and 30
D. Between 30 and 40
E. Greater than 40
_________________

For every positive 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 0 and 20 B. Between 10 and 20 C. Between 20 and 30 D. Between 30 and 40 E. Greater than 40

h(100)+1= 2.4.6.8......100 + 1
for every prime factor from 2, 3, to 47 (47 is the largest prime factor here coz 47*2=94<100. Remember all numbers here are even so to find the prime factor we have to, at least, devide each of them by 2). As we see the product contains prime factor from 2 to 47 ----> these prime numbers can't be factors of h(100)+1!. Thus the possible smallest prime factor of h(100)+1 must be greater than 47. E is correct.

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

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

h(100)= 2*4*6*.....*100 = (2*1)*(2*2)*(2*3) .....*(2*47)*(2*48)*(2*49)*(2*50)
As we observe h(100) is divisible by primes like 2,3,5......,47 ---> h(100)+1 is not divisible by these primes, in other words, these primes can't be factor of h(100)+1. Thus, the smallest prime factor of h(100)+1 must be greater than 47 . E is my choice.

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

1. Between 2 and 10
2. Between 10 and 20
3. Between 20 and 30
4. Between 30 and 40
5. Greater than 40

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

(1) Note first:Let x, y be two pos. integers then xy+1 is not divisible by x or y.
(2) Therefore H(100)+1 = 2*4*6*....*96*98*100 + 1 cannot be divisible by any odd number k smaller than 50 (because k*2 is a factor of H(100)).

Note that n is not the number of terms, the function is defined as the product of all even integer up to n.

Therefore, h(100) + 1 = (taking 2 as the common factor) 2^50 * ( 1 * 2 * 3 * 4 * â€¦* 50) + 1

Note also that the problem is not asking for a value of P, it is only asking you what might be P.

Note that a number divisible by 2, can be written as 2k where k is an integer, similarly, a number divisible by 3 can be written as 3i, where i is an integer.

Note also that h(100) is divisible by all the numbers 2,3,4,5,6,7,...50.

Hence, when h(100) + 1 is divided by the numbers 2,3,4,5,6,7,..50, the remainder is 1.

Hence, Note that h(100) cannot be divisible by any number that is less than or equal to 50. Hence, the smallest number that is a factor of h(100) is 51. Answer E.

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