Problem Solving

Hello Community

Could anyone post an explanation for this question?

Q1) For every positive even integer n,the function h(n) is defined to be the product of all the even integrers from 2 to n,inclusive.If p is the smallest prime factor of h(100) + 1,then p is

a) btw 2 and 10
b) btw 10 and 20
c) btw 20 and 30
d) btw 30 and 40
e) greater than 40

Thanks

soni_pallavi wrote:Hello Community

Could anyone post an explanation for this question?

Q1) For every positive even integer n,the function h(n) is defined to be the product of all the even integrers from 2 to n,inclusive.If p is the smallest prime factor of h(100) + 1,then p is

a) btw 2 and 10
b) btw 10 and 20
c) btw 20 and 30
d) btw 30 and 40
e) greater than 40

Thanks

h(100) + 1 = 2+4+6+...+100 + 1
=> 2*(1+2+3+...+50) + 1 = 50*51 + 1

I think from here you need to check manually

clearly it is not divisible by 2, 3, 5, and 17
now, 7, 11, 13, 19, 23, 29, 31, 37 remains

50*51+1 mod 7 = 1*2 + 1 = 3
50*51+1 mod 11 = 6*7 + 1 = 10
50*51+1 mod 13 = -2*-1 + 1 = 3
50*51+1 mod 19 = -7*-6 + 1 = 5
50*51+1 mod 23 = 4*5 + 1 = 21
50*51+1 mod 29 = -8*-7 + 1 = 28
50*51+1 mod 31 = -11*-10 + 1 = 18
50*51+1 mod 37 = 13*14+1 = 2

so it will be greater than 40

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, the p is

A: Between 2 & 10
B: Between 10 & 20
C: Between 20 & 30
D: Between 30 & 40
E: Greater than 40

Since the difference between them is 1, h(100) and h(100)+1 are consecutive integers.
Consecutive integers are COPRIMES: they share no factors other than 1.

Let's examine why:

If x is a multiple of 2, the next largest multiple of 2 is x+2.
If x is a multiple of 3, the next largest multiple of 3 is x+3.

Using this logic, if we go from x to x+1, we get only to the next largest multiple of 1.
So 1 is the only factor common both to x and to x+1.
In other words, x and x+1 are COPRIMES.

Thus:
h(100) and h(100)+1 are COPRIMES. They share no factors other than 1.

h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100
Factoring out 2, we get:
h(100) = 2^50 (1 * 2 * 3 *... * 47 * 48 * 49 * 50)

Looking at the set of parentheses on the right, we can see that every prime number between 1 and 50 is a factor of h(100).
Since h(100) and h(100)+1 are coprimes, NONE of the prime numbers between 1 and 50 can be a factor of h(100)+1.

Thus, the smallest prime factor of h(100) + 1 must be greater than 50.

The correct answer is E.