Functions

Akansha wrote: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

OA is E

h(100)= 2*4*6*.....100;
h(100)+1= 2^50(1*2*3*4*5.....50)+1;

any prime no. less than 47 when divided by h(100)+1; will leave a remainder of 1; therefore smallest prime factor of h(100)+1; will be greater than 47 hence E

manpsingh87 wrote:
h(100)= 2*4*6*.....100;
h(100)+1= 2^50(1*2*3*4*5.....50)+1;

any prime no. less than 47 when divided by h(100)+1; will leave a remainder of 1; therefore smallest prime factor of h(100)+1; will be greater than 47 hence E

instead of 47, smallest prime factor must be greater than 50.. right??
Answer will remain the same..

It's basically the same thing since there aren't any prime factors between 47 and 50.

mapsingh did a nice job of explaining it.

One way to look at it is through the divisibility rules.

If X is divisible by Y and and Z is NOT divisible by Y, then X+Z will NOT be divisible by Y.

Based on mapsingh's analysis, we know that our (h100) term is divisible by all possible primes up to 50 (or 47). 1 is not divisible by any prime number, therefore our h(100)+1 term will not be divisible by any term that (h100) is divisible by.

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
Options:
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 co-primes: 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. Integers that share no factors other than 1 are called coprimes.

Thus, in the problem above, 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 share no factors other than 1, 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.

manpsingh87 wrote:

Akansha wrote: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

OA is E

h(100)= 2*4*6*.....100;
h(100)+1= 2^50(1*2*3*4*5.....50)+1;

any prime no. less than 47 when divided by h(100)+1; will leave a remainder of 1; therefore smallest prime factor of h(100)+1; will be greater than 47 hence E

hello

could you please elaborate it further, how do i reach it h(100)+1= 2^50(1*2*3*4*5.....50)+1??

i cant get it

regards nafi

thanks in advance

nafiul9090 wrote:

manpsingh87 wrote:

Akansha wrote: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

OA is E

h(100)= 2*4*6*.....100;
h(100)+1= 2^50(1*2*3*4*5.....50)+1;

any prime no. less than 47 when divided by h(100)+1; will leave a remainder of 1; therefore smallest prime factor of h(100)+1; will be greater than 47 hence E

hello

could you please elaborate it further, how do i reach it h(100)+1= 2^50(1*2*3*4*5.....50)+1??

i cant get it

regards nafi

thanks in advance

hi, the function h(n) is defined to be the product of all even integers from 2 to n inclusive;

so, h(10)= 2*4*6*8*10;
=2*1*2*2*2*3*2*4*2*5;
= 2^5(1*2*3*4*5);

similarly h(100)=2*4*6*......100;
=2^50(1*2*3*4*5....50);

i hope it helps..!!!

Cheers for the method!