Oneva wrote: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

Answer is E

What I dont get is why? If it is asking for the smallest prime wouldnt it be 2?

h(100) = 2*4*6*...*98*100 = 2(1*2*3*...*50) = 2*50!

That means h(100) is divisible by all the numbers <= 50 (and hence h(100) is divisible by all prime numbers less than equal to 50).

Now you have to know this rule to solve this problem :

If n is a multiple of every prime < p, n + 1 CANNOT be a multiple of any of those primes < p.
(See below for example of this rule)

So, h(100)+1 can not be a multiple of those primes and the largest prime which divides h(100) is 47, hence
h(100)+1 must be divisible by a prime number greater than 47.

Answer is E.

Test Case for the rule mentioned above :

5! is divisible by primes : 2, 3, 5
So, 5!+1 shouldn't be divisible by 2, 3 or 5

7! is divisible by primes : 2, 3, 5, 7
So, 7!+1 wouldn't be divisible by 2,3,5 or 7.

Hope that clears the rule I mentioned.

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

Here is the rule that is being tested with this problem:

If x is a positive integer, the only factor common both to x and to x+1 is 1. They share no other factors.

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. They share no other factors. (For future reference, integers that share no factors other than 1 are called coprimes.)

Thus, in the problem above, we know that 1 is the only factor common both to h(100) and to h(100) + 1. They share no other factors.

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). This means that NONE of the prime numbers between 1 and 50 is a factor of h(100) + 1, because h(100) and h(100) + 1 share no factors other than 1.

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

The correct answer is E.

Mitch/ Ansu

can u entend this problem to explaining the statement below ?



10!+ 11! is a multiple of any integer from 1 to 10, because every integer between 1 and 10
inclusive is a factor of both 10! and 11!, separately.

bblast wrote:Mitch/ Ansu

can u entend this problem to explaining the statement below ?



10!+ 11! is a multiple of any integer from 1 to 10, because every integer between 1 and 10
inclusive is a factor of both 10! and 11!, separately.

If every term of an expression is a multiple of n, then the entire expression is a multiple of n.

Since 9 is a factor both of 10! and of 11!, 9 is a factor of 10! + 11!.
Since 7 is a factor of 10!, 11! and 93!, 7 is a factor of 10! + 11! + 93!.
Since 10 is a factor of 10!, 11!, 93!, and 154!, 10 is a factor of 10! + 11! + 93! + 154!.

Another way to view this situation:
Looking at 10! + 11!, we can factor out any integer from 1 to 10. For example:
10! + 11! = 2(10*9*8*7*6*5*4*3*1 + 11*10*9*8*7*6*5*4*3*1).
10! + 11! = 6(10*9*8*7*5*4*2*3*1 + 11*10*9*8*7*5*4*3*2*1).

Thus, every integer from 1 to 10 is a factor of 10! + 11!.

Does this help?

By the way, we should should recognize that 10! + 11! = 10!(1+11) = 10! * 12.

thanks Anurag and Mitch

The elaboration certainly helped.

It follows that the largest prime factor of 10!+11! is 7 (and not 11)?

also the largest number which can be a factor of our expression is again 10!(12)