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I'm going to assume your h(100) is actually f(100) or else, f(n) was supposed to be h(n).If p is the smallest prime factor h(100)+1. then p is:
Maybe I am wrong, maybe my calculator is broken, but 32 * 120 = 3840 NOT 600.mp2437 wrote: To visualize it more simply, think of a smaller example, say h(10) = 2 * 4 * 6 * 8 * 10. This is the same as 2^5*(1*2*3*4*5) = 32 * 120 = 600.
This does make it more clear. Thanks for the explanation. Is that a rule or something that all even integers multiplied up to N equals what you wrote in this case 2^50(1x2x...50)? Are there other such rules, what if the question said it was all odd integers up to 100? or just all integers up to 100?mp2437 wrote: Particularly, this may be easier to understand:
"I will elaborate on this more to make it simple.
h(n) = 2 x 4 x .....n
h(100) = 2 x 4 x ...100
Since there are 50 even nos. in 1-100 we can write:
h(100) = 2^50 *(1x2x3x4...50)
now, h(100) will be divisible by each nos. from 1 to 50 as it is in the numerator.
but, h(100) + 1 = (2 x 4 x ...100 + 1) = 2^50 (1 x 2 x...50) +1
with 1 added to it these prime nos. cannot divide the nos. in simpler terms if A can divide B, A cannot divide B+1. say 2 can divide 10, but cannot divide 11.
Thank you for the explanation! That makes a lot more sense now. The answer was E.mp2437 wrote:For consecutive integers, say from 1 to 10, it is the same idea as that with just even integer products. The key here is that any number less than 100 will be divided by the product of 1 to 10, since the product contains a multiple of the number you are dividing it by (i.e - the product of 1 to 10 will be divisible by 7 because you have 1*2*3*...6*7*8*9*10, so dividing by 7 will work). This means that up to 10, you are certain that any number will be divisible in your product. However, if you try dividing by 11, you will see that nothing is divisible by it. So you know that adding 1 to this product, or any number for that matter, could make the number divisible by a prime number, but you have to keep in mind that the prime number will be greater than 10, since every prime number below it can already be divided into the main product.
Now for the example on even integers (same idea as the one above, only you distribute the 2 out)
For every even integer, they all share the "2" in common, so you could just pull it out and distribute accordingly. So the product of the 4 even integers 2 * 4 * 6 * 8 is the same (1*2) * (2*2) *(3*2) * (4*2), so you could pull out a 2 from each one, which could also be written as 1*2*3*4*2*2*2*2 = 2^4*(1*2*3*4). Naturally, you could see this being applied to any product of even integers up to n, where the product will be 2^n * (1*2*...*n), where n is the number of even integers in your set.
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