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- anonymous

If K is an integer, what is the smallest possible value of K such that 1040K is the square of an integer?

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- taramgrant0543664

1040k = 2*2*2*2*5*13*k
to make it a square:
(2*2*5*13) (2*2 *k)
so k must= 5* 13 or 65

- anonymous

wait im not understanding this problem can you explain in details? in other words dumb it down for me

- anonymous

honestly i dont even know what theyre asking for

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- taramgrant0543664

I don't really know how to explain it very well, you have to break it into its counterparts as the prime factors

- anonymous

why is it 65 why muktiply 13 *5

- taramgrant0543664

Because the 2's have counterparts but the 5 and the 13 doesn't so you can multiply them together

- anonymous

so we're looking for the product without duplicate numbers?

- taramgrant0543664

The numbers 5 and 13 can't break down any farther without becoming an irrational number

- anonymous

so what if there was 2x2x2x2x3x3x5x7x8?

- anonymous

then is it 2^2 and 3^3 and 5x7x8?

- anonymous

just multiply the last 3?

- taramgrant0543664

You divide by 2 initially as it is the smallest integer so that's why you have 2^4, at that point you can't divide be two anymore without it becoming a rational number because you are left at 65. 3 and 4 don't divide into that so the next real number that divides into it is 5 so that's why you get 5 and 13 and neither of those can be broken down anymore

- anonymous

can you break down the question for me cuz i dont understand the last part of the question... "such that 1040K is the square of an integer?" What does this mean?

- taramgrant0543664

We need to find a number k such that 1040k = 2^n for some number n.
The prime factorization of 1040 is 2^4 5^1 13^1.
In order for 1040k to be a square number, we need every prime factor in 1040k to be at least raised to the second power. The only numbers that are not are 5 and 13 - thus, we need one more factor of 5 and one more factor of 13. Therefore, k = 5*13 = 65.

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