65 as that is what is left post factoring. All the factors should have 2 values atleast

One should also remember that the exponents of prime factors are even.

or we can solve the problem by taking the prime factorization of 1040 , when we do so we notice the factors of 13 and 5 are not even rather they are only 1 each..

hence we choose the option he because when we multiply 1040 by 65 13 and 5 factors make the exponents of 1040*65 even


hence option is E is correct.

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

A. 2

B. 5

C. 10

D. 15

E. 65

1040 can be broken down into its component prime number as 2^4*5*13
the square of \(2^4\) is \(2^2\) however 65 does not have a integer square root therefore k has to be 65 as only then can we get a integer when a square root is taken

Please explain.

Since we are looking for square, the exponent has to be even..
Similarly if we are looking at a cube, the exponent should be divisible by 3..
.1040k=2*520k=\(2^2*260k=2^2*2*130k=2^4*65=2^4*5*13\)
So 2 has a power of 4, and we require one more of 5 and 13 to make the entire term as square

Questions like this can often be solved by figuring out prime factors.

We know that 1040 * k is a perfect square, so find the prime factorization of 1040 first:

104*10
2*52* 5*2
2*2*26*5*2
2*2*2*13*5*2

So the prime factorization of 1040 = \(2^4*5*13\)

For a number to be a perfect square, each of its prime factors need to be paired with a matching prime factor.

\(2^4\) is 16, a perfect square. Each factor of 2 is paired with another factor of 2. But 5 and 13 don't have matching factors, so we need another 5 * 13 to make a perfect square. That product is k.

k = 5*13 = 65

Answer: E

Prime factorization of 1040 = \(2^4*5*13\)

Since \(\sqrt{2^4}\) = 4

Since \(2^4\) is already a perfect square the other factors not having a perfect square are 5 and 13

Hence the lowest number required to be multiplied to 1040 to make it a square = 5 * 13 = 65

Hence answer is E

Use backsolving: Plug the answer choices in as k using the on-screen calculator and then take the square root of the product.

Hi

Does the on-screen calculator have square-root? If yes, doesn't this make it a non-question!

It has

Keep in mind that the onscreen GRE calculator can handle numbers less than 100,000,000
So, this question could be extended (using bigger numbers) to render the calculator useless.

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