Given that 4704/m = n^2, where m and n are whole numbers and n is as large as possible, find the value of m and of n

n=28,m=6

Step-by-step explanation:

Rearrange to give 4704 = m * n^2

Express 4704 in its prime factorisation:

2^5 * 3^1 * 7^2

We know that n must be as large as possible and, because it is squared, be the even numbers.

We get n^2 = 2^4 * 7^2 so n = 2^2*7 = 28

m is the leftovers

2^1*3 = 6

2nd one in the first row is correct

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step-by-step explanation: