x is a positive integer and x^{3} is a multiple of both 2424 and 400400. What is the largest integer that must be a divisor of x?

@devanshsolani -

What is the largest integer that must be a divisor of x?

Ans- largest divisor of any number will be the number itself. Hence, x.

Now find x.

Breaking 24 and 400 in prime factorization-

24= 3*2^{3}

400= 2^4 * 5^{2}

Find a number that is multiple/divisible by both 24 and 400:

number1 = (3)(2^{4})(5^{2})

if we see this number1, it is not a perfect cube, but a multiple of 24 and 400.

Make the above number1 a perfect cube -as we need a cube that is multiple/divisible by 24 and 400=

x^{3} =number1 * 3^{2} * 2^{2} * 5

x^{3} = 3^{3} * 2^{6} * 5^{3}

Now, find the cube root of x^3 to get x

x= 3 * 2^{2} * 5 = 3 * 4 * 5 = 60

Hope this helps!!

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