For any integer P greater than 1, P! denotes the product of all the in

Question

For any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?

A. 72
B. 36
C. 9
D. 8
E. 4

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$!vakumar.m · Answer

If 10! Is divisible by 10080*h and h is a perfect square, what is the greatest possible value of h?

10080 = 2 * 7 * 8 * 9 * 10
=> h has to be a product of either of these numbers - (3, 4, 5, 6)
the greatest possible value of h is 36 as 36 is a perfect square.
Answer B

nicbo · Answer

KarishmaB  could you please explain this one?

Thank you in advance.

KarishmaB · Answer

10080 = 2^5 * 3^2 *5 * 7

"10! Is divisible by 10080*h"  means

\frac{10!}{10080h}=x  where x is an integer.

10! = 10080 * h * x

1*2*3*4*5*6*7*8*9*10 = 2^5 * 3^2 *5 * 7 * h * x

Look at all that gets cancelled:

1*2*3* 4*5 *6* 7*8 * 9 *10 =  2^5 * 3^2  * 5  *  7  * h * x

We are left with:

2 * 3 * 6 * 10 = h * x = 2^2 * 3^2 * 2 * 5

Since h is a perfect square, its greatest value is 2^2 * 3^2 since a perfect square must have even exponents for all prime numbers.

Hence the greatest value of h is 36.

Answer (B)

Anuramaa · Answer

Question is asking 10!/1008*H where H is a square

Basically, 1008*H can have every factor of 10! for 10! to be able to still be divisible by 1008*H

1008 = 2^5 * 3^2 * 5^1 * 7^1
10! = 2^7 * 3^4 * 5^1 * 7^1

Extra numbers in 10! = 2^2*3^2 = 6^2 = 36 which is a perfect square

For any integer P greater than 1, P! denotes the product of all the in

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For any integer P greater than 1, P! denotes the product of all the in

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