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Bunuel
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Rajaking
Hi Bunuel,

Thanks for posting the question. I have restarted my prep after quite some time. But, I remember questions like these require us to calculate the greatest integer function. In this case, it is asked the highest power of 288 that can divide 55!. So, first, we begin with prime factorizing 288.

288 = 2^5 X 3^2, it has two prime factors

now we use the GIF using the prime factors 2 and 3
Using 2
=[55/2] + [55/2^2] + [55/2^3] + [55/2^4] + [55/2^5]
= 27 + 13 + 6 + 3 + 1
= 50 no of factors of 2 in 55!

Using 3
=[55/3] + [55/3^2] + [55/3^3]
= 18 + 6 + 2
=26 no of factors of 3 in 55!

Since the minimum out of those two factors is 26, does it mean the answer is 26? D?

Not sure if we have any follow-up steps.

Regards,
RK


Rajaking

further part as follws

Since 288 =2^5 *3^2
but we have total of 50's number of 2
& required is pairs of 2^5 from these 50's number of 2
so a total of 10 pairs
Similarly out of 26's number of 3
we get 3^2 only 13 times

So total 10 no's of 2^5 * 3^2 is possible within 55!

Hope it clears

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288 = 2^5 X 3^2
Let the power be k

Now, in 55!,
3 occurs 18+1+2 = 21 times
2 occurs 27+1+2+3+4 = 37 times

So, (2k>=21) && (5k>=37) should be satisfied
Hence, k = max(10, 7)
k = 10 ->Option C
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I am confused. In 288^10 we only get 3^20 but since 55! needs 3^26 don't we need to go to the power of 288^13 to divide evenly?
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Hello from the GMAT Club BumpBot!

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