## Problem of the day 1.07.09

**300! is divisible by (24!)^n. what is the max. possible integral value of n?**

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**300! is divisible by (24!)^n. what is the max. possible integral value of n?**

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Written by Implex

July 1, 2009 at 9:05 am

Posted in Number Thoery, Problem of the week

Tagged with cat, IIM, Number Theory, Online CAT, quant

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%d bloggers like this:

max. possible integral value of n is…3

anupJuly 1, 2009 at 2:14 pm

MAximum value of n is 12

NareshJuly 1, 2009 at 3:59 pm

Both of you have got it wrong

RahulJuly 1, 2009 at 4:15 pm

den is it 13???..plz send me d soln..

anupJuly 2, 2009 at 6:24 am

The highest prime in 24! is 23

so the n will be limited by the power of 23 in 300!

Now solve!

RahulJuly 2, 2009 at 6:27 am

yes the power of the highest prime will determine the value of 23.

hence the answer is 13 + 1 =14.

yodhaJuly 3, 2009 at 9:14 am

Answer is 13 yodha! why do u add 1?

RahulJuly 3, 2009 at 9:16 am

sorry just made a slight error.

i added 1 with the thought 300/13^2 (i.e. considered the power of 13). now u can understand. how big is this error.

sorry for that.

yodhaJuly 4, 2009 at 2:20 pm

hi rahul..it may b too trivial to ask but can u explain as in y the highest power of prime would be the determining factor,also y cant the power of 24 do?

saurabhJuly 28, 2009 at 11:04 am

because 24=2^3×3

there will be many 2s and 3s in 300!

but there will be very few 23s.

I hope u get it!

RahulJuly 28, 2009 at 12:12 pm

thnx a lot

saurabhJuly 28, 2009 at 5:04 pm