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# number theory

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number theory [#permalink]  01 Apr 2004, 10:08
pls could you show full working when you attempt this:

What is the least number that should be multiplied to 100! to make it perfectly divisible by 3^50 ?

(a) 144 (b) 72 (c) 108 (d) 216
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I have no Idea on how to solve this type of question. Is there any other way that we can easily find how many 2's or how many 3's are there in a certain Factorial. Any one willing to share the TRICK???
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..I have gotten it now...

To look for the least numbers, lets try using 2 and 3 and if we don't, we can increase our search....

using 2 and 3 we get:

for 2
we get,..{100/2]+[100/(2^2)]+[100/(2^3)] +[100/(2^4)] +[100/(2^5)] +[100/(2^6)] = 50 + 25 +12 +6 + 3 +1 = 97

therefore 100! has 2^97 as the greatest power of 2 that can divide it

if we do for 3,

we get 3 ^48 as the greatest power of 3 that can divide 100!

but we can simplify 2^97 as 4^48 * 2^1

so now, you have 4^48 and 3^48

Therefore, the largest power of 12 that can divide 100! is 48.

for 3 ^50 to be included in 100!, 100! needs to be multiplied by 3^2 * 2 ^3 = 9 * 8 = 72.

does anyone have a shorter method?
тА║
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