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# number theory

Author Message
Intern
Joined: 31 Mar 2004
Posts: 28
Location: texas

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01 Apr 2004, 11:08
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Manager
Joined: 27 Feb 2004
Posts: 185
Location: illinois

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03 Apr 2004, 20:07
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???
Intern
Joined: 31 Mar 2004
Posts: 28
Location: texas

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03 Apr 2004, 22:56
..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║
03 Apr 2004, 22:56
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