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# Factors

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Senior Manager
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18 Jul 2004, 02:15
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can someone help me slove such problems.

If m is a product of all integers from 1 to 30, inclusive, then what is that greatest interger of n for which 3^n is a factor of m.

a. 10
b. 12
c. 14
d. 16
e. 18

Please explain a method to do such questions. Thanks.
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Last edited by carsen on 18 Jul 2004, 05:55, edited 1 time in total.
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18 Jul 2004, 04:54
My ans is b - 12

Basically one needs to find the number of '3's present in 30!

i.e. 30/3 + 30/3^2 + 30/3^3 + ...
= 10 + 2 + 0 + 0 +....
= 12

-Amith
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18 Jul 2004, 05:09
C. n=14
is it m that the qs was referring to? be careful with your typing, it freaked me out trying to find P

m = 1x2x3x4x...x30
there are 10 multiples of 3 (3x1, 3x2, 3x3,...3x10)
3 mutilples of 3^2 (9x1, 9x2, 9x3)
1 multiple of 3^3 (27)
therefore the number of times factor 3 appears in the product m is 10+3+1 = 14.
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18 Jul 2004, 07:25
Carsen
If I am not mistaken, we discussed same type of problem few days back. Perhaps it was sth like max power of 10 or something.

Here again, we can apply same technique.

We have to look in 30!.

Number contributing to 3 are... and the Power they are contributing....

Numbers : 3 6 9 12 15 18 21 24 27 30
Powers : 1 1 2 1 1 2 1 1 3 1 (Total = 14)

PS: 9 = 3*3, 27 = 3*3*3

If I have not done any calculation mistake... (C) is the answer...

Try to correlate this problem with the previous problem we solved. U will get the approach.
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18 Jul 2004, 08:43
That's the perfect response, JPV.

Carson, this is what I tell my students about any problem that has factors, multiples, products, divisors, or any other reference to perfect integers:

The number in the denominator must disappear completely to make an integer.

If I know that x/y is an integer, than I know that somehow, y will cancel with x.

This problem is broken down essentially to 30!/3^n, and we know it's an integer, because 3^n is a factor of 30!. Since we're just talking about a string of 3's on the bottom, we need to know how many 3's are on top to figure out the maximum number on the bottom. And since the 3's are coming only from the multiples of 3 in 30!, we need only to consider, as jpv and the others said, 3,6,9,13,15,18,21,24,27,30. Prime factor each of these numbers, and you'll find that there are 14 3's among them.

So if n was 15, let's say, then there would be 15 3's on the bottom and only 14 on top, and the bottom wouldn't canel out completely, and you wouldn't have an integer. So it has to be 14.
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18 Jul 2004, 19:46
amithmp wrote:
My ans is b - 12

i.e. 30/3 + 30/3^2 + 30/3^3 + ...
= 10 + 2 + 0 + 0 +....
= 12

oops, my method is right, butI did wrong division

30/3 + 30/3^2 + 30/3^3 + ...
= 10 + 3 + 1 + 0 +....
= 14
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18 Jul 2004, 19:59
thanks carsen for this question!
18 Jul 2004, 19:59
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