Princeton PS

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Princeton PS

by maolivie » Mon Jul 09, 2007 12:41 pm
What is the largest integer value of x that makes 30!/3^x an integer?

Sorry, I don't have the answer choices.

I calculated 10 multiple of 3s in 30! (27, 24, 21,...), but I didn't know what to do from there

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by givemeanid » Mon Jul 09, 2007 3:28 pm
30 / 3 = 10
That means there are 10 numbers <= 30 that have 3 as a factor.
(3,6,9,12,15,18,21,24,27,30)

10 / 3 = 3 (floor of the result)
Of those 10, 3 have atleast 2 3s as factors.
(9,18,27)

3 / 3 = 1
Of those 3, 1 has 3 3s as factors.
(27)

So, total number of 3s in 30! = 10+3+1 = 14
x = 14 for 30!/3^x to be an integer.
So It Goes

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by maolivie » Tue Jul 10, 2007 6:03 am
what made you decide to divide 10/3?

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by givemeanid » Tue Jul 10, 2007 6:18 am
When you divide 30 by 3 first time, you get all the numbers (up to and including 30) that have 3 as a factor once.
Now, the quotient you get is 10. When you divide the quotient again, you get the number of numbers that have two 3s as factors.
Continuing this, if you divide again, you get the number of numbers that have three 3s as factors.

If this is confusing, pick any other pair of numbers. Lets say 44 and 5.

So, 44/5 = 8 (ignore the decimal part). What does 8 mean? It means there are 8 numbers smaller than 44 that divide by 5 ONCE.
You can check this too: 5, 10, 15, 20, 25, 30, 35, 40
Now, divide 8 by 5. 8/5 = 1 (ignore the decimal) What does 1 mean? It means there is 1 number that divides by 5 TWICE.
You can check this too: 25 (the smallest number that will divide by 5 twice is 5*5 = 25. If we go higher, the smallest number that divides by 5 thrice is 5*5*5 = 125)
So, total number of times 5 is a factor in = 8+1 = 9. Hence for 44!/5^x to be an integer, x >= 9


If you picked 130 and 5.
Then 130/5 = 26 (you can count these)
26/5 = 5 (25,50,75,100,125. 5 is a factor twice)
5/5 = 1 (125. 5 is a factor thrice)
Total = 26 + 5 + 1 = 32
So It Goes

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by 800GMAT » Wed Jul 11, 2007 6:44 pm
Is it 13?

10 multiples of 3, so 3^10
9 has 1 additional 3
and 27 has 2 additional 3s

10+1+2=13

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by jayhawk2001 » Wed Jul 11, 2007 9:01 pm
800GMAT wrote:Is it 13?

10 multiples of 3, so 3^10
9 has 1 additional 3
and 27 has 2 additional 3s

10+1+2=13
Actually 18 has to be included in the second category.
So, it has to be 10+1+1+2 = 14