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Dan
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Determine the largest possible integer n such that 942! is 01 Nov 2004, 03:03
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Determine the largest possible integer n such that 942! is divisible by 15^n.

(A) 62 (B) 125 (C) 233 (D) 314 (E) 471



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Determine the largest possible integer n such that 942! is 01 Nov 2004, 08:56
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I'd go with A) 62

942/15 = 62 r 12.
Just think logically, how many times 15^n will be in 15! --> Just once. So divide 942! by 15 and you will get the answer by looking at the integer part.
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Determine the largest possible integer n such that 942! is 01 Nov 2004, 11:13
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I'm going for C as my suspect,
Although i didn't get this exact number..
The key, i guess, is to figure out how many times does "5" appear in 942!.
942/5 = 188 so the answer is surely not A or B.
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Determine the largest possible integer n such that 942! is 01 Nov 2004, 13:49
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Well, I looked at this question again and I see what you mean Dookie. Let's say 15!, we have 1*15 but also 3 factors of 5 in there and a few more of 3. This means that by looking at the factors of 5, we will find the number of factors of 15. 942/5 = 188 is a good start. We know that it is at least 188. What we also need to see is numbers like 25,50,75,100 which each have 5^2 so 2 factors of 5. Next, we have 125 which has 3 factors of 5. And finally, we have 625 which has 4 factors of 5. For number of "extra" 5's we need to consider:
1- Where there is 2* 5's we have 942/25=37
2- Where there is 3*5's we have 942/125=7
3- Where there is 4*5's we have 942/625=1

Hence, max n is 188+37+7+1=233
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Determine the largest possible integer n such that 942! is 01 Nov 2004, 13:55
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Nice Paul,
I got lucky with my estimate
:lol:
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Determine the largest possible integer n such that 942! is 01 Nov 2004, 14:04
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Good calculation Paul !

I tried to apply the possessed method and got 928 :shock:
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Determine the largest possible integer n such that 942! is 02 Nov 2004, 03:47
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To add to Paul's, the number of factors of five is 233, but we also have to conisder the number of factors of 3, since to divide by 15, we need to take care of both 5s and 3s in there..taking 5s alone is not enough unless we consider 3s as well..

so 233 factors of 5s, and 467 factors of 3s. But since the latter makes number 942 not evenly divisible by 15^467, the answer is 233.

Take a smaller example of same question but where ...n such that 20! is divisible by 6^n.
Factors of 6 are 2 and 3. Do we consider number of factors of 2 or 3? We have to calculate both and see...unless there is a shortcut.
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Determine the largest possible integer n such that 942! is 02 Nov 2004, 05:34
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The reason I did not consider factors of 3 is because let's say in 15, there are 5*3s but only 3*5s. This means that in any given number, there are less factors of 5 than there are factors of 3. Hence, looking at factors of 5 will automatically take care for the factors of 3 because the former is more restrictive.
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Determine the largest possible integer n such that 942! is 03 Nov 2004, 15:47
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I just would like to add that the possessed method, for a reason i did not find, seems unuseful for this question while it is for 20! and 2^k for instance. I must admit I am a little bit puzzled with that point...
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Determine the largest possible integer n such that 942! is 03 Nov 2004, 15:58
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twixt
I just would like to add that the possessed method, for a reason i did not find, seems unuseful for this question while it is for 20! and 2^k for instance. I must admit I am a little bit puzzled with that point...
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Are you sure the possessed method does not work? :lol: I found it quite useful and it DOES work. Here it is (remember to keep only the integer part):
942/5 = 188
188/5 = 37
37/5 = 7
7/5 = 1
Add it up: 188+37+7+1 = 233 :P
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Determine the largest possible integer n such that 942! is 05 Nov 2004, 04:11
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Thank you Paul !
My mistake : I used only the remainder but did (942-188)/15... I dont know why I always considered the substraction...



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