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15 Apr 2005, 06:42
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
72% (00:53) correct
28%
(01:40)
wrong
based on 1119
sessions
History
Date
Time
Result
Not Attempted Yet
How many terminating zeroes does 200! have?
(A) 40
(B) 48
(C) 49
(D) 55
(E) 64
(A) 40
(B) 48
(C) 49
(D) 55
(E) 64
03 Oct 2013, 05:11
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To find number of terminating zero's count the number of 5's
\(\frac{200}{5}\) = 40
\(\frac{200}{25}\) = 8
\(\frac{200}{125}\) = 1
If you add them its 49 zero's
Explanation of this concept on this thread in detail...
question-on-terminating-zero-s-154440.html
\(\frac{200}{5}\) = 40
\(\frac{200}{25}\) = 8
\(\frac{200}{125}\) = 1
If you add them its 49 zero's
Explanation of this concept on this thread in detail...
question-on-terminating-zero-s-154440.html
25 Feb 2015, 16:45
Kudos
Bookmarks
Hi All,
It looks like everyone who posted in this thread understands the correct answer. For anyone who doesn't quite get why the math works the way that it does, here are some examples to prove the point.
200! = (200)(199)(198)(197)....(2)(1) so we know that it's a gigantic number. The reason why it will end in a "string" of 0s is because of all of the multiples of 5 involved.
When multiplying integers, there are two ways to get a number that ends in a 0:
1) multiply a multiple of 5 by an even number
2) multiply a multiple of 10 by an integer.
(5)(2) = 10 so we get one 0 for every multiple of 5
However, 25 = (5)(5). It has TWO 5s in it, so there will be two 0s.
eg (25)(4) = 100
With 125 = (5)(5)(5), we have THREE 5s, so there will be three 0s.
eg (125)(8) = 1,000
So when we divide 200 by 5.....200/5 = 40 multiples of 5. SOME of those multiples of 5 are multiples of 25 (or 125) though. Each of those special cases has to be accounted for.
25, 50, 75, 100, 125, 150, 175 and 200 are all multiples of 25, so they each include one extra 0 (and 125 includes two extra 0s).
So we have 40 + 8 + 1 = 49 zeroes.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
It looks like everyone who posted in this thread understands the correct answer. For anyone who doesn't quite get why the math works the way that it does, here are some examples to prove the point.
200! = (200)(199)(198)(197)....(2)(1) so we know that it's a gigantic number. The reason why it will end in a "string" of 0s is because of all of the multiples of 5 involved.
When multiplying integers, there are two ways to get a number that ends in a 0:
1) multiply a multiple of 5 by an even number
2) multiply a multiple of 10 by an integer.
(5)(2) = 10 so we get one 0 for every multiple of 5
However, 25 = (5)(5). It has TWO 5s in it, so there will be two 0s.
eg (25)(4) = 100
With 125 = (5)(5)(5), we have THREE 5s, so there will be three 0s.
eg (125)(8) = 1,000
So when we divide 200 by 5.....200/5 = 40 multiples of 5. SOME of those multiples of 5 are multiples of 25 (or 125) though. Each of those special cases has to be accounted for.
25, 50, 75, 100, 125, 150, 175 and 200 are all multiples of 25, so they each include one extra 0 (and 125 includes two extra 0s).
So we have 40 + 8 + 1 = 49 zeroes.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
