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The number of zeros at the end of m when written in integer

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The number of zeros at the end of m when written in integer [#permalink] New post 16 Sep 2017, 10:35
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\(m=2^{16}3^{17}4^{18}5^{19}\)

\(n=2^{19}3^{18}4^{17}5^{16}\)



Quantity A
Quantity B
The number of zeros at the end of m when
written in integer form
The number of zeros at the end of n when
written in integer form


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: The number of zeros at the end of m when written in integer [#permalink] New post 21 Sep 2017, 08:41
Any help with this one?
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Re: The number of zeros at the end of m when written in integer [#permalink] New post 05 Oct 2017, 18:01
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can be written as
m = 2^16 3^17 4^17 5^16 4 5^3
n = 2^16 3^17 4^17 5^16 3 2^3
now we see in first 4 terms are same for both m and n but it m we see 5^3 * 4 that means two more zeroes
thats why quantity a is greater
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Re: The number of zeros at the end of m when written in integer [#permalink] New post 09 Oct 2017, 22:17
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the number of zeroes will be determined by the number of 5s and 2s. since 5*2 gives us one zero, the number of zeroes will be the number of (5*2) that we get. So in first case it is 19 and in second case it is 16. Hence A is greater than B. :)
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Re: The number of zeros at the end of m when written in integer [#permalink] New post 22 Feb 2018, 04:05
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m=2^{16}3^{17}4^{18}5^{19}

n=2^{19}3^{18}4^{17}5^{16}
We can rewrite it as following:

m= 2^{16}..2^{36}...5^{19} = 2^{52}..5^{19} = 19 zeroes
n= 2^{19}..2^{34}...5^{16} = 2^{53}..5^{16} = 16 zeroes

So A is greater.
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Re: The number of zeros at the end of m when written in integer [#permalink] New post 24 Feb 2018, 17:23
A
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Re: The number of zeros at the end of m when written in integer [#permalink] New post 04 Mar 2018, 02:48
Can someone explain this in detail?
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Re: The number of zeros at the end of m when written in integer [#permalink] New post 21 Mar 2018, 12:12
Answer: A
The parameters that can produce 0 are 2 and 5.
n= 2^19 * 3^18 * 4^17 * 5^16 = 2^16 * 5^16 * 3^18 * 4^17 (there are 16 2s and 16 5s, we didn’t consider 4 because there are only 16 5s, if there were more than 19 5s we used 4s after 2s. So there are 16 zeros in the end of n)
m= 2^16 * 3^17 * 4 ^18 * 5^19= 5^19 * 2^52 * 3^17 = 5^19 * 2^19 * 2^33 * 3^17 (there are 19 multiplication of 2 with 5. So there are 19 zeros in the end of m)
So the number of zeros in m are more than n. And the answer is A.
Re: The number of zeros at the end of m when written in integer   [#permalink] 21 Mar 2018, 12:12
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