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The quantity will end in how many zeros

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The quantity will end in how many zeros [#permalink] New post 09 Aug 2017, 15:41
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Question Stats:

62% (02:03) correct 37% (03:29) wrong based on 24 sessions


The quantity \(3^3 4^4 5^5 6^6\) - \(3^6 4^5 5^4 6^3\) will end in how many zeros ?

A. 3

B. 4

C. 5

D. 6

E. 9
[Reveal] Spoiler: OA

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Re: The quantity will end in how many zeros [#permalink] New post 07 Feb 2018, 01:18
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3^3 4^4 5^5 6^6 - 3^6 4^5 5^4 6^3
=...(2^14)(5^5) - ....(2^13)(5^4)
There should be 4 zeroes ending up this quantity.
The answer is B.
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Re: The quantity will end in how many zeros [#permalink] New post 08 Feb 2018, 22:51
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realize that for a \(0\) to occur there has to be a multiplication of \(5\) and \(2\)

simplify the 1st term \(3^3 4^4 5^5 6^6\)= \(3^3* (2^2)^4*5^5* (2*3)^6\) = \(3^9* 2^1^4* 5^5\)
Similarly simplify the 2nd term that should come out to be \(3^9* 2^1^3* 5^4\)
Subtracting 2nd term from 1st term:take the common term which is whole of the 2nd term

\(3^9*2^1^3*5^4\)\((10-1)\)
now we have to find out the number of zeros in the common term because non common term is 9
2 and 5 multiply to 10. Here the limiting number is 5 which is equal to 4 hence 4 zeros
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Re: The quantity will end in how many zeros [#permalink] New post 23 Feb 2018, 05:21
Hi, I don't quite understand your last sentence. What do you mean by the limiting number?

amorphous wrote:
realize that for a \(0\) to occur there has to be a multiplication of \(5\) and \(2\)

simplify the 1st term \(3^3 4^4 5^5 6^6\)= \(3^3* (2^2)^4*5^5* (2*3)^6\) = \(3^9* 2^1^4* 5^5\)
Similarly simplify the 2nd term that should come out to be \(3^9* 2^1^3* 5^4\)
Subtracting 2nd term from 1st term:take the common term which is whole of the 2nd term

\(3^9*2^1^3*5^4\)\((10-1)\)
now we have to find out the number of zeros in the common term because non common term is 9
2 and 5 multiply to 10. Here the limiting number is 5 which is equal to 4 hence 4 zeros
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Re: The quantity will end in how many zeros [#permalink] New post 26 Feb 2018, 09:34
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gremather wrote:
Hi, I don't quite understand your last sentence. What do you mean by the limiting number?

amorphous wrote:
realize that for a \(0\) to occur there has to be a multiplication of \(5\) and \(2\)

simplify the 1st term \(3^3 4^4 5^5 6^6\)= \(3^3* (2^2)^4*5^5* (2*3)^6\) = \(3^9* 2^1^4* 5^5\)
Similarly simplify the 2nd term that should come out to be \(3^9* 2^1^3* 5^4\)
Subtracting 2nd term from 1st term:take the common term which is whole of the 2nd term

\(3^9*2^1^3*5^4\)\((10-1)\)
now we have to find out the number of zeros in the common term because non common term is 9
2 and 5 multiply to 10. Here the limiting number is 5 which is equal to 4 hence 4 zeros


since for a '\(0\)' to occur \(5\) has to be multiplied by \(2\). The number of zeros will depend on the minimum power raised of either of the two numbers

For eg.
\(100 = 5^2 * 2^2 = 2\) zeros at the end (because both the terms have power raised to 2)
\(125 = 5^3 * 2^0 = 0\) zeros at the end (because 2 is raised to a power of 0 hence 2 becomes the limiting number)
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Re: The quantity will end in how many zeros   [#permalink] 26 Feb 2018, 09:34
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