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The quantity will end in how many zeros

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Carcass
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The quantity will end in how many zeros [#permalink] New post   09 Aug 2017, 15:41
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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
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B
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amorphous
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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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YMAkib
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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   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] New post   06 Sep 2019, 07:34
Hi! looking for another solution to this question.

I've simplified upto 3^9*2^13*5^4(9) and don't know how to proceed to fnd the number of zero's.

@greenlighttestprep, @Carcass - please help!
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Re: The quantity will end in how many zeros [#permalink] New post   06 Sep 2019, 08:04
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The trick is how many "five you do have in the quantity ??

we do have nine five numbers.

Now, you will have a zero whenever you do have \(2 \times 5 = 10\)

In our quantity we do have 4 couples of 5 plus one 5 disparaged

5*5
5*5
5*5
5*5
5

That means two couples of five are 4 zeros.

Hope this helps
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Re: The quantity will end in how many zeros [#permalink] New post   06 Sep 2019, 08:07
Carcass wrote:
The trick is how many "five you do have in the quantity ??

we do have nine five numbers.

Now, you will have a zero whenever you do have 2 \times 5 = 10

In our quantity we do have 4 couples of 5 plus one 5 disparaged

5*5
5*5
5*5
5*5
5

That means two couples of five are 4 zeros.

Hope this helps



Much clearer!! Thank you!
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Re: The quantity will end in how many zeros [#permalink] New post   09 Dec 2019, 02:56
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Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



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


Still if someone needs to be more clear visit here.
https://math.stackexchange.com/questions/2663286/this-quantity-see-question-will-end-with-how-many-zeros
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Re: The quantity will end in how many zeros [#permalink] New post   26 Nov 2020, 12:29
3

More clearly explaining "limiting number"



Let's say you ended up with \(3^{19}2^{13}5^{4}\) and you want to know how many 0's it ends in.

Solution: Just factor out 10s



\(3^{19}2^{13}5^{4} = 3^{19}2^{9}(2*5)^{4} = 3^{19}2^{9}(10)^{4}.\)

\(10^{4}\) tells you there's 4 zeros.
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Re: The quantity will end in how many zeros [#permalink] New post   03 Dec 2020, 23:22
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I do not at all understand how the concept of limiting numbers applies in this case. Yes, the first term has 5 zeroes; the second one has 4. I don't see how subtraction of one from the other then results in 4, and being told "well it's the limiting number" doesn't really illuminate that much.

I can see how it works mathematically, in, say, the case of 100,000-10,000 but I am hesitant to apply this to other situations more broadly with confidence.
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Re: The quantity will end in how many zeros [#permalink] New post   04 Dec 2020, 04:09
I think the right variant is 'C', or am I mistaken?
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Re: The quantity will end in how many zeros [#permalink] New post   04 Dec 2020, 05:25
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DylanGillo wrote:
I think the right variant is 'C', or am I mistaken?


what do you mean for the exact "variant" Sir '?
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Re: The quantity will end in how many zeros [#permalink] New post   14 Sep 2023, 23:00
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