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Founder  Joined: 18 Apr 2015
Posts: 8139
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Kudos [?]: 1710 , given: 7485

The quantity will end in how many zeros [#permalink]
Expert's post 00:00

Question Stats: 67% (02:04) correct 32% (03:02) wrong based on 52 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

_________________ Intern Joined: 08 Dec 2017
Posts: 40
Followers: 1

Kudos [?]: 44  , given: 70

Re: The quantity will end in how many zeros [#permalink]
1
KUDOS
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. Director  Joined: 07 Jan 2018
Posts: 684
Followers: 10

Kudos [?]: 657  , given: 88

Re: The quantity will end in how many zeros [#permalink]
3
KUDOS
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
_________________

This is my response to the question and may be incorrect. Feel free to rectify any mistakes

Manager Joined: 15 Feb 2018
Posts: 53
Followers: 1

Kudos [?]: 18 , given: 33

Re: The quantity will end in how many zeros [#permalink]
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 Director  Joined: 07 Jan 2018
Posts: 684
Followers: 10

Kudos [?]: 657  , given: 88

Re: The quantity will end in how many zeros [#permalink]
2
KUDOS
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)
_________________

This is my response to the question and may be incorrect. Feel free to rectify any mistakes

Manager Joined: 18 Jun 2019
Posts: 86
Followers: 0

Kudos [?]: 6 , given: 41

Re: The quantity will end in how many zeros [#permalink]
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. Founder  Joined: 18 Apr 2015
Posts: 8139
Followers: 157

Kudos [?]: 1710  , given: 7485

Re: The quantity will end in how many zeros [#permalink]
1
KUDOS
Expert's post
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
_________________
Manager Joined: 18 Jun 2019
Posts: 86
Followers: 0

Kudos [?]: 6 , given: 41

Re: The quantity will end in how many zeros [#permalink]
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! Re: The quantity will end in how many zeros   [#permalink] 06 Sep 2019, 08:07
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