It is currently 19 Sep 2019, 22:12

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# The quantity will end in how many zeros

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Founder
Joined: 18 Apr 2015
Posts: 8139
Followers: 157

Kudos [?]: 1710 [0], given: 7485

The quantity will end in how many zeros [#permalink]  09 Aug 2017, 15:41
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 [1] , given: 70

Re: The quantity will end in how many zeros [#permalink]  07 Feb 2018, 01:18
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 [3] , given: 88

Re: The quantity will end in how many zeros [#permalink]  08 Feb 2018, 22:51
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 [0], given: 33

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

Kudos [?]: 657 [2] , given: 88

Re: The quantity will end in how many zeros [#permalink]  26 Feb 2018, 09:34
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 [0], given: 41

Re: The quantity will end in how many zeros [#permalink]  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.

Founder
Joined: 18 Apr 2015
Posts: 8139
Followers: 157

Kudos [?]: 1710 [1] , given: 7485

Re: The quantity will end in how many zeros [#permalink]  06 Sep 2019, 08:04
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 [0], given: 41

Re: The quantity will end in how many zeros [#permalink]  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!
Re: The quantity will end in how many zeros   [#permalink] 06 Sep 2019, 08:07
Display posts from previous: Sort by

# The quantity will end in how many zeros

 Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.