GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 07 Jul 2020, 21:48 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10

Author Message
TAGS:

### Hide Tags

Manager  Joined: 08 Oct 2010
Posts: 174
Location: Uzbekistan
Schools: Johnson, Fuqua, Simon, Mendoza
WE 3: 10
Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10  [#permalink]

### Show Tags

22
1
165 00:00

Difficulty:   95% (hard)

Question Stats: 50% (02:43) correct 50% (02:24) wrong based on 1496 sessions

### HideShow timer Statistics

Find the number of trailing zeros in the product of $$(1^1)*(5^5)*(10^{10})*(15^{15}) *(20^{20})*(25^{25})*...*(50^{50})$$.

A. 150
B. 200
C. 250
D. 245
E. 225

Originally posted by feruz77 on 25 Jan 2011, 05:42.
Last edited by Bunuel on 07 Oct 2017, 07:24, edited 1 time in total.
Edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 65062
Re: Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10  [#permalink]

### Show Tags

40
82
feruz77 wrote:
Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)………. *(50^50).
A) 10^150
B) 10^200
C) 10^250
D) 10^245
E) 10^225

Can someone help me how to solve this question? I think there must be more than one solution method.

We have a trailing zero when we multiplying 2 by 5. Now, in the product: (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)*...*(50^50) there will be obviously more 5-s than 2-s so the # of 2-s will be limiting factor for the # of trailing zeros.

So we should count # of 2-s for even bases, basically we should factor out 2-s: $$10^{10}*20^{20}*30^{30}*40^{40}*50^{50}=$$
$$=2^{10}*(2^2)^{20}*(2)^{30}*(2^3)^{40}*2^{50}*(something)=2^{10+40+30+120+50}*(something)=2^{250}*(something)$$.

So there will be 250 trailing zeros in the above product as there are at least as many 5-s as 2-s.

_________________
Senior Manager  Joined: 13 Aug 2012
Posts: 386
Concentration: Marketing, Finance
GPA: 3.23
Re: Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10  [#permalink]

### Show Tags

9
3
Looking at the numbers it looks like

(1x5)^5
(2x5)^10
...
(10x5)^50

1. Determine the limiting factor. Is it 2 or is it 5? We know that all the numbers are multiple of 5 but not of 2. Thus, the limiting factor in this case is 2. Let's drop all the 5. Then, we count factors of 2 of even multiples.

2^10 = 10
4^20 = 20 + 20
6^30 = 30
8^40 = 40 + 40 + 40
10^50 = 50

250

More examples of Trailing Zeroes: Arithmetic: Trailing Zeroes
##### General Discussion
Manager  Joined: 08 Oct 2010
Posts: 174
Location: Uzbekistan
Schools: Johnson, Fuqua, Simon, Mendoza
WE 3: 10
Re: Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10  [#permalink]

### Show Tags

1
Thanks Bunuel.

My remark:
Because of conditions of the stem, I think, this question is an exclusion from the general approach where one must count 5s a number of which in factorials are usually less or equal to a number of 2s.
Manager  Joined: 28 Dec 2013
Posts: 65
Re: Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10  [#permalink]

### Show Tags

How come (2)^30 had no power inside with the 2?
Math Expert V
Joined: 02 Sep 2009
Posts: 65062
Re: Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10  [#permalink]

### Show Tags

sagnik242 wrote:
How come (2)^30 had no power inside with the 2?

$$30^{30}=(2*15)^{30}=2^{30}*15^{30}$$
_________________
Intern  Joined: 22 Feb 2014
Posts: 24
Re: Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10  [#permalink]

### Show Tags

Bunuel wrote:
feruz77 wrote:
Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)………. *(50^50).
A) 10^150
B) 10^200
C) 10^250
D) 10^245
E) 10^225

Can someone help me how to solve this question? I think there must be more than one solution method.

We have a trailing zero when we multiplying 2 by 5. Now, in the product: (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)*...*(50^50) there will be obviously more 5-s than 2-s so the # of 2-s will be limiting factor for the # of trailing zeros.

So we should count # of 2-s for even bases, basically we should factor out 2-s: $$10^{10}*20^{20}*30^{30}*40^{40}*50^{50}=$$
$$=2^{10}*(2^2)^{20}*(2)^{30}*(2^3)^{40}*2^{50}*(something)=2^{10+40+30+120+50}*(something)=2^{250}*(something)$$.

So there will be 250 trailing zeros in the above product as there are at least as many 5-s as 2-s.

Hi

In this step:
So we should count # of 2-s for even bases, basically we should factor out 2-s: $$10^{10}*20^{20}*30^{30}*40^{40}*50^{50}=$$
$$=2^{10}*(2^2)^{20}*(2)^{30}*(2^3)^{40}*2^{50}*(something)=2^{10+40+30+120+50}*(something)=2^{250}*(something)$$.

Why we counted only 2? what does it mean by limiting factor and whats the importance of it??

Thanks a lot
Math Expert V
Joined: 02 Sep 2009
Posts: 65062
Re: Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10  [#permalink]

### Show Tags

GGMAT730 wrote:
Bunuel wrote:
feruz77 wrote:
Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)………. *(50^50).
A) 10^150
B) 10^200
C) 10^250
D) 10^245
E) 10^225

Can someone help me how to solve this question? I think there must be more than one solution method.

We have a trailing zero when we multiplying 2 by 5. Now, in the product: (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)*...*(50^50) there will be obviously more 5-s than 2-s so the # of 2-s will be limiting factor for the # of trailing zeros.

So we should count # of 2-s for even bases, basically we should factor out 2-s: $$10^{10}*20^{20}*30^{30}*40^{40}*50^{50}=$$
$$=2^{10}*(2^2)^{20}*(2)^{30}*(2^3)^{40}*2^{50}*(something)=2^{10+40+30+120+50}*(something)=2^{250}*(something)$$.

So there will be 250 trailing zeros in the above product as there are at least as many 5-s as 2-s.

Hi

In this step:
So we should count # of 2-s for even bases, basically we should factor out 2-s: $$10^{10}*20^{20}*30^{30}*40^{40}*50^{50}=$$
$$=2^{10}*(2^2)^{20}*(2)^{30}*(2^3)^{40}*2^{50}*(something)=2^{10+40+30+120+50}*(something)=2^{250}*(something)$$.

Why we counted only 2? what does it mean by limiting factor and whats the importance of it??

Thanks a lot

We have a trailing zero when we multiplying 2 by 5. So, each pair of 2 and 5 gives one more 0 at the end of the number. Our expression gives more 5's than 2's, so the number of 2 will determine the number of 0: for each 2 we have a 5, which when multiples will give 0.

Similar questions to practice:
if-n-is-the-greatest-positive-integer-for-which-2n-is-a-fact-144694.html
what-is-the-largest-power-of-3-contained-in-103525.html
if-n-is-the-product-of-all-positive-integers-less-than-103218.html
if-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html
if-n-is-the-product-of-all-multiples-of-3-between-1-and-101187.html
if-p-is-the-product-of-integers-from-1-to-30-inclusive-137721.html
what-is-the-greatest-value-of-m-such-that-4-m-is-a-factor-of-105746.html
if-6-y-is-a-factor-of-10-2-what-is-the-greatest-possible-129353.html
if-m-is-the-product-of-all-integers-from-1-to-40-inclusive-108971.html
if-p-is-a-natural-number-and-p-ends-with-y-trailing-zeros-108251.html
if-73-has-16-zeroes-at-the-end-how-many-zeroes-will-147353.html
find-the-number-of-trailing-zeros-in-the-expansion-of-108249.html
how-many-zeros-are-the-end-of-142479.html
how-many-zeros-does-100-end-with-100599.html
find-the-number-of-trailing-zeros-in-the-product-of-108248.html
if-60-is-written-out-as-an-integer-with-how-many-consecuti-97597.html
if-n-is-a-positive-integer-and-10-n-is-a-factor-of-m-what-153375.html
if-d-is-a-positive-integer-and-f-is-the-product-of-the-first-126692.html
what-is-the-largest-integer-k-such-that-10-is-divisible-by-172488.html

Hope it helps.
_________________
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 10648
Location: Pune, India
Re: Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10  [#permalink]

### Show Tags

8
5
feruz77 wrote:
Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25)………. *(50^50).

A. 150
B. 200
C. 250
D. 245
E. 225

Responding to a pm:

The method discussed in my post is useful while finding the maximum power of a number in a factorial. The given product is not in factorial form and hence the method needs to be suitably modified. That said, it should not be a big problem to modify the method if you understand the basics. Zeroes are produced by multiplying a 2 and a 5. So number of 0s in this product will depend on how many matching 2s and 5s we have here. In factorials, we have more 2s than 5s because we have consecutive numbers so we usually don't bother about finding the number of 2s. Here we have handpicked numbers so we need to ensure that we have both.

From where do we get 2s? From even multiples of 5.
10^10, 20^20, 30^30, 40^40, 50^50
$$(2*5)^{10}, (2^2*5)^{20}, (2*15)^{30}, (2^3*5)^{40}, (2*25)^{50}$$
So number of 2s is 10 + 40 + 30 + 120 + 50 = 250

Now, let's see the number of 5s. Each term of the product has a 5. So the number of 5s is at least 5 + 10 + 15 + 20 + 25 + ... + 50
Then we also need to account for terms that have multiple 5s such as 25 and 50 but let's get to that later.

5 + 10 + 15 + 20 + 25 + ... + 50 = 5(1 + 2 + 3 + ..10) = 5*10*11/2 = 275

Note that number of 5s will be even more than 275 while the number of 2s is only 250. So there will be 250 trailing zeroes.
_________________
Karishma
Veritas Prep GMAT Instructor

Manager  B
Joined: 03 May 2014
Posts: 145
Location: India
WE: Sales (Mutual Funds and Brokerage)
Re: Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10  [#permalink]

### Show Tags

Is the answer 250 or 10²⁵⁰ ?
Math Expert V
Joined: 02 Sep 2009
Posts: 65062
Re: Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10  [#permalink]

### Show Tags

gps5441 wrote:
Is the answer 250 or 10²⁵⁰ ?

___________________
_________________
Non-Human User Joined: 09 Sep 2013
Posts: 15383
Re: Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10   [#permalink] 22 Feb 2020, 21:38

# Find the number of trailing zeros in the product of (1^1)*(5^5)*(10^10  