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# 1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 *

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Intern
Joined: 16 Aug 2006
Posts: 3
1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 * [#permalink]

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26 Sep 2006, 00:20
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1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 *â€¦â€¦â€¦100^100
1) 1200
2) 1300
3) 1050
4) None of these

2) The power of 990 that will divide 1090! Is
1)101
2)100
3)108
4)109
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7440
Location: Pune, India
Re: 1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 * [#permalink]

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22 Jan 2013, 20:46
2
KUDOS
Expert's post
2) The power of 990 that will divide 1090! Is
1)101
2)100
3)108
4)109

The theory for this question has been discussed in detail in this post:
http://www.veritasprep.com/blog/2011/06 ... actorials/
(the same link is given for the 1st question as well)

I would suggest you to check out the post before looking at the solution.

$$990 = 2*3^2*5*11$$

We need one 2, two 3s, one 5 and one 11 to make 990. The number of 11s will certainly be less than the number of 2s and number of 5s. Number of 9s you can make will also be more than the number of 11s you will get. So all we have to worry about is the number of 11s in 1090!

1090/11 = 99
99/11 = 9

Total number of 11s = 99 + 9 = 108

So the power of 990 that will divide 1090! is 108 .
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Karishma
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Senior Manager Joined: 19 Jul 2006 Posts: 358 ### Show Tags 26 Sep 2006, 01:23 1) ans: 1050 number of zeros introduced by ( 10^10 * 20^20 * 30^30 * ...... * 100^100) = 10+20+30+40+50+60+70+80+90+100 = 550 number of zeros introduced due to multiplication of numbers having 5 at unit digit and numbers having 2 at unit digit ( it can be determined by the numbers of 5s as 2 is surplus ( 5^5 * 15^ 15 * ........... * 95 ^ 95) = 5+15+25+35+45+55+65+75+85+95 = 500 total no. of zeros = 550 + 500 = 1050 Intern Joined: 16 Aug 2006 Posts: 3 ### Show Tags 26 Sep 2006, 02:16 But Dear answer for this question is 1300 and for the second one it is 108... Manager Joined: 28 Aug 2006 Posts: 243 Location: Albuquerque, NM ### Show Tags 26 Sep 2006, 02:45 1 This post was BOOKMARKED number of zeros introduced by ( 10^10 * 20^20 * 30^30 * ...... * 100^100) = 10+20+30+40+50+60+70+80+90+200 = 650 (100^100 introduces 200 zeros) number of zeros introduced due to multiplication of numbers having 5 at unit digit and numbers having 2 at unit digit ( it can be determined by the numbers of 5s as 2 is surplus ( 5^5 * 15^ 15 * ........... * 95 ^ 95) = 5+15+25+35+45+55+65+75+85+95 = 500 when 25 gets multiplied by 4 one gets 2 zeros so instead of multiplying by 2, we rather multiply by 4 and count 25 more zeros BAsically you have 25 extra 5's from 25, 50 from 50 and 75 from 75 so add 150 to 650 and 500, answer is 1300 Manager Joined: 28 Aug 2006 Posts: 243 Location: Albuquerque, NM ### Show Tags 26 Sep 2006, 02:50 second question: power of 990 that will divide 1090! 990 = 2 X 5 X 9 X 11 Thus the number of 11s detemine the power Multiples of 11 from 1 to 1090 are 99 (1089 is last), then you have 121 and multiples of that give you extra 11s they are 9 (since 121*9 = 1089) answer 99 + 9 = 108 Intern Joined: 16 Aug 2006 Posts: 3 ### Show Tags 27 Sep 2006, 00:07 thanks jainen24 and AK for your answers Intern Joined: 16 Dec 2012 Posts: 4 Re: [#permalink] ### Show Tags 22 Jan 2013, 11:00 can anybody please give a more detailed explanation for this question. thanks Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7440 Location: Pune, India Re: 1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 * [#permalink] ### Show Tags 22 Jan 2013, 20:34 bhupesh_dadhwal wrote: 1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 *â€¦â€¦â€¦100^100 1) 1200 2) 1300 3) 1050 4) None of these To get a 0 at the end of the product i.e. to have 10 as a factor, you need a 2 and a 5. In the first 100 natural numbers, there will be many more 2s than 5s because every alternate number has a 2. So number of 2s does not pose any restriction. We need to count the number of 5s. That will be the number of 10s we can make. We will obtain 5s from the multiples of 5. $$5^5*10^{10}*15^{15}*20^{20} ... 100^{100}$$ The first term will give five 5s. The second term will give ten 5s. The third term will give fifteen 5s and so on... $$5 + 10 + 15 + 20 + 25 +.. + 100 = 5(1 + 2+3 + ...20) = 5 *20*21/2 = 1050$$ But we missed out on some 5s: $$25^{25} = 5^{2*25}$$ gives another 25 5s, not just the 25 we have already counted above. $$50^{50} = (2*5^2)^{50}$$ gives another 50 5s, not just the 50 we have already counted above. Same for 75 and 100. Adding, we get 25+ 50 + 75 + 100 = 250 Total number of 5s = 1050 + 250 = 1300 Check this link for more discussion on this topic: http://www.veritasprep.com/blog/2011/06 ... actorials/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: 1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 * [#permalink]

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08 Sep 2016, 00:00
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Re: 1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 *   [#permalink] 08 Sep 2016, 00:00
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# 1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 *

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