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

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1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 * [#permalink] New post 25 Sep 2006, 23:20
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1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 *………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
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Re: 1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 * [#permalink] New post 22 Jan 2013, 19:46
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bhupesh_dadhwal wrote:
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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 [#permalink] New post 26 Sep 2006, 00:23
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
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 [#permalink] New post 26 Sep 2006, 01:16
But Dear answer for this question is 1300 :( and for the second one it is 108...
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 [#permalink] New post 26 Sep 2006, 01:45
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
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 [#permalink] New post 26 Sep 2006, 01:50
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
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 [#permalink] New post 26 Sep 2006, 23:07
thanks jainen24 and AK for your answers
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Re: [#permalink] New post 22 Jan 2013, 10:00
can anybody please give a more detailed explanation for this question.
thanks
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Re: 1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 * [#permalink] New post 22 Jan 2013, 19:34
Expert's post
bhupesh_dadhwal wrote:
1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 *………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/
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Re: 1) Find the number of zeros in the product: 1^1 * 2^2* 3^3 *   [#permalink] 22 Jan 2013, 19:34
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