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

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