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by rakeshd347 » Wed Oct 02, 2013 7:21 pm
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

OA is C
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by theCodeToGMAT » Wed Oct 02, 2013 8:30 pm
10^10 = 10
10^20 = 20
10^30 = 30
10^40 = 40
10^50 = 50
2^100 & 5^100 = 100 --> taken from calculating power of two in 20^20 & power of four in 40^40

So, 100 + 150 = 250

Answer [spoiler]{C}[/spoiler]
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by ganeshrkamath » Wed Oct 02, 2013 11:37 pm
rakeshd347 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

OA is C
(1^1)*(5^5)*(10^10)*(15^15) *(20^20)*(25^25).......... *(50^50) = (5^5)*(10^10)*(15^15) *(20^20)*(25^25).......... *(50^50)
All the terms are multiples of 5.
To calculate the number of zeros, we need to find the number of 10's.
10 = 2*5

Power of 2 = 2^10 * 4^20 * 2^30 * 8^40 * 2^50
= 2^10 * 2^40 * 2^30 * 2^120 * 2^50
= 2^(10+40+30+120+50)
= 2^250

Power of 5 = 5^5 * 5^10 * 5^15 * 5^20 * 25^25 * 5^30 * 5^35 * 5^40 * 5^45 * 25^50
= 5^(5+10+15+20+50+30+35+40+45+100)
= 5^(350)

Power of 10 = 10^250

So there are 250 trailing zeros.

Choose C

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