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# How many trailing zeros will 11^50 - 1 will have ?

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How many trailing zeros will 11^50 - 1 will have ?
A) 3

B) 4
C) 5
D) 6
E) 7

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Best answer
(10 + 1)^50 - 1
=>...... 50C2*(100) + 50C1*(10) + 50C0 - 1
=>  25*49*100 + 500 = 3 zeroes
answered Mar 29, 2015 by Senior Associate (452 points)
selected Mar 30, 2015 by
2Comments
Thanks Meenakshi, what does C2, C1 and C0 mean? I am unable to understand the formula - can you please clarify?
commented Apr 2, 2015 by Beginner (1 point)
it is part of the bionomial expansion. It has read as 50C2.
nCr  = n!/ ( r! * ( n - r)!)
it is count of occurence of such terms.
So for example, we want to expand
(x+y)^2 = 2C0 * x^2 * y^0 + 2C1 * x ^1 * y^1 + 2C2 * x^0  * y^2
i.e. = x^2 + 2xy + y ^2

More reading http://www.tutors4you.com/binomialtheoremtutorial.htm

Since we have the power 50 so the expansion, goes like this

(x + y) ^ 50 = 50C0* x^50 * y^0 + 50C1 * x^49 * y ^ 1 .... 50C50 * x^0 * y^50
commented Apr 2, 2015 by Guru (5,808 points)
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