#1**+5 **

I believe that it will end in three zeros for this reason:

11 = 10 + 1

So: 11^100 = (10 + 1)^100

Using the Binomial Theorem:

The last term will be: 100C0·(10^0)·(1^100) = 1

This term will be coancell by the -1 term. So it will end with a zero.

The second last term will be: 100C1·(10^1)·(1^99) = (100)(10)(1) = 1000 (three zeroes)

The third last term will be: 100C2·(10^2)·(1^98) = (4950)(100)(1) = 495.000 (three zeroes)

The fourth last term will be: 100C3·(10^3)·(1^97) = (161700)(1000)(1) = 161700000 (won't change the three zeroes at the end)

The fifth last term will be: 100C4·(10^4)·(1^96) = (3921225)(10000)(1) = whatever, but won't change the three zeroes at the end.

Similarly for all the other terms; the power of ten won't change the three zeroes at the end.

geno3141
Jan 7, 2015

#1**+5 **

Best Answer

I believe that it will end in three zeros for this reason:

11 = 10 + 1

So: 11^100 = (10 + 1)^100

Using the Binomial Theorem:

The last term will be: 100C0·(10^0)·(1^100) = 1

This term will be coancell by the -1 term. So it will end with a zero.

The second last term will be: 100C1·(10^1)·(1^99) = (100)(10)(1) = 1000 (three zeroes)

The third last term will be: 100C2·(10^2)·(1^98) = (4950)(100)(1) = 495.000 (three zeroes)

The fourth last term will be: 100C3·(10^3)·(1^97) = (161700)(1000)(1) = 161700000 (won't change the three zeroes at the end)

The fifth last term will be: 100C4·(10^4)·(1^96) = (3921225)(10000)(1) = whatever, but won't change the three zeroes at the end.

Similarly for all the other terms; the power of ten won't change the three zeroes at the end.

geno3141
Jan 7, 2015