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# With how many zeros does 11^100-1 end? Explain with Binomial theorem.

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With how many zeros does 11^100-1 end? Explain with Binomial theorem.

Jan 7, 2015

#1
+17747
+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.

Jan 7, 2015

#1
+17747
+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