Finding the last two digits of a number by binomial theorem

Question

What are the last two digits of $11^{25}$ to be solved by binomial theorem like $(1+10)^{25}$? If there is any other way to solve this it would help if that is shown too.

Well, what happens when you apply the binomial theorem to $(1+10)^{25}$? — Gerry Myerson, Sep 09 '13 at 09:33
Does, "compute the number" count as any other way? The number is 108347059433883722041830251. — Douglas S. Stones, Sep 09 '13 at 10:21

score 8 · Accepted Answer · answered Sep 09 '13 at 09:59

8

$$11^{25} = (1+10)^{25} = \dbinom{25}{0} + 10\cdot\dbinom{25}{1} + \sum_{k=2}^{25}\dbinom{25}{k}\cdot10^k =$$ $$ 1 + 5\cdot10 + 2\cdot 10^2 + \sum_{k=2}^{25}\dbinom{25}{k}\cdot10^k =$$ $$\underline{1} + \underline{5}\cdot10 + 10^2\cdot \left(2+\sum_{k=2}^{25}\dbinom{25}{k}\cdot10^{k-2}\right), $$ hence the last digits are 5 and 1:

answered Sep 09 '13 at 09:59

Mher

5,011

3

You have deprived the author of the joy of working it out. – Gerry Myerson Sep 09 '13 at 10:13
hahaha...i have not seen the process yet :) – Mustahid Sep 09 '13 at 22:40

score 3 · Answer 2 · answered Sep 09 '13 at 11:06

Here's another way. $11\times11=121$ ends in 21. $11\times21=231$ ends in 31. $11\times31=341$ ends in 41. See the pattern for the last two digits of powers of 11? Now prove that the pattern continues to hold, and then see what that tells you about $11^{25}$.

score 2 · Answer 3 · edited Jun 12 '20 at 10:38

2

A short cut for finding last two digits of any power of a number that ends in 1 .

Last two digits of

$$(\ldots a1)^{\displaystyle\ldots x}$$

TENS DIGIT: unit's digit of ($x$ $\times$ a)

UNITS DIGIT : $1$

so here $$11^{25}$$ last two digits are units digit of $(1*5)$ and $1$ so $$51$$

edited Jun 12 '20 at 10:38

Community

1

answered Sep 09 '13 at 11:45

Harish Kayarohanam

1,980
2
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Finding the last two digits of a number by binomial theorem

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