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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.

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    $\begingroup$ Well, what happens when you apply the binomial theorem to $(1+10)^{25}$? $\endgroup$ – Gerry Myerson Sep 9 '13 at 9:33
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    $\begingroup$ Does, "compute the number" count as any other way? The number is 108347059433883722041830251. $\endgroup$ – Douglas S. Stones Sep 9 '13 at 10:21
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$$11^{25} = (1+10)^{25} = \dbinom{25}{0} + 10\cdot\dbinom{25}{1} + \sum_{k=2}^{25}\dbinom{25}{k}\cdot10^k =$$ $$ 1 + 5\cdot10 + 2\cdot 10^2 + \sum_{k=2}^{25}\dbinom{25}{k}\cdot10^k =$$ $$\underline{1} + \underline{5}\cdot10 + 10^2\cdot \left(2+\sum_{k=2}^{25}\dbinom{25}{k}\cdot10^{k-2}\right), $$ hence the last digits are 5 and 1:

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    $\begingroup$ You have deprived the author of the joy of working it out. $\endgroup$ – Gerry Myerson Sep 9 '13 at 10:13
  • $\begingroup$ hahaha...i have not seen the process yet :) $\endgroup$ – Mustahid Sep 9 '13 at 22:40
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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 $11^{25}$.

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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 $$51$$

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