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This question already has an answer here:

If $11^\text{10}-1$ is divided by $100$, then solve for '$x$' of the below term $$11^\text{10}-1 = x \pmod{100}$$

Whatever I tried:

$11^\text{2} \equiv 21 \pmod{100}$.....(1)

$(11^\text{2})^\text{2} \equiv (21)^\text{2} \pmod{100}$

$11^\text{4} \equiv 441 \pmod{100}$

$11^\text{4} \equiv 41 \pmod{100}$

$(11^\text{4})^\text{2} \equiv (41)^\text{2} \pmod{100}$

$11^\text{8} \equiv 1681 \pmod{100}$

$11^\text{8} \equiv 81 \pmod{100}$

$11^\text{8} × 11^\text{2} \equiv (81×21) \pmod{100}$ ......{from (1)}

$11^\text{10} \equiv 1701 \pmod{100} \implies 11^\text{10} \equiv 1 \pmod{100}$

Hence, $11^\text{10} -1 \equiv (1-1) \pmod{100} \implies 11^\text{10} - 1 \equiv 0 \pmod{100}$ and thus we get the value of $x$ and it is $x = 0$ and $11^\text{10}-1$ is divisible by $100$.

But this approach take a long time for any competitive exam or any math contest without using calculator. Any easier process on how to determine the remainder of the above problem quickly? That will be very much helpful for me. Thanks in advance.

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marked as duplicate by Bill Dubuque divisibility Feb 22 at 14:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I don't understand from which source I suggest the problem concerning number theory, all become duplicate. I think, Bangladesh Math Olympiad is a duplicate and fraud. Shame on it and I blame it. $\endgroup$ – Anirban Niloy Feb 22 at 14:36
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$$11^{10}=(10+1)^{10}=10^{10}+10×10^9+\frac {(10×9)}{2}×10^9+\cdots+(10×10)+1$$(using binomial expansion ). Now note that every term except last one is a multiple of $100$.

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    $\begingroup$ You beat me by less than a minute! :) $\endgroup$ – Deepak Feb 22 at 14:21
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A really quick way to "see" the answer is with binomial theorem.

$11^{10} = (10+1)^{10} = 10^{10} + k_1\cdot 10^9 + k_2 \cdot 10^8 + ... + 10\cdot 10^1 + 1$

where the $k$'s represent various combinatorial constants. The values are unimportant. What's important is that when we take the whole thing modulo $100$, the expression reduces to $1$. Subtracting one, we get the required result $11^{10} - 1 \equiv 0 \pmod{100}$.

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