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I am really confused about this question.

To know if this number is divisible by $1100$ we should check if it's divisible by $5$, $2$, and $11$ since $1100=5^{2} \times 2^{2} \times 11$.

It is easy to show this for both $2$ and $5$, but I got confused when I begin in the case of $11$, since we know that $11^{10}$ is surely divisble by $11$, so how is $11^{10} - 1$ also divisible by $11$? I am sure because I saw the final answer and it's (yes it's divisible).

Maybe the answer is clear but I am not able to catch it. Any hints are welcome !

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  • $\begingroup$ Note that it's not enough to show that it's divisible by $2$ and $5$. You need to show that it's divisible by $4$ and $25$. $\endgroup$ – Arthur Dec 2 '16 at 9:41
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    $\begingroup$ The answer is wrong. For no integers $a>1$ and $b>0$ you can have $a^b-1$ is divisible by $a$. $\endgroup$ – egreg Dec 2 '16 at 10:21
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    $\begingroup$ The question $$\text{Determine whether $11^{10} - 1$ is divisible by $100$.}$$ would make sense (and its answer is positive). The question $$\text{Determine whether $11^{10} - 1$ is divisible by $1000$.}$$ would make sense (and its answer is negative). $\endgroup$ – Did Dec 2 '16 at 11:11
  • $\begingroup$ Where does this problem come from? As pointed by others, the prolem might be incorrectly copied. $\endgroup$ – Martin Sleziak Dec 2 '16 at 15:49
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For no integers $a>1$ and $b>0$ you can have $$ a\mid a^b-1 $$ because $a\mid a^b$, so you'd get $a\mid (a^b-(a^b-1))$, so $a\mid 1$.

Then you need no computation to show $11^{10}-1$ is not divisible by $1100$, because it is not divisible by $11$.

The answer you were given is wrong.

I tried some variations, in order to identify a possible typo, but $11^{10}-11$ is not divisible by $1100$, nor is $10^{11}-1$.

If the numbers are in binary notation, it is wrong as well, because $$ \mathtt{11}_2^{\mathtt{10}_2}-\mathtt{1}_2=8,\qquad \mathtt{1100}_2=12 $$ Interpreting only the base as binary and the exponent as decimal, $$ {\mathtt{11}_2}^{10}-\mathtt{1}_2=59048 $$ which is not divisible by $12$.

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  • $\begingroup$ "I tried some variations, in order to identify a possible typo" See comment on main for another direction than the ones you searched in. $\endgroup$ – Did Dec 2 '16 at 11:23
  • $\begingroup$ @Did Possibly 100 was the number, yes. $\endgroup$ – egreg Dec 2 '16 at 11:29
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The final answer is wrong, $11^{10} - 1$ is not divisible by $11$ because the remainder when dividing it by $11$ is $10$.

Therefore, the number is also not divisible by any multiple of $11$, and in particular, by $1100$.

You can verify it yourself using any calculator with enough digits that $11^{10}-1 = 25937424600$ and that this number is not divisible by $11$.

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$11^{10}-1$ is clearly not divisible by $11$, in fact it leaves a remainder of $10$, with quotient $11^9 -1$.

To see that $11^{10}-1$ is divisible by $100$, we can do the following: To check divisbility by four, see that $11 \equiv -1 \mod 4$ (read up congruence notation and rules from Wikipedia, if you are unaware),hence $11^{10} \equiv (-1)^{10} \equiv 1 \mod 4$, so that $4$ divides $11^{10}-1$.

See that $11^2 =121 \equiv -4 \mod 25$, so that $11^{10} \equiv (-4)^5 \equiv -1024 \equiv 1 \mod 25$. Hence,$11^{10}-1$ is a multiple of $25$, and hence of $25 \cdot 4=100$.

In fact, $11^{10}-1 = 25937424600 = 2^3 \cdot 3 \cdot 5^2 \cdot 3221 \cdot 13421$, as a confirmation.

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