# Finding out the remainder of $\frac{11^\text{10}-1}{100}$ using modulus [duplicate]

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.

• 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. – Anirban Niloy Feb 22 '19 at 14:36

$$11^{10}=(10+1)^{10}=10^{10}+10×10^9+\frac {(10×9)}{2}×10^9+\cdots+(10×10)+1$$(using binomial expansion ). Now note that every term except last one is a multiple of $$100$$.
$$11^{10} = (10+1)^{10} = 10^{10} + k_1\cdot 10^9 + k_2 \cdot 10^8 + ... + 10\cdot 10^1 + 1$$
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 $$11^{10} - 1 \equiv 0 \pmod{100}$$.