If $(27)^{999}$ is divided by $7$, then the remainder is :

Question

If for $ 3 \leq r \leq 30 $, $ ^{30}C_{30-r} + 3 \left( ^{30}C_{31-r} \right) + 3 \left( ^{30}C_{32-r} \right) + ^{30}C_{33-r} = ^m C_r $, then $ m $ equals to_________

Answer 1

To find the remainder when $(27)^{999}$ is divided by 7, we can use Fermat's Little Theorem, which states that if $p$ is a prime number and $a$ is an integer not divisible by $p$, then $a^{p-1} \equiv 1 \pmod{p}$.

Since 7 is a prime number, by Fermat's Little Theorem, we have:

$(27)^{6} \equiv 1 \pmod{7}$

First, we calculate $27 \mod 7$:

$27 \div 7 = 3$ remainder $6$, so $27 \equiv 6 \pmod{7}$.

Now, using the result from Fermat's Little Theorem, we know:

$(6)^{6} \equiv 1 \pmod{7}$

To find $(6)^{999} \mod 7$, we write $999$ as a multiple of $6$ plus a remainder:

$999 = 6 \times 166 + 3$, hence $(6)^{999} = ((6)^{6})^{166} \times (6)^{3}$.

From Fermat's theorem, $(6)^{6} \equiv 1 \pmod{7}$, thus:

$((6)^{6})^{166} \equiv 1^{166} \equiv 1 \pmod{7}$.

We still need to calculate $(6)^{3} \mod 7$:

$(6)^{3} = 6 \times 6 \times 6 = 216$.

Now calculate $216 \mod 7$:

$216 \div 7 = 30$ remainder $6$, so $216 \equiv 6 \pmod{7}$.

Therefore, $(6)^{999} \equiv 1 \times 6 \equiv 6 \pmod{7}$.

Thus, the remainder when $(27)^{999}$ is divided by 7 is 6.

Answer 2