# Remainder=?

## Mar 7, 2018

This can be calculated in a number of ways. One way using brute force is

${27}^{1} / 7$ has a remainder $= 6$ .....(1)
${27}^{2} / 7 = \frac{729}{7}$ has a remainder $= 1$ .....(2)
${27}^{3} / 7 = \frac{19683}{7}$ has a remainder $= 6$ …….. (3)
${27}^{4} / 7 = \frac{531441}{7}$ has a remainder $= 1$ ….. (4)
${27}^{5} / 7 = \frac{14348907}{7}$ has a remainder $= 6$ …..(5)
${27}^{6} / 7 = \frac{387420489}{7}$ has remainder $= 1$ …. (6)

As as per emerging pattern we observe that the remainder is $= 6$ for an odd exponent and the remainder is $= 1$ for an even exponent.

Given exponent is $999 \to$ odd number. Hence, remainder $= 6.$

Mar 7, 2018

Alternate solution

#### Explanation:

Given number needs to be divided by $7$. Hence it can be written as

${\left(27\right)}^{999}$
$\implies {\left(28 - 1\right)}^{999}$

In the expansion of this series, all terms which have various powers of $28$ as multiplicants will be divisible by $7$. Only one term which is $= {\left(- 1\right)}^{999}$ now needs to be tested.

We see that this term ${\left(- 1\right)}^{999} = - 1$ is not divisible by $7$ and therefore, we are left with remainder $= - 1.$
Since remainder can not be $= - 1$, we will have to stop division process for remaining terms of expansion when the last $7$ remains.

This will leave remainder as $7 + \left(- 1\right) = 6$