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1. Find the remainder when 67⁹⁹ is divided by 7.

A. 4

B. 6

C. 1

D. 2

Solution:

\begin{aligned} Remainder of \frac{67^{99}}{7} \\ or, R = \frac{{(63 + 4)}^{99}}{7} \end{aligned}

63 is divisible by 7 for any power, so required remainder will depend on the power of 4
Required remainder

\begin{aligned} \frac{4^{99}}{7} == R == \frac{4^{(96 + 3)}}{7} \\ \frac{4^{3}}{7} \Rightarrow \frac{64}{7} \Rightarrow \frac{(63 + 1)}{7} == R \Rightarrow 1 \\ Note : \\ \frac{4}{7} remainder = 4 \\ \frac{(4 \times 4)}{7} = \frac{16}{7} remainder = 2 \\ \frac{(4 \times 4 \times 4)}{7} = \frac{64}{7} = 1 \\ \frac{(4 \times 4 \times 4 \times 4)}{7} = \frac{256}{7} remainder = 4 \\ \frac{(4 \times 4 \times 4 \times 4 \times 4)}{7} = 2 \end{aligned}

If we check for more power we will find that the remainder start repeating themselves as 4, 2, 1, 4, 2, 1 and so on. So when we get A number having greater power and to be divided by the other number B, we will break power in (4n + x) and the final remainder will depend on x i.e.

\frac{A^{x}}{B}

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