Question

What is the remainder left out when \[{8^{2n}} - {(62)^{2n + 1}}\] is divided by 9?

Answer 1

Hint: Try to express 8 and 62 in terms of 9 and then evaluate the expression \[{8^{2n}} - {(62)^{2n + 1}}\] to find the terms that does not contain a multiple of 9. Add them to get the remainder.

Complete step by step answer:
We need to find out the remainder when \[{8^{2n}} - {(62)^{2n + 1}}\] is divided by 9. So, we can express \[{8^{2n}} - {(62)^{2n + 1}}\] in multiples of 9 plus some number which is the remainder.

We can express 8 as 9 – 1 and 62 as 63 – 1, which is \[9 \times 7 - 1\].

Hence, we have the following:

\[{8^{2n}} - {(62)^{2n + 1}} = {(9 - 1)^{2n}} - {(9 \times 7 - 1)^{2n + 1}}..............(1)\]

We know the binomial expansion formula as given below:

\[{(a + b)^n} = {}^n{C_0}{a^n} + {}^n{C_1}{a^{n - 1}}{b^1} + ...... + {}^n{C_{n - 1}}{a^1}{b^{n - 1}} + {}^n{C_n}{b^n}...........(2)\]

Using formula (2) for the first term is equation (1), that is \[{(9 - 1)^{2n}}\], we get:

\[{(9 - 1)^{2n}} = {}^{2n}{C_0}{9^{2n}} + {}^{2n}{C_1}{9^{2n - 1}}{( - 1)^1} + ...... + {}^{2n}{C_{2n - 1}}{9^1}{( - 1)^{2n - 1}} + {}^{2n}{C_{2n}}{( - 1)^n}\]

Simplifying, we get:

\[{(9 - 1)^{2n}} = {}^{2n}{C_0}{9^{2n}} - {}^{2n}{C_1}{9^{2n - 1}} + ...... - {}^{2n}{C_{2n - 1}}9 + {}^{2n}{C_{2n}}\]

The term which is not a multiple of 9 is \[{}^{2n}{C_{2n}}\], that is, 1.

Now, using formula (2) for the second term in equation (1), that is \[{(9 \times 7 - 1)^{2n + 1}}\], we get:

\[{((9 \times 7) - 1)^{2n + 1}} = {}^{2n + 1}{C_0}{(9 \times 7)^{2n + 1}} + {}^{2n + 1}{C_1}{(9 \times 7)^{2n}}{( - 1)^1} + ...... + {}^{2n + 1}{C_{2n}}{(9 \times 7)^1}{( - 1)^{2n}} + {}^{2n + 1}{C_{2n + 1}}{( - 1)^{2n + 1}}\]

Simplifying, we get:

\[{((9 \times 7) - 1)^{2n + 1}} = {}^{2n + 1}{C_0}{(9 \times 7)^{2n + 1}} - {}^{2n + 1}{C_1}{(9 \times 7)^{2n}} + ...... + {}^{2n + 1}{C_{2n}}(9 \times 7) - {}^{2n + 1}{C_{2n + 1}}\]

The term which is not a multiple of 9 is \[ - {}^{2n + 1}{C_{2n + 1}}\], that is, –1.
Hence, equation (1) can be re-written as follows:

\[{8^{2n}} - {(62)^{2n + 1}} = 9m + ({}^{2n}{C_{2n}} - ( - {}^{2n + 1}{C_{2n + 1}}))\]

Simplifying it, we get:

\[{8^{2n}} - {(62)^{2n + 1}} = 9m + (1 - ( - 1))\]

\[{8^{2n}} - {(62)^{2n + 1}} = 9m + (1 + 1)\]

\[{8^{2n}} - {(62)^{2n + 1}} = 9m + 2\]

Where m is some integer.

Any number can be expressed in the form \[p = dq + r\], where d is the divisor, p is the dividend, q is the quotient and r is the remainder.

Hence, the remainder when \[{8^{2n}} - {(62)^{2n + 1}}\] is divided by 9 is 2.

Note: Odd powers of –1 is –1 itself and even powers of –1 is 1. You can also write \[{8^{2n}}\] as \[{64^n}\] which can be written as \[{(9 \times 7 + 1)^n}\] and can expand further.