Any number which is a multiple of 6 will give remainder 0.
Now 7^2 = 49 = 48 + 1
7^2 divided by 6 will give a remainder of 1
and 8^2 = 64 = 60 + 4
8^2 divided by 6 will give a remainder of 4

(7^2)(8^2) divided by 6 = 1*4 divided by 6

4 divided by 6 gives a remainder of 4.

OA - D

what if the question was to find the remainder when (7³)(8³)is divided by 6.
What is the rule of thumb? Is it the remainder of 56 divided by 6 raised to the power 3 (and then divide the answer by 6 to find the solution if remainder³ is greater than 6)?

Since, in this specific question, the dividend and divisor have a common factor 2. so, you can apply another method.
write 8^2 = 2^6.
Now, divide both numerator and denominator by 2.
so, it leaves us with (7^2)*(2^5) and this has to be divided by 3.
Now, 7 = 6+1 and 2 = 3-1
7^2 divided by 3 leaves a remainder of 1.
and 2 has an odd power so 2^5 divided by 3 leaves a remainder of -1. because remainders are always positive we need to add the divisor. so the remainder becomes: -1+3 = 2.

so (7^2)*(2^5) divided by 3 gives us a remainder of 2.
but, remember we earlier divided it by 2 so we need to multiply this remainder by 2.

So, the answer to the question (7^2)*(2^5) divided by 6 gives us a remainder of 4.

With this method 8^3 translates to 2^9. now when you divide by 2 you will have 2^8. and 2 leaves a remainder of -1 but it is raised to an even power so it becomes 1.

(7^3)*(2^8) divided by 3 will give a remainder of 1 and then you need to multiply it by 2 to give a remainder of 2 when divided by 6.

attempted this more simplistically -- 7 * 7 * 8*8/ 6 => 6 can divide 7 & 8 leaving => (1)(1)(2)(2) / 6 => 6 cannot divide 1 & 2 so the remainder is 4

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