mouda wrote:
Carcass wrote:
What is the remainder when (7²)(8²) is divided by 6?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
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.
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