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Re: When a number a is divided by 6, the remainder is 3 and when another
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14 Jul 2023, 13:06
To find the remainder when a^2+b^2 is divided by 12, we need to consider the remainders when a and b are divided by 6 and 12, respectively.
Given that the remainder when a is divided by 6 is 3, we can express a as: a=6k+3, where k is an integer.
Similarly, the remainder when b is divided by 12 is 9, so we can express b as: b=12m+9, where m is an integer.
Now, let's substitute these expressions into a^2+b^2 and simplify:
a2+b2=(6k+3)2+(12m+9)2
=36k^2+36k+9+144m^2+216m+81
=36k^2+144m^2+36k+216m+90
Notice that 36 is divisible by both 6 and 12, so we can ignore those terms. Thus, the remainder when a^2+b^2 is divided by 12 will be the remainder when
36k+216m+90 is divided by 12.
Let's simplify further: 36k+216m+90=12(3k+18m+7)+6
From this expression, we can see that when 36k+216m+90 is divided by 12, the remainder is 6.
Therefore, the correct answer is (C) 6.