What is the remainder when 7^2 * 8^2 is divided by 6?

\((48+1)(60+4)=48*60+48*4+60+4\)=> All of them except 4 are divisible by 6, so the answer is 4 (D)

the solution from chetan2u was mind blowing and most simplifying.

However I have a remainder rule I know.
"Multiply the remainders of the dividends and divide the product by the divisor"

49/6 is remainder 1. 64/6 is remainder 4.
Therefore 4 * 1 equals 4.
4/6 is 0 remainder 4.
Answer is D

you know this rule, you do this math in 4 seconds.

7^2*8^2 = (56)^2 = (54+2)^2 = 54^2+2*2*54+2^2 = multiple of 6+multiple of 6 + 4 and this will give you a remainder of 4.

D is the correct answer.

Hope this helps.

\(\frac{(49 * 64)}{6}\) gives remainer x
49 is remainder 1. 64 is remainder 4. 4*1 = 4. So x = 4
OR
7 is remainer 1. 8 is remainder 2
\(1^2 * 2^2 = 4\)
D

Why is it wrong to transform 49*64/6 into 49*32/3 (divide numerator and denominator by 2)
Then remainder would be 2..
(48+1)(30+2)/3, 48 and 30 are multiples of 3, remainder=2
Why is it not possible to simplify the equation?

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Changing the divisor will lead to different remainders.
For example, 27 divided by 15 = 1 with remainder 12
If we simplify 27/15 to get 9/5, then we see that 9 divided by 5 = 1 with remainder 4

Yes, it would, and the remainder will actually get reduced by the divisor times. If divisor(common factor) is 5, then you will get actual remainder by multiplying the remainder you get in simplified form by 5.

Why does it happen?

Say, we are looking for remainder when 100 is divided by 15.
This will give us 10 as the remainder.

But if we reduce \(\frac{100}{15}\) to \(\frac{20}{3}\), that is we divide both by common term 5, the remainder is 2. The actual remainder is nothing but 2*5.

Now 100 is 15*6+10, so \(\frac{15*6+10}{15}=\frac{15*6}{15}+\frac{10}{15}\). So remainder is given by 5/15. But we are reducing it to simplified term 2/3, thereby reducing the remainder from 10 to 10/5 or 2

Great Question.
Here is what i did=>
7=6k+1
and 8=6k+2
so 7^2=> 6k+1
8^2=> 6k+4
Hence there product would leave the remainder 4

Hence D

\(\frac{7^2}{6}\) = Remainder 1
\(\frac{8^2}{6}\) = Remainder 4

Thus, remainder of \(\frac{7^2 * 8^2}{6}\) will be 1*4 , ie 4

Hence remainder will be (D) 4

I had come to an answer of 2.

This was my approach:
7^2*8^2
Simplify: 7*7*8*8
Then divide by 6: (7*7*8*8)/6
We get: (7*7*8*4)/3
We get 1568/3
Which gives a quotient of 522 and remainder of 2

What did I do incorrectly here?

Another way of looking at the problem:

(7^2 * 8^2)/6 = (56^2)/6

Now, since the remainder of 56/6 is 2 remainder of 56^2/6 will be 4.

Example: M/7 we get remainder r. What will be the remainder of M^2/7?

M = 7a+r. where a is a quotient and r is remainder.

M^2= (7a + r)(7a +r)
= 49a^2 + 7ar + 7ar + r^2
M^2/7 = 49a^2/7 + 7ar/7 +7ar/7 + r^2 => Remainder is r^2.

(7^2*8^2)/6=
(56^2)/6=
((54+2)^2)/6=
(2^2)/6=4

D = 4

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6/3= 2 so reminder will be
2*(reminder of 1568/3 ) = 4

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BrentGMATPrepNow chetan2u

Please confirm if it's wrong to reduce the fraction for such remainder problems. Are we required to transform the numerator only while retaining the divisor of the question stem?

The logic used in the above solution is the same as if the denominator were to be 6 instead of 3, yet the remainders we get are different. Please help.

What is the remainder when \(7^2 * 8^2\) is divided by 6

Remainder of Product of two numbers is same as Product of their remainders

=> Remainder of \(7^2 * 8^2\) by 6 = Remainder of \(7^2\) by 6 * Remainder of \(8^2\) by 6
= Remainder of 49 by 6 * Remainder of 64 by 6 = 1 * 4 = 4

So, Answer will be D
Hope it helps!

Watch the following video to learn the Basics of Remainders

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