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Q.

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

 A.

2

 B.

4

 C.

6

 D.

8

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Solution:
Option(B) is correct

$7^2 × 8^2 = (7×8)^2 = 56^2$

The number immediately before 56 that is divisible by 6 is 54.

Now, writing $56^2$ as $(54 + 2)^2$, we have,

$56^2 = (54 + 2)^2$

$= 54^2 + 2(2)(54) + 2^2$ by the formula $(a + b)^2 = a^2 + 2ab + b^2$

$= 54[54 + 2(2)] + 2^2$

$= 6 × 9[54 + (2×2)] + 4$ here, the remainder is 4

Edit: For an alternative solution, check comment by Shobia.

Edit 2: For yet another alternative solution, check comment by Sravan Reddy.


(4) Comment(s)


Sravan Reddy
 ()

$7^2 = 49 = 48+1$

$7^2 * 8^2 = (48+1)*64$

$48*64 + 64$. When divided by $6$, $48*64$ is divisible by 6 and 64 divided by 6 leaves reminder 4.

$\textbf{'Modulus' approach:}$

$7^2 =49 \text{mod}6 = 1\text{mod}6$

$7^2 * 8^2 = 1*64\text{mod}6 = 64\text{mod}6 = 4\text{mod}6$

Hence, the remainder is 4.



Shobia
 ()

$7^2=49$, $8^2=64$

$49*64=3136$

$\dfrac{3136}{6}$ will give remainder 4



Amar
 ()

well $\frac{3136}{6} = \frac{1568}{3}$

this gives remainder of 2