Question

If 7A6146B is divisible by 88 then what is the value of A² + B²?​

Answer 1

The value of $A^{2} +B^{2}$ is 25.

Explanation:

Given: 7A6146B is divisible by 88

To find: The value of $A^{2} +B^{2}$

Solution:

We have, 7A6146B is divisible by 88. For this to be happen, 7A6146B needs to be divisible by 8 and 11 both.

Now, for 7A6146B to be divisible by 8, the last three digit of the number needs to be divisible by 8.

So, we have 46B to be divisible by 8.

Therefore, B can have values 0,1,2,3,4,5,7,8,9

Out of all these values, 46B is divisible by 8 only when B is 4.

So, we have $B=4$

Now, we have 7A61464

For this number to be divisible by 11, the difference of the sum of digit at odd places of the number to the even places of the number needs to be 0 or multiple of 11.

So, we have

Sum of digits at odd places$=7+6+4+4=21$

Sum of the digits at even places$=A+1+6=A+7$

Now, for difference to be 0 or multiple of 11, the value of A should be 3.

So, we have $A=3$

Now, $A^{2} +B^{2}=3^{2} +4^{2}=25$

Hence, the value of $A^{2} +B^{2}$ is 25.

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Divisibility by 88

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