Question : If a 10-digit number $620x976y52$ is divisible by 88, then the least value of $(x^2+y^2)$ will be:
Option 1: 8
Option 2: 7
Option 3: 11
Option 4: 10
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Correct Answer: 10
Solution :
Given number = $620x976y52$
Co-prime factors of $88 = 8 × 11$
For the smallest possible value of $y$,
$y52$ will be divisible by 8
So, $y$ will be 1 for 152, which is divisible by 8.
As $y = 1$, the possible value of x for which $620x976152$ is divisible by 11.
Sum of even places in number $= 2 + x + 7 + 1 + 2 = 12 + x$
Sum of odd places in number $= 6 + 0 + 9 + 6 + 5 = 26$
For divisibility rule 11: The difference between the sum of digits at odd places and the sum of digits at even places is equal to zero or a multiple of 11.
AT $x=3$, difference = 26 – 15 = 11
$\therefore$ Value of $x^2 + y^2=(3)^2 + 1^2=9 + 1=10$
Hence, the correct answer is 10.
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