If a six-digit number 53a29b is divisible by | vedclass

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(C) For the number to be divisible by $88$,it must be divisible by both $8$ and $11$.For the number to be divisible by $8$,the number formed by the la

Vedclass · Accepted Answer

$8$ and $6$

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(C) For the number to be divisible by $88$,it must be divisible by both $8$ and $11$.For the number to be divisible by $8$,the number formed by the last three digits $(29b)$ must be divisible by $8$.$290 + b$ divided by $8$: $296 / 8 = 37$. Thus,$b = 6$.For the number to be divisible by $11$,the difference between the sum of digits at odd places and the sum of digits at even places must be $0$ or a multiple of $11$.Sum of digits at odd places (from right): $b + 2 + 3 = 6 + 2 + 3 = 11$.Sum of digits at even places (from right): $9 + a + 5 = 14 + a$.Difference: $(14 + a) - 11 = a + 3$.For divisibility by $11$,$a + 3$ must be $0$ or $11$. Since $a$ is a digit,$a + 3 = 11 \Rightarrow a = 8$.Therefore,$a = 8$ and $b = 6$.

If a six-digit number $53a29b$ is divisible by $88$,then find the value of $a$ and $b$.

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