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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $6$

Vedclass · Answer

(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 $b$.

Similar Questions

$\frac{17}{29} \times \frac{87}{102} \times \frac{48}{27} \times \frac{3}{2} = ?$

The difference between two positive integers is $3$. If the sum of their squares is $369$,then the sum of the numbers is

$6894.986 + 5025.005 + 600.020 = ?$

What least number must be added to $7231$ so that the resulting number is exactly divisible by $5$ and $9$ together?

Sum of three numbers is $132$. The first number is twice the second,and the third number is one-third of the first. Find the second number.

Vedclass Test Series

Exam Paper Generator

Online Exam Module