Correct Answer - Option 3 : 8
Given:
When number 136 is added to number 5B7, the result obtained is equal to 7A3 with A and B both being integers.
Number 7A3 is divisible by 3.
Concepts used:
Divisibility test of 3 is: The sum of digits of number must be divisible by 3.
Calculation:
136 + 5B7 = 7A3
⇒ (100 + 30 + 6) + (500 + 10B + 7) = 700 + 10A + 3
⇒ 600 + (3 + B)10 + 13 = 700 + 10A + 3
⇒ 600 + (3 + B)10 + 10 + 3 = 700 + 10A + 3
⇒ 600 + (3 + 1 + B)10 + 3 = 700 + 10A + 3
⇒ 3 + 1 + B = A
⇒ 4 + B = A
In sum 136 + 5B7 = 7A3, at hundred’s place, the sum is 6 but one carry has been taken,
⇒ B ≥ 6, 0 ≤ A ≤ 3.
Number 7A3 is divisible by 3.
Sum of digits of number 7A3 = 7 + A + 3 = A + 10
As 7A3 is divisible by 3,
⇒ A + 10 = 12
⇒ A = 2
⇒ 4 + B = A
When the value of B is 8, the value of A is 8 possible.
∴ The only possible value of B is 8.
