Question

Number 136 is added to 5B7 and the sumobtained is 7A3, where A and B are integers.It is given that 7A3 is exactly divisible by 3.The only possible value of B is(a) 2(b) 5(c) 7(d)8​

Answer 1

Answer:

8

Step-by-step explanation:

Given,

136 is added to 5B7 and the sum obtained is 7A3

  1 3 6

+ 5 B 7

(1) (1)     (The number (1) represents the carry.)

 --------

  7 A  3

So, the number B should be such that when it is added to 3 and 1, it should give the carry 1 and the left number i.e, A is written to the sum.

Now, when we check from the options given, if we take 2 or 5 as the B value, it does not produce any carry. So, let us check using 7 and 8.

When B = 7, 136 + 577 = 713 is not divisible by 3.

When B = 8, 136 + 587 = 723 is divisible by 3.

Hence, the only possible value of B is 8.