Question

Number 136 is added to 5B7 and the sum
obtained 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​

Answer 1

Answer:

☞ 136+5B7 = 7A3

☞ 7A3÷3 leaves no remainder.

☞ Divisibility rule of 3:

☞ If the sum of all the digits of a number is a multiple of 3 then, the number is also exactly divisible by 3.

☞ So, 7+A+3 = ?

☞ So, 10+A = 12, which nearest to 10 and a multiple of 3.

☞ Therefore, A = 12–10 = 2

☞ Therefore, 136+5B7 = 723

☞ 5B7 = 723 – 136

☞ 5B7 = 587

☞ therefore, the value of B = 8

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Answer by :

Amrutha Varshini