A number 34A is exactly divisible by 2 and leaves a remainder 1, when divided by 5, find A.

Given, a number 34A

Need to find out it must be divisible by 2 and when it divides with 5 it leaves the remainder 1

⇒ For a number to be exactly divisible by 2 the unit place of a number must be divided by 2.

So, by trial and error method number ending with 2,4, 6 are divisible by 2 but it should be divisible by 5 to get a remainder 1.

∴ if A = 6 then the number will be 346 which is divisible by 2 and also by 5 by leaving a remainder 1.

Hence, A = 6 and the number is 346.