study wrote:

If \(a\) and \(b\) are positive integers, is \(10^a + b\) divisible by 3?

1. \(\frac{b}{2}\) is an odd integer

2. the remainder of \(\frac{b}{10}\) is \(b\)

what positive integer divided by 10 gives a remainder of itself?

Hi all, let me try to explain this one.

So we have 10^a + b /3 what is the remainder?

First of all note that you are told that a,b are positive integers. THEY ALWAYS TELL YOU THIS FOR SOME REASON, DON'T IGNORE IT

So we have that 10^a = (10,100 etc...)

Now, what is the divisibility rule for multiples of 3? That's right the digits of the number must add to a multiple of 3

We already have the 1 on the 10^a. Let's see what we can get out of 'b'

Statement 1: b/2 is odd integer. So b could be (2,6,10,14 etc...)

If b is 2 then remainder is 0

If b is 6 then remainder is 1

If b is 10 then remainder is 2

Not good enough

Statement 2: b/10 remainder is b. Now wait a second. What this is telling us is that b<10. But that could be (1,2,3,4,5 etc...)

This is stil not enough

Statement (1) and (2) - Now going back to statement 1. We still have 2 or 6 as possible answer giving different remainders

Hence (E) is the correct answer

Kudos if you like!