sampath wrote:If x and y are positive integers, is the integer 10^x-y divisible by 9 ?
1. y is divisible by 3
2. (10^x + y ) is not divisible by 9

If you add or subtract two integers a and b, and the result is divisible by 3, there are two possibilities: either a and b are *both* divisible by 3, or a and b are *both* not divisible by 3. If one of a or b is divisible by 3, and the other is not, you will never get a multiple of 3 when you add or subtract a and b. So, looking at Statement 1, if x is a positive integer, then 10^x is never divisible by 3. So if y *is* divisible by 3, 10^x - y cannot be divisible by 3, and thus cannot be divisible by 9. So Statement 1 is sufficient to give a 'no' answer to the question.

Statement 2 is not sufficient. It might be that y=1, for example, and in that case 10^x - 1 will be a number in which every digit is 9, so must be divisible by 9, or y might be 2, in which case 10^x - y will also not be divisible by 9.

shayam wrote:Hey Ian,

In the below concept - divisible by 3, does it hold good for other integers?
Your concept to the problem-
If you add or subtract two integers a and b, and the result is divisible by 3, there are two possibilities: either a and b are *both* divisible by 3, or a and b are *both* not divisible by 3. If one of a or b is divisible by 3, and the other is not, you will never get a multiple of 3 when you add or subtract a and b.

If these properties are unfamiliar, it's best to explore them first using specific numerical examples, since the 'proofs' are a bit abstract. But we can see why all of the above facts are true algebraically. First, if you add or subtract two multiples of, say, 7, you must get a multiple of 7. There's nothing special about 7 here - that's true for any number at all. You can see this by factoring. If we have two multiples of 7, we can write our numbers as 7a and 7b. Then if we add these numbers, we get

7a + 7b = 7(a+b)

and as you can see, we can factor out a 7 from the sum.

If, however, you add a multiple of 7 to a number which is *not* a multiple of 7, you can never get a multiple of 7 as a result. Again, we can see this by factoring. We can again write our multiple of 7 as 7a. The other number is not a multiple of 7, so it gives some remainder, r, when we divide it by 7. So we can write our second number as 7q + r, where 0 < r < 7. Now if we add (or similarly subtract), we get

7a + 7q + r = 7(a + q) + r

and the result is r greater than a multiple of 7. So the remainder will be r when we divide this sum by 7, and our number is not a multiple of 7.

Finally, if you have two numbers neither of which is a multiple of 7, then anything can happen if you add or subtract. For example, if you add 10+11, you get a multiple of 7, but if you add 10+12, you do not. You could look at this algebraically again; we have two numbers 7q + r and 7s + t, where r and t are nonzero remainders. When we add, we get:

7q + r + 7s + t = 7(q + r) + s + t

Now this number is divisible by 7 only if s+t is divisible by 7 - that is, only if the remainders of our two numbers add to 7. So when we add two non-multiples of 7, we can sometimes get a multiple of 7, and sometimes not.

Again, there's nothing special about 7 here - this is true for any divisor at all - but it's easier to use a concrete number for illustration.