Question number one. The difference between a positive integer and the sum of its digits is always divisible by what number? Given that this is a multiple-choice question, let's test a two-digit number and a three-digit number before we prove what the correct answer is. With a two-digit number, let's look at 13. The sum of its digits is 4, so 13 minus 4 is 9. And with a three-digit number, for example a 120, the sum of its digits is 3. Subtracting, we have 117, which is equal to 13 times 9. So we would guess at this point that the answer is 9. And in general, we would have to consider a n-digit positive integer, so let the digits be denoted by d0, d1, all the way up to dn-1. And in base-10 representation, we would have a d0 times 10 to the 0 power plus d1 times 10 to the 1st power plus d2 times 10 squared, all the way up to dn-1 times 10 to the power n-1. And then we subtract from this the sum of the digits, d0 plus d1 plus d2, all the way up to dn-1. And our goal is to show that this number is divisible by 9, so we simplify as much as possible. We can group the terms and then look at each term and decide if it's indeed divisible by 9. So, one more step. We have a d0 times 0, a d1 times 9, a d2 times 99, and a dn-1 times, we will have a number consisting of all nines, and there is exactly n-1 of them here. So we see that the first term vanishes, and then the first term after that, d1 times 9 is divisible by 9, d2 times 99 will be divisible by 9, and likewise, the last term here is divisible by 9, and each term in between is also like that. So we see that we have obtained a positive integer that's equal to 9 times some positive integer, and that is divisible by 9. So it is indeed always true that the difference between a positive integer and the sum of its digits will be divisible by 9, and so E is the correct answer.

positive integer

sum of digits

divisible by 9

base-10 representation

mathematical proof