2017 AMC 10 A

Note that . This can be seen from the fact that . Thus, if , then , and thus . The only answer choice that is is .

One divisibility rule for division that we can use in the problem is that a multiple of has its digit always add up to a multiple of . We can find out that the least number of digits the number is , with and , assuming the rule above. No matter what arrangement or different digits we use, the divisor rule stays the same. To make the problem simpler, we can just use the and . By

randomly mixing the digits up, we are likely to get: . By adding to this number, we get: . Knowing that this number is divisible by when is subtracted,

we can subtract from every available choice, and see if the number is divisible by afterwards. After subtracting from every number, we can conclude that (originally ) is the only number divisible by . So our answer is .

The number can be viewed as having some unique digits in the front, following by a certain number of nines. We can then evaluate each potential answer choice.

If is correct, then must be some number , because when we add one to we get . Thus, if is the correct answer, then the equation must have an integer solution (i.e. must be divisible by ). But since it does not, is not the correct answer.

If is corret, then must be some number , because when we add one to , we get .Thus, if is the correct answer, then the equation must have an integer solution. But since it does not, is not the correct answer.

Based on what we have done for evaluating the previous two answer choices, we can create an equation we can use to evaluate the final three possibilities.Notice that if , then must be a number whose initial digits sum to , and whose other, terminating digits, are all . Thus, we can evaluate the three final possibilities by seeing if the equation has an integer solution.

The equation does not have an integer soutionfor , so is not correct. However, the equation does have an integer solutionfor so is the answer.