AMC 10 Daily Practice Round 3

Set . We can rearrange as follows: and so .

We need to determine for which values of , , and the integer is divisible by .

Observe that , and since and have no prime factors in common, will be divisible by exactly when it is divisible by both and .

Since is odd, it is not a multiple of , and so is a multiple of exactly when is a multiple of .

Factoring gives , and since is not a multiple of , we get that has a factor of but not a factor of .

Therefore, is a multiple of exactly when is a multiple of .

We now have that is a multiple of exactly when is a multiple of and a multiple of , or equivalently, when is a multiple of .

Since is a multiple of , is a multiple of exactly when is a multiple of .

The digits , , and are each between and inclusive, so is at least and at most . The only multiples of between and are and , so we must have that or .

To answer the question, we count the number of triples of integers, each between and inclusive, for which or .

First, suppose . If , then . In this case, we can have and , or and , or and , and so on to and . This gives a total of triples. If , then . In this case, we can have and , and , and so on to and for a total of triples. Continuing in this way, there are triples when , triples when , triples when , triples when , triples when , triples when , and there is one triple when . The number of triples when is .

Now assume . If , then , which is impossible given that and . There are no triples with . Similarly, if , then there are no triples. If , then , which means and , so there is only one triple. If , then , so either and or and . There are triples in this case. If , then , and there are triples. Continuing in this way, when A=, there are triples, when , there are triples, when , there are triples, when , there are triples, and when , there are triples. The number of triples when is . The number of triples is .