AMC 10 Daily Practice Round 3

There are no -digit integers with the given property. This is because the product of the digits of a -digit number is at most .

Observe that . We are essentially looking for products of , , or digits that equal . The only digit that is a multiple of is itself, so no matter how many digits has, exactly two of its digits must be . The number of -digit numbers with the given property is , since if two of the digits are , the third must be . There are three -digit numbers with one digit equal to and two digits equal to . They are , , and .

If has four digits, then two digits are and the other two digits have a product of . The only ways to express as a product of two integers is and . In either case, there are six ways to choose where the two digits equal to go. They are

55_ _

5_5_

5__5

_55_

_5_5

__55

In each of these cases, there are ways to place the remaining digits. There are also two choices for the remaining two digits ( and or and ), so the number of -digit numbers with the given property is .

Using a similar method of counting, we conclude that if has digits, then there are ways to place the two digits that are equal to . The remaining three digits must have a product of , so they must be , , and or , , and or , , and . If the remaining digits are , , and , then there are choices of where to place the remaining digits. This is because once the is placed, the last two digits must be (there is no choices). If the remaining digits are , , and , then there are ways to place the remaining digits. This is because there are ways to order the digits , , and . Finally, if the remaining digits are all , then there is only one way to place the digits. Thus, there are five-digit numbers with the given property.

The total number of integers with with the property that the product of the digits is is . The integer formed by the rightmost digits of is .