+174 Using the digits 1, 2, 3, 4, 5, how many even three-digit numbers less than 500 can be formed if each digit can be used more than once?
+2440 Let's consider properties of integers that fit the following conditions.
It is probably easiest to consider the possibilities for the final digit first. Because one criterion is an even number, the final digit must be either a 2 or a 4.
Eliminating possibilities for the first digit is relatively simple, as well. The numbers must not exceed 500, so the first digit of this three-digit integer cannot be a 5 because it would violate the aforementioned criterion. Otherwise, though, no restriction exists for the beginning digit, so the numbers 1 to 4 are all candidates for the first digit
Excluding the digit restriction, the middle digit cannot affect the ability for a number to qualify for any of the predetermined conditions. Therefore, 1 to 5 are all possibilities.
There are 4 possibilities for the first digit, 5 possibilities for the middle digit, and 2 for the last digit. The product is the number of integers that satisfy the preset conditions. \(4*5*2=40\text{ total possibilities}\)
+36915 First digit cannot be 5 so yo have 4 * 5 * 5 possible numbers = 100
2 out of 5 will be even as 2 and 4 are the only evens 2/5 x 100 = 40 possibilities.
