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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?

Jan 11, 2021

#1
+1

first digit    4   choices

next digit   5 choices

last digit    5 choices

4*5*5=......   possibles

Jan 11, 2021
#2
+1

* * * edit  * * *

first digit    4   choices

next digit   5 choices

last digit    2 choices   (only 2 or 4 will make even)

4*5*2=......   possibles

Guest Jan 11, 2021
#3
+303
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Here there are two way to do it in the link https://web2.0calc.com/questions/maths_68552 but they got 40

Here is one way by TheXSquaredFactor:

Let's consider properties of integers that fit the following conditions.

Even three-digit number

Less than 500

Comprised of the digits 1 to 5

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. \(4*5*2= 40 \)

Here is the second way by ElectricPavlov:

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.

Jan 11, 2021