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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?

wiskdls  Apr 7, 2018
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 #1
avatar+1895 
+2

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. \(4*5*2=40\text{ total possibilities}\)

TheXSquaredFactor  Apr 7, 2018
 #2
avatar+12248 
+1

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.

ElectricPavlov  Apr 7, 2018

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