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How many even, three-digit positive integers have the property that two of the digits are equal?

 Feb 6, 2022
 #1
avatar+189 
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118 numbers.

 Feb 6, 2022
 #2
avatar+23200 
+1

This is just to give an explanation for I's answer:

 

If the two equal digits are both 0, they must be found in the last two positions, while there are nine

possibilities for the first digit:  100, 200, 300, 400, 500, 600, 700, 800, 900.

 

If the two equal digits are both 1, they must be found in the first two position, while there are five

possibilities for the last digit:  110, 112, 114, 116, 118.

(The same is true if the two equal digits are both 3, 5, 7, and 9; for a total of twenty five.)

 

If the two equal digits are both 2:

-- if they are in the last two positions, there are 8 possibilities for the first position:

          122, 322, 422, 522, 622, 722, 822, and 922.

(The same is true if the two equal digits are both 4, 6, and 8; for a total of thirty two.)

-- if they are in the first and third position, there are 9 possibilities for the second position:

          202, 212, 232, 242, 252, 262, 272, 282, and 292.

(The same is true if the two equal digits are both 4, 6, and 8; for a total of thirty six.)

-- if they are in the first and second position, there are 4 possibilites for the third position:

          220, 224, 226, and 228.

(The same is true if the two eaual digits are both 4, 6, and 8; for a total of sixteen.)

 

When I total these, I get:  9 + 25 + 32 + 36 + 16  =  118.

 Feb 6, 2022

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