2011 AMC 10A Problem 13

Below is the professionally curated solution for Problem 13 of the 2011 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 10A solutions, or check the answer key.

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Concepts:digitscaseworkmultiplication principle

Difficulty rating: 1280

13.

How many even integers are there between 200200 and 700700 whose digits are all different and come from the set {1,2,5,7,8,9}?\{1,2,5,7,8,9\}?

1212

2020

7272

120120

200200

Solution:

Since the hundreds digit can only be a 22 or 5,5, we can case on this value.

Case 1: hundreds digit is 22

The only option for the units digit is 8,8, since the number must be even. This leaves 44 options for the tens digit.

This gives us 14=41 \cdot 4 = 4 numbers for this case.

Case 2: hundreds digit is 55

Similarly to above, 22 and 88 are the only options for the units digit, leaving 44 options for the tens digit.

This gives us 24=82 \cdot 4 = 8 numbers for this case.

The total number of integers is then 4+8=12.4 + 8 = 12.

Thus, A is the correct answer.

Problem 13 in Other Years