2013 AMC 10A Problem 13

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Concepts:digitscasework

Difficulty rating: 1370

13.

How many three-digit numbers are not divisible by 5,5, have digits that sum to less than 20,20, and have the first digit equal to the third digit?

5252

6060

6666

6868

7070

Solution:

Note that for the number to not be divisible by 5,5, the units digits cannot be either 00 or 5.5.

Let xx be the hundreds and units digit and yy be the tens digit. Then we want 2x+y<20. 2x + y \lt 20. Casing on the 99 options of x,x, we get:

If xx is 1,2,3,1, 2, 3, or 4,4, then yy can be anything since y<10.y \lt 10.

If x=6,x = 6, then y<8,y \lt 8, which gives us 88 solutions.

If x=7,x = 7, then y<6,y \lt 6, which gives us 66 solutions.

If x=8,x = 8, then y<4,y \lt 4, which gives us 44 solutions.

If x=9,x = 9, then y<2,y \lt 2, which gives us 22 solutions.

This gives us a total of 410+8+6+4+2=60 4 \cdot 10 + 8 + 6 + 4 + 2 = 60 solutions.

Thus, B is the correct solution.

Problem 13 in Other Years