2013 AIME I Problems/Problem 2
Contents
Problem 2
Find the number of five-digit positive integers, , that satisfy the following conditions:
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(a) the number is divisible by
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(b) the first and last digits of are equal, and
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(c) the sum of the digits of is divisible by
Solution 1
The number takes a form of , in which . Let and be arbitrary digits. For each pair of , there are exactly two values of that satisfy the condition of . Therefore, the answer is
Solution 2
For a number to be divisible by , the last digit of the number must be or . However, since the first digit is the same as the last one, the last (and first) digits can not be , so the number must be in the form , where is divisible by 5 Since there is a chance that sum of digits of a randomly selected positive integer is divisible by , This gives us a answer of - mathleticguyyy
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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