1988 AIME Problems/Problem 9
Problem
Find the smallest positive integer whose cube ends in .
Solution
A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of ; using the binomial theorem gives us . Since we are looking for the tens digit, we get . This is true if the tens digit is either or . Casework:
- : Then our cube must be in the form of . Hence the lowest possible value for the hundreds digit is , and so is a valid solution.
- : Then our cube is . The lowest possible value for the hundreds digit is , and we get . Hence, since , the answer is
See also
1988 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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