Difference between revisions of "1988 AIME Problems/Problem 9"
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Revision as of 18:11, 4 July 2013
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 
, which is our minimum.
. Hence the lowest possible value for the hundreds digit is 
is a valid solution.
. The lowest possible value for the hundreds digit is 
, and we get 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
