Difference between revisions of "1988 AIME Problems/Problem 9"
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A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of <math>(10k + 2)^3</math>; using the [[binomial theorem]] gives us <math>1000k^3 + 600k^2 + 120k + 8</math>. Since we are looking for the tens digit, <math>\mod{100}</math> we get <math>20k + 8 = 88 \pmod{100}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework: | A little bit of checking tells us that the units digit must be 2. Now our cube must be in the form of <math>(10k + 2)^3</math>; using the [[binomial theorem]] gives us <math>1000k^3 + 600k^2 + 120k + 8</math>. Since we are looking for the tens digit, <math>\mod{100}</math> we get <math>20k + 8 = 88 \pmod{100}</math>. This is true if the tens digit is either <math>4</math> or <math>9</math>. Casework: | ||
− | *<math>4</math>: Then our cube must be in the | + | *<math>4</math>: Then our cube must be in the form of <math>(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}</math>. Hence the lowest possible value for the hundreds digit is <math>4</math>, and so <math>442</math> is a valid solution. |
*<math>9</math>: Then our cube is <math>(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}</math>. The lowest possible value for the hundreds digit is <math>1</math>, and we get <math>192</math>, which is our minimum. | *<math>9</math>: Then our cube is <math>(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}</math>. The lowest possible value for the hundreds digit is <math>1</math>, and we get <math>192</math>, which is our minimum. | ||
Revision as of 12:56, 24 October 2007
Problem
Find the smallest positive integer whose cube ends in 888.
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.
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 |