Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1988 AIME Problems/Problem 9"

m (→‎See also)
(solution)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
Find the smallest positive integer whose [[perfect cube|cube]] ends in <tt>888</tt>.
  
 
== Solution ==
 
== Solution ==
 +
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 from 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.
 +
 +
The answer is <math>192</math>.
  
 
== See also ==
 
== See also ==
−
* [[1988 AIME Problems]]
+
{{AIME box|year=1988|num-b=8|num-a=10}}
  
−
{{AIME box|year=1988|num-b=8|num-a=10}}
+
[[Category:Intermediate Number Theory Problems]]

Revision as of 20:08, 26 September 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 $(10k + 2)^3$; using the binomial theorem gives us $1000k^3 + 600k^2 + 120k + 8$. Since we are looking for the tens digit, $\mod{100}$ we get $20k + 8 = 88 \pmod{100}$. This is true if the tens digit is either $4$ or $9$. Casework:

  • $4$: Then our cube must be in the from of $(100k + 42)^3 \equiv 3(100k)(42)^2 + 42^3 \equiv 200k + 88 \pmod{1000}$. Hence the lowest possible value for the hundreds digit is $4$, and so $442$ is a valid solution.
  • $9$: Then our cube is $(100k + 92)^3 \equiv 3(100k)(92)^2 + 92^3 \equiv 200k + 688 \pmod{1000}$. The lowest possible value for the hundreds digit is $1$, and we get $192$, which is our minimum.

The answer is $192$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1988_AIME_Problems/Problem_9&oldid=17028"