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 \equiv 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 \equiv 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 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>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>. Hence, since <math>192 < 442</math>, the answer is <math>\fbox{192}</math>
 
 
The answer is <math>192</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 01:42, 9 September 2016

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 $20k + 8 \equiv 88 \pmod{100}$. This is true if the tens digit is either $4$ or $9$. Casework:

  • $4$: Then our cube must be in the form 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 $192$. Hence, since $192 < 442$, the answer is $\fbox{192}$

See also

1988 AIME (ProblemsAnswer KeyResources)
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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