1988 AIME Problems/Problem 9

Revision as of 01:42, 9 September 2016 by Ayushk (talk | contribs) (Solution)

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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