2001 AIME II Problems/Problem 1

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Problem

Let $N$ be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of $N$ forms a perfect square. What are the leftmost three digits of $N$?

Solution

The two-digit perfect squares are $16, 25, 36, 49, 64, 81$. We try making a sequence starting with each one:

  • $16 - 64 - 49$. This terminates since none of them end in a $9$, giving us $1649$.
  • $25$.
  • $36 - 64 - 49$, $3649$.
  • $49$.
  • $64 - 49$, $649$.
  • $81 - 16 - 64 - 49$, $81649$.

The largest is $81649$, so our answer is $\boxed{816}$.

See also

2001 AIME II (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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