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Difference between revisions of "2001 AIME II Problems/Problem 1"

 
 
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== Problem ==
 
== Problem ==
 +
Let <math>N</math> be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of <math>N</math> forms a perfect square. What are the leftmost three digits of <math>N</math>?
  
 
== Solution ==
 
== Solution ==
 +
The two-digit perfect squares are <math>16, 25, 36, 49, 64, 81</math>. We try making a sequence starting with each one:
 +
 +
*<math>16 - 64 - 49</math>. This terminates since none of them end in a <math>9</math>, giving us <math>1649</math>.
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*<math>25</math>.
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*<math>36 - 64 - 49</math>, <math>3649</math>.
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*<math>49</math>.
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*<math>64 - 49</math>, <math>649</math>.
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*<math>81 - 16 - 64 - 49</math>, <math>81649</math>.
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 +
The largest is <math>81649</math>, so our answer is <math>\boxed{816}</math>.
  
 
== See also ==
 
== See also ==
−
* [[2001 AIME II Problems]]
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{{AIME box|year=2001|n=II|before=First question|num-a=2}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 20:34, 4 July 2013

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 (Problems • Answer Key • Resources)
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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