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>. | ||
+ | *<math>25</math>. | ||
+ | *<math>36 - 64 - 49</math>, <math>3649</math>. | ||
+ | *<math>49</math>. | ||
+ | *<math>64 - 49</math>, <math>649</math>. | ||
+ | *<math>81 - 16 - 64 - 49</math>, <math>81649</math>. | ||
+ | |||
+ | The largest is <math>81649</math>, so our answer is <math>\boxed{816}</math>. | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2001|n=II|before=First question|num-a=2}} | |
+ | |||
+ | [[Category:Introductory Algebra Problems]] |
Revision as of 16:59, 30 October 2007
Problem
Let be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of
forms a perfect square. What are the leftmost three digits of
?
Solution
The two-digit perfect squares are . We try making a sequence starting with each one:
. This terminates since none of them end in a
, giving us
.
.
,
.
.
,
.
,
.
The largest is , so our answer is
.
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 |