AIME 2001 II · 第 1 题

AIME 2001 II — Problem 1

专题: Contest Math
难度: L4
来源: AIME

题目详情

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 $N$ ?

解析

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 $\boxed{816}$ .

Video Solution

https://youtu.be/rOUG1tXqnfU