两个连续的正整数相乘。其乘积四舍五入到最接近的百位数是 202100。求这两个正整数的和。

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完整解法

1. Translate the rounding statement into a possible range for the product.

Since the product rounds to $202100$ to the nearest hundred,
$202050 \leq n(n+1) < 202150.$

2. Find the consecutive integers whose product lies in the range.

Test consecutive integers near $\sqrt{202100}$ .

$449 \times 450 = 202050$

This lies in the range, so the integers are $449$ and $450$ .

Their sum is
$449+450=899.$

答案

$899$

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