#
Problem
of the Week

Problem
C and Solution

Where
Does the Year Go?

## Problem

The positive integers are written consecutively in rows, with seven
integers in each row. That is, the first row contains the integers \(1\), \(2\), \(3\), \(4\), \(5\), \(6\), and \(7\). The second row contains the integers
\(8\), \(9\), \(10\), \(11\), \(12\), \(13\), and \(14\). The third row contains the integers
\(15\), \(16\), \(17\), \(18\), \(19\), \(20\), and \(21\), and so on.

Determine the row and the column that the integer \(2024\) is in.

## Solution

The last number in row \(1\) is
\(7\), the last number in row \(2\) is \(14\), and the last number in row \(3\) is \(21\). Observe that the last number in each
row is a multiple of \(7\).
Furthermore, the last number in row \(n\) is \(7\times
n\). So, we will find the largest multiple of \(7\) that is less than \(2024\).

We solve the equation \(7\times n =
2024\) to get \(n \approx
289.14\).

Therefore, the largest multiple of \(7\) that is less than 2024 is \(289 \times 7 = 2023\). This means that
\(2023\) is the last number in row
\(289\). Thus, \(2024\) will be the first number in in row
\(290\).

Therefore, \(2024\) is in row \(290\) and column \(1\).