#
Problem
of the Week

Problem
E and Solution

One
More Coin

## Problem

A coin collecting club has between \(12\) and \(30\) members attend its monthly meeting.
For one such meeting, they noticed that all of the members present each
had the same number of coins except one member who had one more coin
than each of the other members. Between them, the members had a total
number of \(1000\) coins.

How many members attended the meeting?

## Solution

Let \(n\) represent the number of
members present at the meeting. We know that \(12 < n <30\) and \(n\) is an integer. Let \(c\) represent the number of coins that all
but one member had. That member had \(c+1\) coins. It follows that \((n-1)\) members had \(c\) coins each and one member had \(c+1\) coins, producing a total of \(1000\) coins.

Thus, $$\begin{align}
(n-1)\times c+1\times (c+1)&=1000\\
nc-c+c+1&=1000\\
nc&=999
\end{align}$$ We are looking for two positive integers with a
product of \(999\), with one of the
numbers between \(12\) and \(30\). The prime factorization of \(999\) is \(3\times 3\times 3\times 37\). We can
combine the factors to produce pairs of positive integers whose product
is \(999\). The possibilities are \(1\) and \(999\), \(3\) and \(333\), \(9\) and \(111\), and \(27\) and \(37\). The only possible product which gives
one factor between \(12\) and \(30\) is \(27\times 37\).

It follows that there were \(27\)
members present at the last meeting, and \(26\) of the members had \(37\) coins each and \(1\) member had \(38\) coins. This can be easily verified, as
\(26\times 37+1\times 38=1000\).