# Problem of the Week

Problem
C and Solution

Five
Magnets

## Problem

Harlow has five magnets, each with a different number from \(1\) to \(5.\) They arranged these magnets to create
a five digit number \(ABCDE\) such
that:

the three-digit number \(ABC\)
is divisible by \(4\),

the three-digit number \(BCD\)
is divisible by \(5\), and

the three-digit number \(CDE\)
is divisible by \(3.\)

Determine the five-digit number that Harlow created.

## Solution

Since \(ABC\) is divisible by \(4\), it follows that \(C\) must be even, so \(C=2\) or \(C=4.\)

Since \(BCD\) is divisible by \(5\), it follows that \(D=0\) or \(D=5.\) However, there is no magnet with a
\(0\), so it follows that \(D=5.\)

We also know that \(CDE\) is
divisible by \(3.\) We can consider the
following two cases.

**Case 1:** \(C=2\).

If \(C=2\), then the three-digit number
\(CDE\) is \(25E.\) The only possibilities for \(E\) are \(1,\) \(3,\) or \(4\). However, none of \(251,\) \(253\) and \(254\) are divisible by \(3.\) It follows that \(C\) cannot equal \(2.\)

**Case 2:** \(C=4\).

If \(C=4\) then the three-digit number
\(CDE\) is \(45E.\) The only possibilities for \(E\) are \(1,\) \(2,\) or \(3.\) Since \(451\) and \(452\) are not divisible by \(3,\) but \(453\) is divisible by \(3,\) it follows that \(C=4\) and \(E=3.\)

Thus, the three-digit number \(ABC\)
is \(AB4.\) The only magnets not used
yet are numbered \(1\) and \(2,\) so this number is \(124\) or \(214.\) Since \(214\) is not divisible by \(4,\) but \(124\) is divisible by \(4,\) it follows that \(A=1\) and \(B=2.\)

Therefore, the five-digit number must be \(12453.\)