# Dozens Poster

Dozens Poster

## Problem

## Student Solutions

54312

**Why?**

It must be a multiple of 3

1 + 3 + 4 + 5 = 13 so the remaining digit could be 2, 5, 8

__Try 8__. It must be a multiple of 4 so the last two digits must be divisible by 4.

853**14**

2-digit multiples of 4 which end in 4 (or 8) have second-last digit even: 04, 24, 44, 64, 84

Best option with 8: 53184

__Try 5__: impossible since we can't make a multiple of 4

__Try 2.__

Using previous pattern, 53124

But 2-digit multiples of 4 which end in 2 have second-last digit odd: 12, 32, 52, ...

54312

This is the largest possibility