CEMC Pascal, 2021

Number theory Difficulty 3.7 AMC 10/12 Find the answer Canada

Suppose that $R$ , $S$ and $T$ are digits and that $N$ is the four-digit positive integer $8RST$ . That is, $N$ has thousands digit 8, hundreds digit $R$ , tens digits $S$ , and ones (units) digit $T$ , which means that $N = 8000+100R+10S+T$ . Suppose that the following conditions are all true:

The two-digit integer $8R$ is divisible by 3.
The three-digit integer $8RS$ is divisible by 4.
The four-digit integer $8RST$ is divisible by 5.
The digits of $N$ are not necessarily all different.

The number of possible values for the integer $N$ is

Pick one

Solution

We make a chart of the possible integers, building their digits from left to right. In each case, we could determine the required divisibility by actually performing the division, or by using the following tests for divisibility:

An integer is divisible by 3 when the sum of its digits is divisible by 3.
An integer is divisible by 4 when the two-digit integer formed by its tens and units digits is divisible by 4.
An integer is divisible by 5 when its units digit is 0 or 5.

$8R$	$8RS$	$8RST$
$81$	$812$	$8120$
$8125$
$816$	$8160$
$8165$
$84$	$840$	$8400$
$8405$
$844$	$8440$
$8445$
$848$	$8480$
$8485$
$87$	$872$	$8720$
$8725$
$876$	$8760$
$8765$

In the first column, we note that the integers between 80 and 89 that are multiples of 3 are 81, 84 and 87. In the second column, we look for the multiples of 4 between 810 and 819, between 840 and 849, and between 870 and 879. In the third column, we add units digits of 0 or 5.

This analysis shows that there are 14 possible values of $N$ .

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