FP2 5.5 - Divisibility tests | Edexcel A-Level Further Mathematics

Derive and use the decimal divisibility tests for $2,3,4,5,6,9,10,11$ . Choose a suitable test, justify it with congruences, and use digit restrictions without confusing a divisibility test with ordinary equality.

Replace a number by a smaller congruent expression

Why is $584\,584$ divisible by $11$ , and why would replacing $584$ by any other three-digit block still work? You do not need to divide a six-digit number each time. Decimal place value turns the question into a short remainder calculation.

You already know

For a positive modulus $m$ , $a\equiv b\pmod m$ means $m\mid(a-b)$ . Addition, multiplication and positive integer powers preserve congruences. In particular, $m\mid a$ exactly when $a\equiv0\pmod m$ . Coprime factors can be cancelled while keeping the modulus fixed.

For example, $8420=8(1000)+4(100)+2(10)+0$ . Reducing the place-value weights modulo a chosen divisor can make most terms vanish, make all weights $1$ , or make them alternate between $1,-1$ .

Divisibility test

A divisibility test replaces an integer by a simpler quantity that is divisible by the proposed divisor exactly when the original integer is. If $N\equiv R\pmod m$ , then

m\mid N\quad\Longleftrightarrow\quad m\mid R

The replacement need not equal the original integer.

The tests below apply to non-negative decimal integers, including zero. For a negative integer, use its positive magnitude: $m\mid N$ exactly when $m\mid(-N)$ . Also $0$ is divisible by each nonzero divisor, since $0=m\times0$ ; the tests correctly allow this boundary case.

Key idea

Look at the remainder of $10$ , or a convenient power of $10$ , modulo the divisor. That remainder explains which digits matter.

The decimal tests and their congruence derivations are not in the formula booklet. Each family below states the rule you need to know.

Use the last digit or the last two digits

Write $N=10q+d$ , where $d$ is the units digit. Because $10$ is divisible by $2,5,10$ , the whole term $10q$ disappears modulo each of those divisors. Only $d$ remains.

Tests for two, five and ten

A non-negative decimal integer is divisible by

$2$ exactly when its last digit is $0,2,4,6,8$ ;
$5$ exactly when its last digit is $0$ or $5$ ;
$10$ exactly when its last digit is $0$ .

Not in the formula booklet: learn it.

For $4$ , one digit is not enough: $10$ is not a multiple of $4$ . Use $N=100q+r$ , with $0\le r<100$ , since $100\equiv0\pmod4$ . The two-digit ending may begin with zero; for example, an ending $04$ represents $r=4$ .

Test for four

A non-negative decimal integer is divisible by $4$ exactly when the integer formed by its last two digits is divisible by $4$ :

100q+r\equiv r\pmod4

For an integer with fewer than two digits, use the whole integer as $r$ .

Not in the formula booklet: learn it.

Test all four divisors together, keeping the reason for each decision visible.

Worked example 1: four tests from the ending

Question. Determine which of $2,4,5,10$ divide $8420$ , without carrying out four long divisions.

Plan. Use the last digit for $2,5,10$ , and the last two digits for $4$ .

\begin{aligned} \text{last digit}&=0 && \text{passes the }2,5,10\text{ tests}\\ \text{last two digits}&=20=4\times5 && \text{passes the }4\text{ test} \end{aligned}

Why this step? Earlier digits contribute multiples of the divisor in each respective test, so they do not affect these decisions.

Answer. All four divisors $2,4,5,10$ divide $8420$ .

Check. $8420/2=4210$ , $8420/4=2105$ , $8420/5=1684$ , $8420/10=842$ , all integers.

Change only the tens digit, from $2$ to $3$ . The last digit stays zero.

Worked example 2: the last digit cannot decide divisibility by four

Question. Which of $2,4,5,10$ divide $8430$ ?

Plan. Retain the three units-digit decisions, then test the changed two-digit ending.

\begin{aligned} \text{last digit}&=0 && 2,5,10\text{ still divide}\\ 30&=4\times7+2 && \text{nonzero remainder modulo }4\\ 8430&\equiv2\pmod4 && \text{earlier hundreds vanish} \end{aligned}

Why this step? An even ending proves divisibility by $2$ , but does not by itself prove divisibility by $4$ .

Answer. $2,5,10$ divide $8430$ , while $4$ does not.

Check. $8430=4(2107)+2$ , confirming the failed test.

A short ending is enough because of the power of ten that vanishes, not because the leading digits are unimportant for every possible divisor.

Watch out

Wrong: “The number is even, so it is divisible by $4$ .” Right: test its last two digits; $30$ is even but leaves remainder $2$ modulo $4$ .

Change the tens digit again.

Replace every power of ten by one for three and nine

Modulo either $3$ or $9$ , $10\equiv1$ , hence $10^k\equiv1$ for every nonnegative integer $k$ . A decimal expansion therefore has the same remainder as the sum of its digits.

For digits $d_k$ , with $d_0$ the units digit,

N=\sum_{k=0}^{s}d_k10^k \equiv\sum_{k=0}^{s}d_k\pmod m, \qquad m=3\text{ or }9

This proves both implications of each test; it is not just a pattern noticed in a few examples.

Tests for three and nine

A non-negative decimal integer is divisible by $3$ exactly when its digit sum is divisible by $3$ , and divisible by $9$ exactly when its digit sum is divisible by $9$ .

Not in the formula booklet: learn it.

The digit sum need not equal the divisor. A sum of $18$ , for example, passes both tests. You may apply the digit-sum rule again to a large sum, because it preserves the same congruence.

Worked example 3: a digit sum divisible by both

Question. Test $5832$ for divisibility by $3$ and $9$ , justifying the digit-sum method.

Plan. Reduce the decimal place values to $1$ , then test the sum.

\begin{aligned} 5832&=5(10^3)+8(10^2)+3(10)+2 && \text{decimal expansion}\\ &\equiv5+8+3+2=18\pmod9 && 10^k\equiv1\pmod9\\ 18&=9\times2=3\times6 && \text{passes both tests} \end{aligned}

Why this step? The same place-value argument works modulo $3$ . Divisibility by $9$ also implies divisibility by $3$ , since $9$ is a multiple of $3$ .

Answer. $5832$ is divisible by both $3$ and $9$ .

Check. $5832=9(648)=3(1944)$ .

Change only the units digit, from $2$ to $5$ .

Worked example 4: three does not guarantee nine

Question. Test $5835$ for divisibility by $3$ and $9$ .

Plan. Use the changed digit sum to distinguish the two tests.

\begin{aligned} 5+8+3+5&=21 && \text{digit sum}\\ 21&=3\times7 && \text{divisible by }3\\ 21&=9\times2+3 && \text{not divisible by }9 \end{aligned}

Why this step? A multiple of $3$ need not be a multiple of $9$ . The digit sum tests the chosen divisor, rather than just whether it has a factor of three.

Answer. $3$ divides $5835$ , but $9$ does not.

Check. $5835=3(1945)=9(648)+3$ .

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FP2 5.5 - Divisibility tests

Replace a number by a smaller congruent expression

Use the last digit or the last two digits

Replace every power of ten by one for three and nine

Continue this lesson.

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