Grade 6

Divisibility, factors and multiples

When one integer divides another exactly, and the factors and multiples this relationship produces.

IntuitionWhat does it mean to divide evenly?

Splitting $12$ candies into $3$ equal bags leaves nothing over — $3$ divides $12$ . Splitting $13$ candies into $3$ bags always leaves a remainder. Divisibility is exactly this "leaves nothing over" relationship, and it is the seed from which factors, multiples, and prime numbers all grow.

Hasse divisibility diagram

D_{12}

of the divisors of

12

(

1, 2, 3, 4, 6, 12

): each upward edge multiplies by a prime factor (

2

3

SchoolDefinitions and the division algorithm

Definition: Divisibility

An integer $b$ divides an integer $a$ , written $b \mid a$ , if there exists an integer $k$ with $a=bk$ . When $b \mid a$ , we call $b$ a factor (or divisor) of $a$ , and $a$ a multiple of $b$ .

a = bq + r, \qquad 0 \le r < b \quad (b>0)

This is the division algorithm: every integer $a$ divided by a positive integer $b$ leaves a unique quotient $q$ and remainder $r$ with $a = bq + r,\ 0\le r<b$ . Divisibility is the special case $r=0$ .

n = \sum_{i=0}^{k} d_i\,10^i

Writing a number in ordinary digits $d_k d_{k-1}\cdots d_1 d_0$ means exactly $n = \sum_{i=0}^{k} d_i\,10^i$ — this place-value expansion is the tool behind every digit-based divisibility rule below.

Divisibility rules by last digits or digit sums
Divisor	Rule	Example
$2$	Last digit is even	$1{,}234$
$5$	Last digit is $0$ or $5$	$1{,}235$
$3$	Digit sum $\sum_i d_i$ divisible by $3$	$3+3+3=9$
$9$	Digit sum $\sum_i d_i$ divisible by $9$	$4+5+3+6=18$
$11$	Alternating sum $\sum_i (-1)^i d_i$ divisible by $11$	$2-9+1-4=-10$

UndergraduateTwo digit-based divisibility theorems

Theorem: Digit-sum test for divisibility by 9 (and 3)

Let $n$ have digits so that $n = \sum_{i=0}^{k} d_i\,10^i$ . Then $9 \mid n$ if and only if $9$ divides the digit sum $\sum_i d_i$ ; the same statement holds with $3$ in place of $9$ .

Why is it true?

Every power of $10$ leaves remainder $1$ when divided by $9$ (since $10=9+1$ ), so shifting a digit to a higher place value never changes its contribution modulo $9$ — the whole number is congruent to the plain sum of its digits.

Proof

Step 1. Show by induction that $10^i \equiv 1 \pmod 9$ for every $i\ge0$ . Base case $i=0$ : $10^0=1\equiv1\pmod9$ . Inductive step: if $10^i\equiv1\pmod9$ , then $10^{i+1}=10\cdot10^i\equiv10\cdot1=10\equiv1\pmod9$ (using $10 \equiv 1 \pmod 9$ ). So $10^i\equiv1\pmod9$ for all $i$ .

Step 2. Substitute into the place-value expansion: $n=\sum_i d_i 10^i \equiv \sum_i d_i\cdot1 = \sum_i d_i \pmod9$ . So $n$ and its digit sum always leave the same remainder upon division by $9$ .

Step 3. Conclude: $9\mid n$ exactly when $n\equiv0\pmod9$ , which by Step 2 happens exactly when $\sum_i d_i\equiv0\pmod9$ , i.e. exactly when $9$ divides the digit sum. Since $9=3\times3$ and the same congruence $10\equiv1\pmod3$ also holds, the identical argument with $3$ in place of $9$ throughout proves the divisibility-by- $3$ version.

Theorem: Alternating digit-sum test for divisibility by 11

Let $n$ have digits so that $n = \sum_{i=0}^{k} d_i\,10^i$ . Then $11 \mid n$ if and only if $11$ divides the alternating digit sum $\sum_i (-1)^i d_i$ .

Why is it true?

Unlike $9$ , powers of $10$ do not stay congruent to $1$ modulo $11$ — instead they flip sign at every step, because $10$ itself is congruent to $-1$ modulo $11$ . Digits at even positions contribute normally, odd positions contribute negatively.

Proof

Step 1. Show by induction that $10^i \equiv (-1)^i \pmod{11}$ for every $i\ge0$ . Base case $i=0$ : $10^0=1=(-1)^0$ . Inductive step: if $10^i\equiv(-1)^i\pmod{11}$ , then $10^{i+1}=10\cdot10^i\equiv(-1)\cdot(-1)^i=(-1)^{i+1}\pmod{11}$ (using $10 \equiv -1 \pmod{11}$ ). So $10^i\equiv(-1)^i\pmod{11}$ for all $i$ .

Step 2. Substitute into the place-value expansion: $n=\sum_i d_i10^i \equiv \sum_i d_i(-1)^i \pmod{11}$ , which is precisely the alternating digit sum $\sum_i (-1)^i d_i$ .

Step 3. Conclude: $11\mid n$ exactly when $n\equiv0\pmod{11}$ , which by Step 2 happens exactly when $11$ divides $\sum_i (-1)^i d_i$ .

UndergraduateReal-World Applications and Worked Examples

Divisibility checks quietly run in the background of everyday systems: check digits catch typing errors in ISBNs and barcodes, and the calendar's leap-year rule keeps our seasons aligned with the astronomical year.

Example: ISBN-10 check digit

A $10$ -digit ISBN $d_1d_2\cdots d_{10}$ is valid exactly when $\sum_{i=1}^{10} i\cdot d_i \equiv 0 \pmod{11}$ (with $d_{10}=X$ standing for $10$ if needed). Verify that the ISBN $0\,306\,40615\,2$ (digits $0,3,0,6,4,0,6,1,5,2$ ) is valid.

Solution

Step 1. List positions $i=1,\ldots,10$ with their digits: $(1,0),(2,3),(3,0),(4,6),(5,4),(6,0),(7,6),(8,1),(9,5),(10,2)$ .

Step 2. Multiply each digit by its position and sum: $1(0)+2(3)+3(0)+4(6)+5(4)+6(0)+7(6)+8(1)+9(5)+10(2) = 0+6+0+24+20+0+42+8+45+20 = 165$ .

Step 3. Check divisibility by $11$ : $165 = 11\times15$ , so $165\equiv0\pmod{11}$ . The check-digit formula holds, so this ISBN is valid — and if a librarian mistyped a single digit while cataloguing it, the weighted sum would almost certainly no longer be a multiple of $11$ , immediately flagging the error.

Example: The leap year rule

A year $Y$ of the Gregorian calendar is a leap year exactly when $(4\mid Y \text{ and } 100\nmid Y)\ \text{or}\ 400\mid Y$ . Using this rule, determine whether $1900$ , $2000$ , and $2024$ are leap years.

Solution

Step 1. Check $1900$ : $4\mid1900$ (since $1900=4\times475$ ), but $100\mid1900$ (since $1900=100\times19$ ) — the exception applies, and $400\nmid1900$ (since $1900/400=4.75$ , not an integer), so the exception is not overridden. $1900$ is not a leap year.

Step 2. Check $2000$ : $4\mid2000$ and $100\mid2000$ (so the exception would apply), but $400\mid2000$ (since $2000=400\times5$ ), which overrides the exception. $2000$ is a leap year.

Step 3. Check $2024$ : $4\mid2024$ (since $2024=4\times506$ ) and $100\nmid2024$ , so the century exception does not even apply. $2024$ is a leap year. Astronomically, this rule (added by Pope Gregory XIII in $1582$ ) exists because Earth's orbit takes about $365.2425$ days, not exactly $365.25$ ; skipping $3$ leap days every $400$ years keeps the calendar close to the true solar year.

Is $1{,}234{,}567$ divisible by $9$ ?

An ISBN-10's weighted digit sum $\sum_{i=1}^{10} i\cdot d_i \equiv 0 \pmod{11}$ comes out to $187$ . Is the ISBN valid?

Which is the correct alternating-sum test for divisibility by $11$ ?

$84$ is divisible by both $4$ and $6$ . Is $84$ necessarily divisible by $24$ ?

References

David M. Burton (2010). Elementary Number Theory
John H. Conway, Richard K. Guy (1996). The Book of Numbers · DOI:10.1007/978-1-4612-4072-3