Divisibility Rules and Multi-Digit Multiplication | Grade 6 Math | LevelKite

In this lesson you will test when a division leaves nothing over, and you will multiply large whole numbers. First, the names.

When you divide, the number you split up is the dividend, the number you divide by is the divisor, and the result is the quotient. Every divisor here is nonzero, meaning it is not 0. You cannot divide by zero: no number times $0$ gives $24$ .

Share 24 marbles equally among 6 friends. Each friend gets 4 marbles and none are left, so $24 \div 6 = 4$ . A whole number (0, 1, 2, 3 and so on) is divisible by a nonzero whole number when the quotient is a whole number and nothing is left over. So 24 is divisible by 6.

Now share 14 marbles among 4 friends. You cannot cut a marble. Each friend gets 3 marbles and 2 are left over. That leftover amount is the remainder.

Share 14 feet of ribbon among 4 friends instead. You can cut ribbon, so each friend gets $3\tfrac{1}{2}$ feet and nothing is left. That is a fractional share, not a remainder. Either way, 14 is not divisible by 4: you never get a whole-number quotient with nothing left over.

Key concept

Divisible means an exact whole-number quotient

Call the dividend $a$ , the divisor $d$ and the quotient $q$ . The divisor $d$ is never zero. A whole number $a$ is divisible by a whole number $d$ when $q \times d = a$ for some whole number $q$ , with no remainder. Place value, the value a digit gets from its position, makes this relationship quick to use in both directions. Divisibility rules read the digits of $a$ to tell you whether a whole-number $q$ exists. Multi-digit multiplication splits a number into place-value parts to work out $q \times d$ .

q \times d = a \quad (d \neq 0)

Multiplying by the divisor undoes dividing by it. $24 \div 6 = 4$ says 24 marbles make 6 equal shares of 4. $4 \times 6 = 24$ puts those shares back together. So you can check a division by multiplying. The right check depends on the kind of quotient, and the divisor is never zero.

Exact quotient. Quotient times divisor equals the dividend: $4 \times 6 = 24$ and $3\tfrac{1}{2} \times 4 = 14$ .
Whole quotient with leftovers (things you cannot cut). Divisor times whole quotient counts what you handed out. Add the remainder to get back the dividend: $4 \times 3 + 2 = 14$ .
Rounded quotient. Multiplying back gives a number that is only close to the dividend, so the check is approximate. $20 \div 3$ is about $7$ , and $7 \times 3 = 21$ , which is near $20$ but not equal to it.

Divisible means the first case with a whole-number quotient: no remainder to add and no fraction in the share.

A divisibility rule is a shortcut. It tells you whether a whole number is divisible by a certain nonzero divisor before you divide. The rules come from place value.

For 2, 5 and 10. Tens, hundreds and every larger place split into twos, fives or tens with nothing left. Only the ones digit can leave something over.
For 4. Every hundred is $25 \times 4$ , and larger places are made of hundreds. Only the last two digits, the tens and ones together, can leave something over.
For 3 and 9. A ten is $9 + 1$ , a hundred is $99 + 1$ and a thousand is $999 + 1$ . The 9s, 99s and 999s split into threes or nines with nothing left, so each ten, hundred or thousand leaves 1 extra. For 318 that leaves $3 + 1 + 8 = 12$ ones to split: 3 from the hundreds, 1 from the ten, and the 8 ones. That is the sum of the digits. Since $12 = 4 \times 3$ , 318 is divisible by 3. Since $12 = 9 \times 1 + 3$ , 318 is not divisible by 9.
For 6. Six is $2 \times 3$ , so every group of 6 is 3 twos and also 2 threes. A number divisible by 6 passes the test for 2 and the test for 3, and passing both is enough for 6.

Divisibility rules for whole numbers. Every divisor is nonzero.

Divisor	A whole number is divisible by it exactly when	Divisible: exact check	Not divisible: remainder check
$2$	its ones digit is $0$ , $2$ , $4$ , $6$ or $8$	$318$ : $159 \times 2 = 318$	$735$ : $2 \times 367 + 1 = 735$
$3$	the sum of its digits is divisible by $3$	$318$ : $3 + 1 + 8 = 12$ ; $106 \times 3 = 318$	$736$ : $7 + 3 + 6 = 16$ ; $3 \times 245 + 1 = 736$
$4$	the number made by its last two digits is divisible by $4$	$1{,}524$ : $24 = 6 \times 4$ ; $381 \times 4 = 1{,}524$	$318$ : $18 = 4 \times 4 + 2$ ; $4 \times 79 + 2 = 318$
$5$	its ones digit is $0$ or $5$	$735$ : $147 \times 5 = 735$	$318$ : $5 \times 63 + 3 = 318$
$6$	it is divisible by both $2$ and $3$	$318$ : passes the tests for $2$ and $3$ ; $53 \times 6 = 318$	$735$ : fails the test for $2$ ; $6 \times 122 + 3 = 735$
$9$	the sum of its digits is divisible by $9$	$738$ : $7 + 3 + 8 = 18$ ; $82 \times 9 = 738$	$318$ : $3 + 1 + 8 = 12$ ; $9 \times 35 + 3 = 318$
$10$	its ones digit is $0$	$730$ : $73 \times 10 = 730$	$735$ : $10 \times 73 + 5 = 735$

Worked example

Test one number with the rules

Is $4{,}716$ divisible by $2$ , $3$ , $4$ , $5$ , $6$ , $9$ or $10$ ? Test each divisor.

Start with the ones digit, because tens, hundreds and thousands always split into twos, fives and tens with nothing left. The ones digit is 6. It is on the list 0, 2, 4, 6, 8, so 4,716 is divisible by 2. It is not 0 or 5, so 4,716 is not divisible by 5 or by 10.
Add the digits, because each ten, hundred and thousand leaves 1 extra when you make groups of 3 or groups of 9. The sum is 18. Since 18 is 6 threes and also 2 nines, 4,716 is divisible by 3 and by 9.
$4 + 7 + 1 + 6 = 18 = 6 \times 3 = 2 \times 9$
Read the last two digits, because every hundred is 25 fours and a thousand is 10 hundreds. The last two digits make 16, and 16 is 4 fours, so 4,716 is divisible by 4.
$16 = 4 \times 4$
Use the rule for 6. The number passed the test for 2 and the test for 3, and passing both is enough for 6. So 4,716 is divisible by 6.
Check one yes answer by multiplying. Dividing 4,716 by 9 gives 524. The divisor 9 is not zero and the quotient 524 is exact, so quotient times divisor must equal the dividend.
$524 \times 9 = 4{,}500 + 180 + 36 = 4{,}716$
Check one no answer with a remainder. Dividing 4,716 by 5 gives the whole quotient 943 with 1 left over. Divisor times whole quotient, plus remainder, must equal the dividend.
$5 \times 943 + 1 = 4{,}715 + 1 = 4{,}716$

Answer

$4{,}716$ is divisible by $2$ , $3$ , $4$ , $6$ and $9$ . It is not divisible by $5$ or $10$ .

Now use place value to multiply. The numbers you multiply are factors, and the result is the product.

$126 \times 34$ means 34 groups of 126. You can count 34 groups as 30 groups plus 4 groups, so $126 \times 34 = 126 \times 30 + 126 \times 4$ . Both sides count the same 34 groups, so they are equal. This is the distributive property: multiplying a number by a sum gives the same result as multiplying it by each part and adding.

Each smaller product, such as $126 \times 4$ , is a partial product. For any whole-number factors, you can split one factor into its place-value parts, multiply by each part, and add the partial products.

To multiply by 30, multiply by 3 and then by 10, because $30 = 3 \times 10$ . Multiplying a whole number by 10 moves each digit one place to the left, and a 0 fills the ones place.

Worked example

Multiply by a two-digit number

Find $126 \times 34$ .

Split the factor 34 into its place-value parts, 30 and 4. Thirty-four groups of 126 are 30 groups plus 4 groups, so the two expressions have the same value.
$126 \times 34 = 126 \times 30 + 126 \times 4$
Multiply 126 by the 4 ones. Use 126 as 100 + 20 + 6 and multiply each part by 4, then add.
$126 \times 4 = 400 + 80 + 24 = 504$
Multiply 126 by the 3 tens. Multiply by 3 first, then by 10, because 30 is 3 times 10. Multiplying by 10 moves each digit one place left and puts 0 in the ones place.
$126 \times 3 = 300 + 60 + 18 = 378 \quad\text{then}\quad 378 \times 10 = 3{,}780$
Add the two partial products. Together they count all 34 groups of 126.
$504 + 3{,}780 = 4{,}284$
Check with a divisibility rule. The digits of 126 add to 9, so 126 is 14 nines. Then 34 groups of 126 make a whole number of nines, so the product must pass the test for the nonzero divisor 9. A pass does not prove the product is right, but a fail would prove a mistake.
$4 + 2 + 8 + 4 = 18 = 2 \times 9$

Answer

The product of $126$ and $34$ is $4{,}284$ .

On paper, you can build each partial product one digit at a time, starting from the ones. For $126 \times 4$ : 4 times 6 ones is 24 ones, which is 2 tens and 4 ones, so write 4 and carry the 2 tens. Carrying like this is regrouping. Next, 4 times 2 tens is 8 tens, and the 2 carried tens make 10 tens, so write 0 and carry 1 hundred. Last, 4 times 1 hundred is 4 hundreds, and the 1 carried hundred makes 5 hundreds. The row is 504, the same as $400 + 80 + 24$ .

Stacking the work keeps the places lined up. Below, the first row under the line is $126 \times 4$ . The second row is $126 \times 30$ , so it ends in 0. The bottom row is the sum of the two partial products.

\begin{array}{r} 126 \\ \times\ 34 \\ \hline 504 \\ 3{,}780 \\ \hline 4{,}284 \end{array}

Worked example

Multiply two three-digit numbers

Find $248 \times 315$ .

Split 315 into its place-value parts: 300, 10 and 5. Counting 315 groups of 248 as 300 groups, 10 groups and 5 groups keeps the total the same.
$248 \times 315 = 248 \times 300 + 248 \times 10 + 248 \times 5$
Multiply 248 by the 5 ones. Use 248 as 200 + 40 + 8 and multiply each part by 5.
$248 \times 5 = 1{,}000 + 200 + 40 = 1{,}240$
Multiply 248 by the 1 ten. Each digit moves one place left and 0 fills the ones place.
$248 \times 10 = 2{,}480$
Multiply 248 by the 3 hundreds. Multiply by 3, then by 100, because 300 is 3 times 100. Each digit moves two places left.
$248 \times 3 = 600 + 120 + 24 = 744 \quad\text{then}\quad 744 \times 100 = 74{,}400$
Add the three partial products. Together they count all 315 groups of 248.
$1{,}240 + 2{,}480 = 3{,}720 \quad\text{then}\quad 3{,}720 + 74{,}400 = 78{,}120$
Check with divisibility rules. 315 ends in 5 and its digits add to 9, so 315 is 63 fives and also 35 nines. Read the product as 248 groups of 315: it is a whole number of fives and of nines, so it must pass the tests for the nonzero divisors 5 and 9. A fail would prove a mistake. A pass does not prove the product is right.
$78{,}120 \text{ ends in } 0 \quad\text{and}\quad 7 + 8 + 1 + 2 + 0 = 18 = 2 \times 9$

Answer

The product of $248$ and $315$ is $78{,}120$ .

Takeaway. Multiplying and dividing describe one relationship. With a nonzero divisor, an exact quotient times the divisor equals the dividend. A whole number is divisible when that quotient is a whole number with no remainder, and the digit rules for 2, 3, 4, 5, 6, 9 and 10 test this for you. To multiply multi-digit whole numbers, split one factor by place value, find each partial product, and add.

Match the check to the quotient. An exact quotient, whole or fractional, checks exactly, but only a whole-number quotient means divisible. Whole-number sharing with leftovers checks as divisor times whole quotient plus remainder. A rounded quotient gives only an approximate check.

Next, you will divide whole numbers by two- and three-digit divisors and estimate products and quotients.