Adjust Quotients and Fix Division Mistakes | Grade 5 Math | LevelKite

When you divide by a 2-digit number, you build the answer one digit at a time. Look at $152 \div 26$ . The dividend, $152$ , is the amount you start with. The divisor, $26$ , is the size of one group. The quotient is how many whole groups you can make. The remainder is what is left over when no more whole groups fit.

You find each digit of the quotient with an estimate, which is a careful first guess. A guess can be a little off, and that is normal. What matters is that you test each guess and fix it.

In this lesson, you will learn two quick tests for a quotient digit. Then you will use them to find and fix mistakes in finished work.

Each time you choose a quotient digit, you are saying how many groups of the divisor you can take out. Say you try the digit $6$ for $152 \div 26$ . You are saying, “I can take $6$ groups of $26$ out of $152$ .” To take them out, you multiply $6 \times 26$ and then subtract. The answer to a multiplication is called a product.

A guess can miss in two ways:

Too many groups. The product is more than the amount you have, so you cannot take it away.
Too few groups. You can subtract, but enough is left to make at least one more group.

The table shows three tries for $152 \div 26$ . Watch what happens each time.

Three tries for 152 ÷ 26

Digit you try	Multiply by 26	Try to subtract from 152	What it tells you
$6$	$6 \times 26 = 156$	You cannot. $156$ is greater than $152$ .	$6$ is too large.
$4$	$4 \times 26 = 104$	$152 - 104 = 48$ . But $48$ is greater than $26$ , so one more group fits.	$4$ is too small.
$5$	$5 \times 26 = 130$	$152 - 130 = 22$ , and $22$ is less than $26$ .	$5$ is just right.

Only the digit $5$ works. You can subtract its product, and what is left is less than $26$ . So $152 \div 26 = 5$ R $22$ . The letter R is short for remainder.

Here is the same answer in words and in numbers. Five groups of $26$ make $130$ , and $22$ more makes $152$ . That is $5 \times 26 + 22 = 152$ .

When a digit does not work, you adjust it. To adjust a digit means to change it and then test again.

Key concept

Two tests for every quotient digit

Multiply the digit you try by the divisor. Then run two tests.

Is it too large? If the product is greater than the number you are dividing at that step, you cannot subtract it. Try the digit that is $1$ less.
Is it too small? If you can subtract, look at what is left. This is the remainder for that step. If it is equal to or greater than the divisor, at least one more group fits. Try the digit that is $1$ more.

Test again after each change. Sometimes you must change the digit more than once. A digit is right only when you can subtract its product and what is left is less than the divisor.

Worked example

Adjust two estimates in one division

Divide $1{,}063 \div 27$ . Test each estimated digit, and adjust it if it does not work.

Find the first number to divide. The first two digits make 10, and 10 is less than 27. So use three digits: 106 tens. The first quotient digit will count tens. Now estimate. Think of 27 as 30 and count by 30s: 90, 120. The number 106 is nearer to 120, and 120 divided by 30 is 4. Try 4.
$120 \div 30 = 4$
Test 4. Multiply 4 by 27 to get the product, 108. You have only 106, so you cannot take 108 away. Taking out 4 groups of 27 is too many, so the digit 4 is too large. Try the digit that is 1 less, which is 3.
$4 \times 27 = 108 \quad\text{and}\quad 108 > 106$
Test 3. The product is 81. You can subtract 81 from 106, and 25 is left. Since 25 is less than 27, no more groups of 27 fit. The digit 3 passes both tests. Write 3 in the tens place of the quotient.
$3 \times 27 = 81 \quad\text{and}\quad 106 - 81 = 25$
Bring down the 3 ones. The 25 tens and 3 ones make 253 ones. Estimate again by counting by 30s: 240, 270. The number 253 is nearer to 240, and 240 divided by 30 is 8. Try 8. The product is 216, and 37 is left. But 37 is greater than 27, so one more group of 27 fits. The digit 8 is too small. Try the digit that is 1 more, which is 9.
$8 \times 27 = 216 \quad\text{and}\quad 253 - 216 = 37$
Test 9. The product is 243. You can subtract 243 from 253, and 10 is left. Since 10 is less than 27, the digit 9 passes both tests. Write 9 in the ones place. The quotient is 39, and the remainder is 10.
$9 \times 27 = 243 \quad\text{and}\quad 253 - 243 = 10$

Answer

$1{,}063 \div 27 = 39$ R $10$ . The estimate $4$ was too large, so it went down to $3$ . The estimate $8$ was too small, so it went up to $9$ . As a check, $39$ groups of $27$ make $1{,}053$ , and $10$ more makes $1{,}063$ .

Now be a mistake detective. You will look at finished work, find what went wrong, and fix it.

A good first move is to multiply back. Multiply the quotient by the divisor, and then add the remainder. A right answer always gives back the dividend. If you get a different number, there is a mistake somewhere.

Be careful, though. Getting the dividend back is not enough by itself. The remainder must also be less than the divisor. You will see why near the end of this lesson.

Next, look at where the digits sit. Each quotient digit sits in a place, such as the tens place or the ones place, just like each digit in the dividend. In the last example, you divided $106$ tens first. So the digit $3$ counted tens, and it went in the tens place.

How do you know where to start? When the dividend is greater than the divisor, as in every problem here, take just enough digits from the front of the dividend to make a number that is equal to or greater than the divisor. That is the first number you divide.

The table lists three mistakes and the questions that catch them.

Three mistakes to look for

Mistake	What right work looks like	Question to ask
First digit in the wrong place	The first quotient digit sits above the last digit of the first number you divide.	Which digits did I divide first? Is my first digit above the last of them?
Missing $0$ in the quotient	Each digit you bring down gets one quotient digit above it. If the new number is less than the divisor, write $0$ .	Did I bring down two digits in a row without writing a quotient digit?
Remainder that is too big	The remainder is less than the divisor.	Is my remainder equal to or greater than the divisor?

Worked example

Find a first digit in the wrong place

Mia divides $1{,}984 \div 32$ . She divides $198$ by $32$ and gets $6$ , but she writes the $6$ above the $9$ , in the hundreds place. She writes her next digit, $2$ , in the tens place. Then she puts a $0$ in the ones place. Her answer is $620$ . Find her mistake and fix it.

Multiply back to see if 620 can be right. The product of 620 and 32 is 19,840. The dividend is 1,984, so Mia's answer is much too big. There is a mistake.
$620 \times 32 = 19{,}840$
Find the first number Mia divided. The numbers 1 and 19 are both less than 32, so she used 198. The 198 ends at the 8, and the 8 is in the tens place. So 198 means 198 tens, and the 6 counts tens. It belongs above the 8, not above the 9. This is her mistake.
$6 \text{ tens} \times 32 = 192 \text{ tens}$
Redo the work with 6 in the tens place. Subtract 192 tens from 198 tens, and 6 tens are left. Since 6 is less than 32, the digit 6 passes both tests. Bring down the 4 ones. The 6 tens and 4 ones make 64 ones.
$198 - 192 = 6$
Divide the 64 ones. Two groups of 32 make 64, and nothing is left. Write 2 above the 4, in the ones place. Now each digit from the 8 on has one quotient digit above it, so there is no place for an extra 0.
$2 \times 32 = 64 \quad\text{and}\quad 64 - 64 = 0$
Multiply back again. This time you get the dividend. The remainder is 0, which is less than 32, so both checks pass.
$62 \times 32 = 1{,}984$

Answer

Mia put her first digit in the hundreds place, but it belongs in the tens place. The correct answer is $1{,}984 \div 32 = 62$ .

Worked example

Find a missing 0 in the quotient

Leo divides $2{,}461 \div 24$ and writes $12$ R $13$ . Here is what he did. He divided $24$ hundreds by $24$ and wrote $1$ . He brought down the $6$ and saw that $6$ is less than $24$ . So he brought down the $1$ right away to make $61$ . He divided $61$ by $24$ , wrote $2$ , and had $13$ left. Find his mistake and fix it.

Multiply back. The product of 12 and 24 is 288, and adding the remainder 13 gives 301. The dividend is 2,461, so there is a mistake.
$12 \times 24 + 13 = 288 + 13 = 301$
Check the first digit. The front digits make 24, which is equal to the divisor. So Leo was right to divide 24 hundreds first. His 1 counts hundreds and sits above the 4, in the hundreds place. Nothing is left.
$1 \times 24 = 24 \quad\text{and}\quad 24 - 24 = 0$
Look at the next digit. Leo brought down the 6 tens. Since 6 is less than 24, no group of 24 fits, so the quotient digit for the tens place is 0. Leo wrote nothing there and brought down another digit. This is his mistake. Write 0 above the 6. The digit 0 passes both tests: you can subtract its product, 0, and the 6 that is left is less than 24.
$0 \times 24 = 0 \quad\text{and}\quad 6 - 0 = 6$
Bring down the 1 to make 61 ones. Leo's last step is right. Two groups of 24 make 48, and 13 is left. Since 13 is less than 24, the digit 2 passes both tests. It goes above the 1, in the ones place.
$2 \times 24 = 48 \quad\text{and}\quad 61 - 48 = 13$
Read the quotient with the 0 in place: 1 hundred, 0 tens, 2 ones. That is 102. Multiply back, and you get the dividend. The remainder 13 is less than 24, so both checks pass.
$102 \times 24 + 13 = 2{,}448 + 13 = 2{,}461$

Answer

Leo left out the $0$ in the tens place of the quotient. The correct answer is $2{,}461 \div 24 = 102$ R $13$ .

Here is the big idea. Every quotient digit is a guess that you can test. If you cannot subtract its product, the digit is too large, so go down. If what is left is equal to or greater than the divisor, the digit is too small, so go up.

Use the same thinking on finished work. Multiply back. Check that the first digit is in the right place and that no $0$ is left out. Then make sure the remainder is less than the divisor. A guess that needs fixing is not a failure. It is part of dividing well.