The Forgiveness Method & Partial Quotients Division The Partial Quotients Algorithm uses a series of “at least, but less than” estimates of how many b’s in a. Students might begin with multiples of 10 – they’re easiest. This method builds towards traditional long division. It removes difficulties and errors associated with simple structure mistakes of long division. Based on EM resources

Partial Quotients Division Easy step by step directions to help with long division…..look at the picture below. What game does it remind you of? Answer: HANGMAN!!!

Partial Quotients Division Start by setting up the problem like this. It looks just like the traditional long division method, except for the long line that is drawn to the right of the divisor. (Just like in the Hangman Game.) Discuss benchmark numbers… X 1 X 10 X 100 177 8

Ask - How many [8s] are in 177? There are at least 10, so that will be the first partial quotient.. 80 10 Multiply 10 * 8, write the produce under the dividend in the problem. Then subtract! Write on the side 8 x 1 = 8 8 x 10 = 80 8 x 100 = 800

Subtract 177 minus 80. Now check, is 97 less than your divisor, 8? If yes, then you have finished dividing. If not….. 8 177 - 80 10 97 8 x 1 = 8 8 x 10 = 80 8 x 100 = 800

Start the process over again. Ask - how many [8s] are in 97? Again, there are at least 10. 8 177 - 80 10 97 80 10 8 x 1 = 8 8 x 10 = 80 8 x 100 = 800

Subtract 97 minus 80. 8 177 - 80 10 Now check, is 17 less than your divisor, 8? If yes, then you have finished dividing. If not….. 97 - 80 10 8 x 1 = 8 8 x 10 = 80 8 x 100 = 800 17

Start the process again. Ask - how many [8s] are in 17 Start the process again. Ask - how many [8s] are in 17. There are at least 2. 8 177 - 80 10 97 - 80 10 8 x 1 = 8 8 x 10 = 80 8 x 100 = 800 Subtract 17 minus 16. 17 - 16 2 1

Since the 1 is less than 8, you are finished dividing. Now add up the partial quotients - 10 plus 10 plus 2. 22 R1 8 177 - 80 10 97 - 80 10 8 x 1 = 8 8 x 10 = 80 8 x 100 = 800 17 - 16 2 1 Write the answer above with the remainder. You are finished. 22

Partial Quotients Division Start by setting up the problem like this. It looks just like the traditional long division method, except for the long line that is drawn to the right of the divisor. (Just like in the Hangman Game.) Now, let’s try to same problem using basic multiplication facts! 177 8

Now, let’s try to same problem using basic multiplication facts! 8 177 Ask - How many [8s] are in 17? There are at least 2, so 2 will be the first partial quotient.. Now, let’s try to same problem using basic multiplication facts! 8 177 160 20 Multiply 2 * 8, write the product under the dividend in the problem. Now, you will notice that there is an empty space under the last 7 in the dividend. We will place a “0” to occupy the empty space and add a “0” to the 2 in the partial quotient column. Then subtract!

Start the process over again. Ask - how many [8s] are in 17? Again, there are at least 2. 8 177 - 160 20 17 16 2 8 x 1 = 8 8 x 10 = 80 8 x 100 = 800

Subtract 17 minus 16. 8 177 - 160 20 Now check, is 1 less than your divisor, 8? If yes, then you have finished dividing. If not…..keep going. 17 - 16 2 8 x 1 = 8 8 x 10 = 80 8 x 100 = 800 1

22 R. 1 8 177 - 160 20 1 is less than your divisor, 8, so you are finished dividing. Now, add up the partial quotients, 20 and 2 and write their sum with the remainder at the top of the problem. 17 - 16 2 8 x 1 = 8 8 x 10 = 80 8 x 100 = 800 1 22

Since the 1 is less than 8, you are finished dividing. Now add up the partial quotients - 10 plus 10 plus 2. 22 R1 8 177 - 80 10 97 - 80 10 8 x 1 = 8 8 x 10 = 80 8 x 100 = 800 17 - 16 2 1 Write the answer above with the remainder. You are finished. 22

Let’s try another one….. 843 ÷ 4 Set up the problem

Ask - How many [4s] are in 843? There are at least 100, so that will be the first partial quotient.. Write on the side 4 x 1 = 4 4 x 10 = 40 4 x 100 = 400

843 4 - 400 100 4 x 1 = 4 4 x 10 = 40 4 x 100 = 400 443 100 - 400 43 Start the process over again. Ask - how many [4s] are in 443? There are at least 100 more.

843 4 - 400 100 4 x 1 = 4 4 x 10 = 40 4 x 100 = 400 443 100 - 400 43 10 - 40 3 Start the process over again. Ask - how many [4s] are in 43? There are at least 10 more.

210 r 3 843 4 - 400 100 4 x 1 = 4 4 x 10 = 40 4 x 100 = 400 443 100 - 400 43 10 - 40 3 Since the 3 is less than 4, you are finished. Now add up the partial quotients: 100 +100 + 10 = 210.

First, underline the 8 in the dividend First, underline the 8 in the dividend. Then ask yourself- How many [4s] are in 8? There are at least 2. Now, for the empty spaces under the 43, add a 0 in the empty spaces in both the problem and the partial quotients comlumn. So 200 will be the first partial quotient.. Let’s look at solving the same problem in a different way!!!! 843 4 Write on the side 4 x 1 = 4 4 x 10 = 40 4 x 100 = 400

843 4 - 800 200 4 x 1 = 4 4 x 10 = 40 4 x 100 = 400 43 Now ask yourself, is 43 less than 4? If not, start the process over again. Ask - how many [4s] are in 40? There are at least 10 more.

843 4 -800 200 4 x 1 = 4 4 x 10 = 40 4 x 100 = 400 43 10 - 40 3 Start the process over again. Ask - how many [4s] are in 43? There are at least 10 more. Is 3 less than 4? If so, then you are done dividing.

210 r 3 843 4 - 800 200 4 x 1 = 4 4 x 10 = 40 4 x 100 = 400 43 10 - 40 3 Since the 3 is less than 4, you are finished. Now add up the partial quotients: 200 + 10 = 210.

Hangman Division (Partial Quotient) See, dividing with The Forgiveness Method or Partial Quotients Method is EASY!!!