Introduction
How can you apply partial quotients to problems with remainders?
Goal:
Goal:
Partial quotients are a very helpful division strategy that breaks down a dividend into pieces, using repeated subtraction of equal groups. This strategy works well for all types of division problems. Watch the video to learn more about how to apply the method and strategy of partial quotients to division problems with remainders.
Partial Quotients with Remainders
Our first problem is 342 ÷ 5.
500 is bigger than our dividend, so we are going to stick with 50. Let’s repeatedly subtract 50. 342 – 50 = 292. 292 – 50 = 242. 242 – 50 = 192. 192 – 50 = 142. 142 – 50 = 92. 92 – 50 = 42. 50 does not fit into 42, so now we find out how many times 5 fits into 42. 42 ÷ 5 = 8 remainder 2. Now I add up the partial quotients. 10, 20, 30, 40, 50, 60. 60 + 8 R2 is 68 R2. The quotient of 342 ÷ 5 = 68 R2.
Our next problem is 569 ÷ 4.
4,000 does not fit into our dividend. Let’s repeatedly subtract 400. 569 – 400 is 169. 400 does not fit into 169, so now we repeatedly subtract 40. 169 – 40 = 129. 129 – 40 = 89. 89 – 40 = 49. 49 – 40 = 9. 40 does not fit into 9, so now we find out how many times 4 fits into 9. 9 divided by 4. 4 fits into 9 two times. 2 times 4 is 8. We subtract. 4 fits into 9 two times, with 1 remainder. Now I add my partial quotients. 100, 110, 120, 130, 140. 140 plus 2 remainder 1, is 142 R1. The quotient of 569 ÷ 4 is 142 R1.
Our last example is 463 divided by 3. 3,000 is larger than our dividend, so we are going to start with 300. 463 – 300 = 163. 300 does not fit into 163, so we repeatedly subtract 30. 163 – 30 is 133. 30 does not fit into 13, so now we divide 13 by 3. 3 does not fit into 1. 3 fits into 13 4 times. 4 x 3 is 12. 13 divided by 3 is 4 remainder 1. I add my partial quotients. 100, 110, 120, 130, 140, 150. 150 + 4 R1 is 154 R1. 463 ÷ 3 is 154 R1.
What powers of 10 would you use to solve 2,653 ÷ 14 using partial quotients?