Unit 3, Lesson 18
Using Long Division
Accelerated 6
Practice
Problem 1
Andre and Jada both found \(657 \div 3\) by using partial quotients, but they did the calculations differently, as shown here.
Andre and Jada's calculation of a partial product problem, 657 divided by 3. Andre's calculation. 657 inside the division bar, divided by 3. 3 divides into 657, 200 times. Subtract 600 from 657 for a remainder of 57. 3 divides into 57, 10 times. Subtract 30 from 57 for a remainder of 27. 3 divides into 27, 9 times. Subtract 27 from 27 for a remainder of 0. Add all of Andre's partial quotients of 200, 10 and 9 for a total of 219. Jada's calcluation. 657 inside the division bar, divided by 3. 3 divides into 657, 50 times. Subtract 150 from 657 for a remainder of 507. 3 divides into 507, 100 times. Subtract 300 from 507 for a remainder of 207. 3 divides into 207, 60 times. Subtract 180 from 207 for a remainder of 27. 3 divides into 27, 9 times. Subtract 27 from 27 for a remainder of 0. Add all of Jada's partial quotients of 50, 100, 60 and 9 for a total of 219.
- How is Jada's work the same as Andre’s work? How is it different?
- Explain why they have the same answer.
to access Practice Problem Solutions.