Grade 4 · MathematicsAbout 90 minutes
MathematicsGrade 4About 90 minutes

Use equal groups and place value to divide larger numbers

Suggested multi-session teaching plan.

Ohio learning standards this draft targets

math-2017:4.NBT.6

Read the standard text

math-2017:4.NBT.6 Find whole number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models.

Provisional targets. Clause-by-clause educational review is still pending.

Section 1 · LearnLearn the idea
For 1,735 ÷ 6, partial quotients 200, 80 and 9 use 1,200, 480 and 54 items. They make 289 groups, with 1 item remaining.For 1,735 ÷ 6, partial quotients 200, 80 and 9 use 1,200, 480 and 54 items. They make 289 groups, with 1 item remaining.1,735 ÷ 6 · track groups and itemsGroups madeItems usedItems left2001,20053580480559541200 + 80 + 9 = 289 groups6 × 289 + 1 = 1,735
For 1,735 ÷ 6, partial quotients 200, 80 and 9 use 1,200, 480 and 54 items. They make 289 groups, with 1 item remaining.

Learning goal: Divide whole numbers with up to four-digit dividends by one-digit divisors, finding a quotient and remainder and explaining the process. Division asks how many equal groups fit or how much belongs in each equal share. Multiplication can build or check the relationship throughout the calculation.

For 1,735 ÷ 6, imagine placing 1,735 items into groups of six. Begin with two hundred groups: 6 × 200 = 1,200 items. Subtract 1,200 from 1,735 to leave 535. The number two hundred counts groups, while 1,200 counts items. Keep those units distinct when recording the partial quotient.

Next make eighty groups, using 6 × 80 = 480 items. Subtract 480 from 535 to leave 55. Then make nine groups, using 6 × 9 = 54 items. One item remains. Add the partial group counts: 200 + 80 + 9 = 289. The result is 289 remainder 1.

Check the original quantity: 6 × 289 + 1 = 1,735. The remainder is less than six, so it cannot form another complete group. This check confirms both the total and that the grouping has finished. A result with remainder seven would need another group and a smaller remainder.

Different valid partial quotients can lead to the same result. One learner may take one hundred groups at a time before using smaller groups. Another may recognize two hundred groups immediately. Both methods work if each subtraction matches a multiple of six, the partial quotients are added correctly and the remaining amount is tracked without loss.

Place value also supports equal sharing. To divide 4,368 among seven equal shares, first allocate six hundred to each share, using 4,200. There are 168 left. Allocate twenty to each share, using 140, leaving 28. Allocate four more to each, using the final 28. Each share receives 600 + 20 + 4 = 624, with no remainder.

Zeros in a quotient need attention. For 2,408 ÷ 4, two thousand contributes five hundred to each share, four hundred contributes one hundred, zero tens contributes no tens and eight ones contributes two. The quotient is 602. Writing sixty-two would place the six in the tens position and fail the multiplication check, since 62 × 4 = 248.

Use estimation to choose a sensible starting partial quotient and assess the answer's size. Since 1,800 ÷ 6 = 300, the quotient for 1,735 ÷ 6 should be a little below three hundred. Since 4,200 ÷ 7 = 600, the quotient for 4,368 ÷ 7 should be a little above six hundred. Estimation guides the process while exact multiplication and subtraction determine the result.

Section 2 · ModelWork through the examples

Worked example: Calculate 3,947 ÷ 5. Seven hundred groups use 3,500, leaving 447. Eighty groups use 400, leaving 47. Nine groups use 45, leaving 2. Add 700 + 80 + 9 = 789, so the quotient is 789 remainder 2. Check 5 × 789 + 2 = 3,947.

Worked error analysis: A learner adds the used amounts 1,200 + 480 + 54 and calls 1,734 the quotient for 1,735 ÷ 6. That sum counts items placed into groups, not the groups themselves. Add 200 + 80 + 9 to find the quotient and keep the one leftover item separate.

Section 3 · DoPractice, demonstrate, and explain

Guided practice: Record a table with partial group count, items used and items remaining. Explain every row as a multiplication and subtraction. Try a different first partial quotient and show that it leads to the same final grouping. Use an equation or labeled area representation to connect the parts.

Independent evidence: Solve unfamiliar four-digit-by-one-digit examples with and without remainders, including a zero in the quotient. Show a valid representation and check divisor times quotient plus remainder. Explain which quantities count groups and which count items. The standard division algorithm is not required for this lesson's place-value strategy.

Section 4 · Check6-question lesson practice test

An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.

Problem 1

What do the partial quotients 200, 80 and 9 count in the example?

Show worked solution

Groups of six, not individual items.

Problem 2

Find 2,594 ÷ 4.

Show worked solution

648 remainder 2; 4 × 648 + 2 = 2,594.

Problem 3

Why is a remainder of seven not finished when dividing by six?

Show worked solution

Six of those items can make one more complete group, leaving one.

Problem 4

Find 3,852 ÷ 6.

Show worked solution

642 exactly: 3,600 + 240 + 12 equals six groups of 600 + 40 + 2.

Problem 5

Find 4,535 ÷ 5 and explain the zero.

Show worked solution

907. Each share has nine hundreds, zero tens and seven ones: 4,500 gives 900 to each share and the remaining 35 gives 7.

Problem 6

Find 5,239 ÷ 8 and check.

Show worked solution

654 remainder 7; 8 × 654 + 7 = 5,239.