Unit 7 · Place Value and Multi-Digit Arithmetic

Picture base-ten blocks on a desk. A tiny cube is 1. Ten cubes make a rod, worth 10. Ten rods make a flat, worth 100. Ten flats make a big cube, worth 1,000. Each block is ten of the block before it. Our number system works the same way. A digit is one of the symbols 0 through 9. The place value of a digit is how much it is worth because of where it sits.

Look at the digit 4 in two numbers. In 40, the 4 sits in the tens place, so it is worth 4 tens, or 40. In 400, the 4 sits in the hundreds place, so it is worth 400. Now compare: 400 ÷ 40 = 10. Moving a digit one place to the left makes it worth ten times as much. Moving it one place to the right divides its value by 10. So 3,000 ÷ 10 = 300, and 300 ÷ 10 = 30.

A common mistake is to say 500 is "five more" than 50. It is not. It is ten times as much, because 50 × 10 = 500. Two places make a bigger jump. The 7 in 7,000 is worth 7,000. The 7 in 70 is worth 70. Since 7,000 ÷ 70 = 100, two places to the left means 100 times as much. Count the places between the digits, and you know how many tens to multiply.

The 2020 Census counted 2,746,388 people in Chicago. That number has seven digits. Nobody reads seven digits one at a time. Instead, we split the number into groups of three, called periods. The commas show where the periods are. From the right, the periods are ones, thousands, and millions. Each period has its own ones, tens, and hundreds place inside it.

To read 2,746,388, read each period as a small number, then say the period name. The millions period is 2, so say "two million." The thousands period is 746, so say "seven hundred forty-six thousand." The ones period is 388, so say "three hundred eighty-eight." Put it together: two million, seven hundred forty-six thousand, three hundred eighty-eight. The word "ones" is never said.

Zeros cause the most trouble. Take 4,005,060. A student might read it as "four million, five thousand, six." That is wrong. The thousands period is 005, which is 5. The ones period is 060, which is 60. So the number is four million, five thousand, sixty. Every zero holds a place. When you write a number from words, fill any empty places with zeros so each period keeps three digits.

A librarian counted 4,306 books in the children’s room. The number 4,306 is written in standard form, which just means the usual way with digits. The same number in expanded form shows what each digit is worth: 4,000 + 300 + 6. The 0 in the tens place is worth nothing, so it does not appear as an addend. In word form the number is four thousand, three hundred six.

Larger numbers work the same way, place by place. Take 2,050,317. The 2 is worth 2,000,000. The 0 in the hundred thousands place is worth nothing. The 5 is in the ten thousands place, worth 50,000. The next 0 is worth nothing. Then 300, 10, and 7. So the expanded form is 2,000,000 + 50,000 + 300 + 10 + 7. Add those five values back together and you get 2,050,317 again.

A common mistake is writing the digits instead of their values, like 4 + 3 + 6 for 4,306. That sum is 13, not 4,306. Another mistake is putting the wrong number of zeros on a value, writing 5,000 when the 5 sits in the ten thousands place. To avoid it, say the place name out loud before you write: "five ten thousands, that is 50,000."

Two high schools reported their attendance for the year. One had 47,215 student-days and the other had 47,198. Which is greater? Both numbers have five digits, so start at the left. The ten thousands digits are both 4. The thousands digits are both 7. The hundreds digits are 2 and 1. Since 2 is greater than 1, 47,215 is greater. We write 47,215 > 47,198. The symbol > means greater than, and < means less than.

When the numbers have different numbers of digits, the longer whole number is greater. Compare 100,000 and 99,999. It might look like 99,999 wins because it is full of nines. But 100,000 has six digits and 99,999 has five. Any six-digit number is greater than any five-digit number. So 100,000 > 99,999, and 99,999 < 100,000.

A common mistake is to compare the first digits without lining up the places. A student sees 9,999 and 10,000 and thinks 9 beats 1. That only works when both numbers have the same number of digits. Always count digits first. If they match, compare from the left, one place at a time. Stop at the first place where the digits are different. That place decides the answer.

Four towns in Illinois reported these populations: 24,318; 24,183; 23,981; 24,320. To order them from least to greatest, first look at the number of digits. All four have five digits. Then compare the ten thousands. All are 2. Compare the thousands. Three towns have 4 and one has 3, so 23,981 is the smallest. It goes first.

Now compare the other three: 24,318; 24,183; 24,320. The hundreds digits are 3, 1, and 3. The 1 is smallest, so 24,183 comes next. That leaves 24,318 and 24,320. The hundreds match. The tens digits are 1 and 2, so 24,318 is less. The full order from least to greatest is 23,981; 24,183; 24,318; 24,320. From greatest to least, it is just the reverse.

Two mistakes show up often. One is mixing up which direction you were asked for. Read the question again before you write the list. The other is stopping too early, like deciding 24,318 and 24,320 are equal because they start the same way. Keep comparing until you find a place that differs. Writing the numbers in a column, ones under ones, makes every place easy to see.

A concert drew 46,381 people. A news report does not need every digit. It can round the number to the nearest thousand. First, find the thousands digit: 6. Next, look at the digit just to its right, the hundreds digit: 3. Since 3 is less than 5, the 6 stays a 6. Every digit to the right becomes 0. So 46,381 rounds to 46,000. The word nearest means the closest number that ends in the zeros you want.

Now round 46,381 to the nearest hundred. The hundreds digit is 3. The digit to its right is 8. Since 8 is 5 or more, the 3 goes up to 4, and the tens and ones become zeros: 46,400. To the nearest ten thousand, the ten thousands digit is 4 and the digit to its right is 6. Round up: 50,000. The same number can round to 46,400, 46,000, or 50,000, depending on the place you choose.

One common mistake is changing the wrong digit. If you are rounding to the hundreds, only the hundreds digit can change. Another mistake is forgetting to round up: writing 46,300 instead of 46,400. A third is keeping the lower digits, writing 46,381 with a 4 in the hundreds. After you round, every digit to the right of your place must be a zero. Check that before you move on.

You are at the store with $150 and want three tickets that cost $48 each. You do not need a pencil. Round $48 to $50, and 3 × $50 = $150. That is an estimate, a quick answer that is close but not exact. It tells you that you have just about enough. The exact cost is $144, so you have $6 left. The estimate got you an answer in seconds.

Rounding also checks bigger sums. Suppose you add 38,214 + 41,895 and get 80,109. Is that reasonable? Round each number to the nearest thousand: 38,000 + 42,000 = 80,000. The estimate is close to 80,109, so the answer makes sense. If you had gotten 70,109, the estimate would warn you that something went wrong.

Sometimes only an exact number will do. A cashier counting a register at closing cannot round. A nurse measuring medicine cannot round. A bank cannot round your balance. The rule is simple. Round when you need a quick idea or a check. Use exact numbers when someone will count every unit. A common mistake is rounding both numbers in the same direction so the estimate drifts far from the real answer.

A museum had 27,458 visitors in March and 15,796 in April. To find the total, write the numbers in a column with the ones under ones. Start at the right. Ones: 8 + 6 = 14. Write 4 and regroup the 1 ten into the tens column. Tens: 5 + 9 + 1 = 15. Write 5 and regroup 1 hundred. Hundreds: 4 + 7 + 1 = 12. Write 2 and regroup 1 thousand.

Keep going. Thousands: 7 + 5 + 1 = 13. Write 3 and regroup 1 ten thousand. Ten thousands: 2 + 1 + 1 = 4. Write 4. The sum is 43,254. Check with an estimate: 27,000 + 16,000 = 43,000, which is close. Each regroup works because ten of any place equals one of the next place. Fourteen ones is one ten and four ones, so the 1 belongs in the tens column.

The most common mistake is dropping a regrouped 1. Without the regroups, the column sums 14, 14, 11, 12, 3 give a wrong answer of 32,144. That is far from the estimate of 43,000, so the estimate catches the error. Another mistake is lining up the numbers by their left ends instead of their right ends. Always line up the ones places first, then add each column.

A stadium holds 52,304 seats and 18,657 tickets have been sold. How many seats are left? Line up the ones and start at the right. Ones: 4 − 7 is not possible with whole numbers. The tens place has a 0, so go to the hundreds. Take 1 hundred from the 3, leaving 2, and make it 10 tens. Then take 1 ten from those 10, leaving 9 tens, and give the ones 10 more: 14 − 7 = 7.

Tens: 9 − 5 = 4. Hundreds: 2 − 6 is not possible, so regroup from the thousands. The 2 thousands becomes 1, and the hundreds become 12: 12 − 6 = 6. Thousands: 1 − 8 is not possible. Regroup from the ten thousands: 5 becomes 4, and thousands become 11: 11 − 8 = 3. Ten thousands: 4 − 1 = 3. The difference is 33,647. Check by adding: 18,657 + 33,647 = 52,304.

The biggest mistake is flipping the digits. A student sees 4 − 7, cannot do it, and writes 7 − 4 = 3 instead. That gives a wrong answer every time. When the top digit is smaller, you must regroup. Zeros make it trickier, because you may need to regroup twice, as we did above. Work slowly, cross out the old digits, and write the new ones above. Then add your answer to the smaller number to check.

A city recorded 45,678 riders on one train line and 23,456 on another. Before adding, estimate: 46,000 + 23,000 = 69,000. Now add exactly. Ones: 8 + 6 = 14, write 4, regroup 1. Tens: 7 + 5 + 1 = 13, write 3, regroup 1. Hundreds: 6 + 4 + 1 = 11, write 1, regroup 1. Thousands: 5 + 3 + 1 = 9. Ten thousands: 4 + 2 = 6. The sum is 69,134, and it is close to the estimate, so it is reasonable.

Reasonable is not the same as right. To be sure, use the inverse operation, the operation that undoes what you did. Subtraction undoes addition. So compute 69,134 − 23,456. Ones: 14 − 6 = 8. Tens: 12 − 5 = 7. Hundreds: 10 − 4 = 6. Thousands: 8 − 3 = 5. Ten thousands: 6 − 2 = 4. You get 45,678, the number you started with. The sum is exactly right.

Suppose a student drops a regroup and writes 59,134. The estimate of 69,000 shows the answer is 10,000 too small. That is the mistake a dropped thousand makes. Now suppose the student writes 69,124, off by only 10. The estimate cannot catch that, but the inverse operation can. Use both tools. Estimate first to catch big errors, then undo the problem to catch small ones.

A pack of gum has 6 pieces. A box holds 40 packs. How many pieces are in a box? Think of 40 as 4 tens. Then 6 × 40 is 6 × 4 tens, which is 24 tens. Twenty-four tens is 240. So a box holds 240 pieces. Every number that ends in zero is a multiple of ten, and this trick works for all of them. Multiply the front digits, then count the zeros back on.

Try 30 × 70. The basic fact is 3 × 7 = 21. The two factors have one zero each, so the product gets two zeros: 2,100. Try 8 × 600. The basic fact is 8 × 6 = 48. There are two zeros, so the product is 4,800. Try 50 × 400. The fact is 5 × 4 = 20, and there are three zeros to attach, so the product is 20,000. Say it as "20 thousand" to hear why.

A common mistake happens when the basic fact itself ends in zero. For 50 × 40, a student finds 5 × 4 = 20 and writes 200. But 200 has only one zero attached to 20. The factors have two zeros, so the answer must be 2,000. Write the basic fact first, then draw one zero for each zero in the factors. Zeros from the fact and zeros from the factors both stay.

A garden is 23 feet long and 14 feet wide. Its area is 23 × 14 square feet. Draw the garden as a rectangle. Cut the long side into 20 and 3. Cut the short side into 10 and 4. Now the garden is four smaller rectangles. This drawing is an area model. Each small rectangle has an area you can find with a basic fact and some zeros.

Find the four pieces. The 20 by 10 piece is 200. The 20 by 4 piece is 80. The 3 by 10 piece is 30. The 3 by 4 piece is 12. Add them: 200 + 80 + 30 + 12 = 322. So 23 × 14 = 322 square feet. Check another way: 23 × 10 = 230 and 23 × 4 = 92, and 230 + 92 = 322. When you decompose the factors this way, nothing is lost.

The most common mistake is to multiply only the matching parts: 20 × 10 = 200 and 3 × 4 = 12, giving 212. That skips two whole pieces of the garden. Every part of one factor must meet every part of the other. Two parts times two parts always makes four pieces. Draw the rectangle and label every piece before you add. If a piece is missing, the drawing will show a hole.

A theater has 46 rows with 27 seats in each row. To find 46 × 27 without a drawing, write the pieces in a column. Each piece is a partial product, part of the whole answer. Ones times ones: 6 × 7 = 42. Tens times ones: 40 × 7 = 280. Ones times tens: 6 × 20 = 120. Tens times tens: 40 × 20 = 800. These are the same four pieces the area model gives.

Now add the partial products: 42 + 280 + 120 + 800 = 1,242. The theater has 1,242 seats. Check with a different route: 46 × 20 = 920 and 46 × 7 = 322, and 920 + 322 = 1,242. The method also works for three-digit numbers. For 213 × 4, the partial products are 3 × 4 = 12, 10 × 4 = 40, and 200 × 4 = 800. The sum is 852.

The most common mistake is forgetting the zeros on a partial product. A student writes 4 × 2 = 8 instead of 4 × 200 = 800, and the answer comes out about 800 too small. Say the full value of each digit before multiplying: "two hundred times four." Another mistake is skipping a piece. For a two-digit times two-digit problem, count your partial products. There must be four.

Back to the 27 classrooms with 34 desks each. The standard algorithm is a short way to write the partial products. Write 27 on top and 34 below it. First multiply 27 by the ones digit, 4. Since 7 × 4 = 28, write 8 and regroup 2. Then 2 × 4 = 8, plus the 2 is 10. The first line is 108. That line is 27 × 4.

Now multiply 27 by the tens digit, 3. But the 3 stands for 30, so first write a placeholder zero in the ones place of the second line. Then 7 × 3 = 21, write 1, regroup 2. Next 2 × 3 = 6, plus 2 is 8. The second line is 810, which is 27 × 30. Add the lines: 108 + 810 = 918. The four partial products 28, 80, 210, and 600 are all hidden inside those two lines.

The most common mistake is skipping the placeholder zero. Then the second line reads 81 instead of 810, and the sum is 189 instead of 918. An estimate catches it: 30 × 30 = 900, and 189 is nowhere near 900. The method works for larger numbers too. For 146 × 23, the lines are 146 × 3 = 438 and 146 × 20 = 2,920, and 438 + 2,920 = 3,358.

A shop packs 372 muffins into boxes of 12. How many boxes? This is 372 ÷ 12. The number being divided, 372, is the dividend. The number you divide by, 12, is the divisor. The answer is the quotient. Instead of finding it all at once, take away easy groups. Ten boxes hold 120 muffins. Take 120 from 372, leaving 252. Ten more boxes: 252 − 120 = 132. Ten more: 132 − 120 = 12.

Now only 12 muffins are left, which is exactly one more box. The groups you took were 10, 10, 10, and 1. Each of those is a partial quotient. Add them: 10 + 10 + 10 + 1 = 31 boxes. Check: 12 × 31 = 372. With one-digit divisors you can take bigger groups. For 856 ÷ 4, take 200 groups (800), leaving 56. Take 10 groups (40), leaving 16. Take 4 groups (16), leaving 0. The quotient is 200 + 10 + 4 = 214.

The most common mistake is forgetting to add the partial quotients at the end. A student stops at the last group and writes 4 instead of 214. Keep a column on the side and add it when the leftover is smaller than the divisor. Another mistake is a subtraction slip in the middle. Check each subtraction before moving on. There is no single right way to choose groups. Bigger groups mean fewer steps.

A charity splits 2,596 cans of food evenly among 4 food pantries. Long division finds 2,596 ÷ 4 one place at a time. Start at the left. Does 4 go into 2? No. Into 25? Yes, 6 times, since 4 × 6 = 24. Write 6 above the 5. Subtract: 25 − 24 = 1. Bring down the 9 to make 19. Then 4 goes into 19 four times, since 4 × 4 = 16. Write 4. Subtract: 19 − 16 = 3.

Bring down the 6 to make 36. Then 4 goes into 36 nine times, since 4 × 9 = 36. Write 9. Subtract: 36 − 36 = 0. The quotient is 649 cans per pantry. Check: 4 × 649 = 2,596. Two-digit divisors follow the same steps. For 1,505 ÷ 35, does 35 go into 150? Yes, 4 times, 4 × 35 = 140, and 150 − 140 = 10. Bring down 5 to make 105. Then 35 goes into 105 three times. The quotient is 43.

The most common mistake is losing a zero in the quotient. For 3,015 ÷ 3, the steps are 3 into 3 once, 3 into 0 zero times, 3 into 1 zero times, then 3 into 15 five times. The quotient is 1,005. A student who skips the zeros writes 15 or 105. Every time you bring down a digit, you must write a digit in the quotient, even when that digit is 0. Estimate first: 3,000 ÷ 3 = 1,000, so 15 cannot be right.

A school sends 125 students on a trip. Each bus holds 40. Divide: 125 ÷ 40 = 3 with 5 left over. The 5 is the remainder, the part that does not fill a whole group. Write it as 3 R 5. But "3 R 5 buses" is not an answer. Three buses leave 5 students at school. The school needs 4 buses. When everyone or everything must fit, round the quotient up.

The same division can mean different things. Suppose 125 cookies are shared by 40 kids. Each kid gets 3 cookies, and 5 cookies are left on the tray. Here the remainder is the answer to "how many are left?" Now suppose you have $125 and tickets cost $40. You can buy 3 tickets. The $5 left is not enough for another, so drop the remainder. To interpret the remainder means deciding which of these fits the story.

Ask three questions. Does everything have to fit? Then round the quotient up. Is the question about the leftover? Then the remainder is the answer. Can you only count whole items? Then drop the remainder. The common mistake is writing "3 R 5" and stopping. A remainder is never the final answer to a word problem. Read the last sentence of the problem again and decide what the leftover means.

A theater sells 48 tickets for each of 23 shows. About how many tickets is that? To estimate, round each factor. 48 is close to 50, and 23 is close to 20. Then 50 × 20 = 1,000. The exact product is 48 × 23 = 1,104, so the estimate is close. Rounding one factor up and the other down often keeps the estimate near the real answer.

For division, rounding is not always enough. Take 1,789 ÷ 6. Rounding gives 1,800 ÷ 6, and 18 ÷ 6 = 3 is a basic fact, so the estimate is 300. Numbers that make a basic fact together, like 1,800 and 6, are called compatible numbers. For 4,182 ÷ 7, choose 4,200, since 42 ÷ 7 = 6. The estimate is 600. The exact quotient, 597 R 3, is very close.

A common mistake is rounding both factors up. For 48 × 23, rounding to 50 × 30 gives 1,500, which is almost 400 too high. Another mistake is choosing a rounded dividend that does not match the divisor, like 1,800 ÷ 7. Look for a number the divisor goes into evenly. Always write your estimate down before you compute. Then compare the two. If they are far apart, look for the error.

A field trip uses 4 buses that each hold 48 people. There are 175 students and 12 adults going. How many seats will be empty? This is a multi-step problem. No single operation answers it. First make a plan. Step 1: find the seats, 4 × 48. Step 2: find the people, 175 + 12. Step 3: subtract the people from the seats.

Now do the steps. Step 1: 4 × 48 = 192 seats. Step 2: 175 + 12 = 187 people. Step 3: 192 − 187 = 5 empty seats. Read the question again: "How many seats will be empty?" The answer is 5 seats. Here is another. A school has 918 desks and gets 96 more. The desks are shared evenly among 26 classrooms. First 918 + 96 = 1,014. Then 1,014 ÷ 26 = 39 desks per room.

The most common mistake is stopping too soon. A student finds 192 seats and writes it down as the answer. But 192 is not what the question asked. Another mistake is doing the steps in the wrong order, like subtracting 12 from 192 before adding the students. Write the plan first, number each step, and put a box around the final answer. Then check that the boxed number answers the actual question.

Earlier we found 1,505 ÷ 35 = 43. Is it right? Multiplication is the inverse operation of division, so it can undo the division. Multiply 35 × 43. Using the standard algorithm: 35 × 3 = 105, and 35 × 40 = 1,400. Add: 105 + 1,400 = 1,505. That is the dividend we started with, so the quotient is right. If the product had been anything else, the division had an error somewhere.

When there is a remainder, add it after multiplying. Earlier we found 200 ÷ 7 = 28 R 4. Check: 7 × 28 = 196, then 196 + 4 = 200. Good. The remainder also has a rule of its own. It must always be smaller than the divisor. Suppose a student writes 200 ÷ 7 = 27 R 11. The check works, since 7 × 27 = 189 and 189 + 11 = 200. But 11 is bigger than 7, so one more group of 7 fits. The correct answer is 28 R 4.

Division can check multiplication too. If you found 24 × 7 = 168, then 168 ÷ 7 should be 24, and it is. The most common mistake is checking with the wrong operation, like adding the divisor to the quotient. Adding will never undo dividing. Use the operation that goes the other way, and make sure the answer you get is reasonable by comparing it to an estimate.