The 153 r 3 Papers and a Decimal Point in the Wrong Place
Ms. Whitfield's Grade 5 class has just finished a division unit. Students solve 3,672 ÷ 24 using partial quotients and must verify the answer with multiplication. In the 24 papers she reviews:
- 15 students wrote 153.
- 6 students wrote 153 r 3. Their work shows 24 × 100 = 2,400 and 24 × 50 = 1,200 subtracted, leaving 72; they then found 72 ÷ 24 = 3 but recorded that 3 as a remainder.
- 3 students wrote 15.3.
The same quiz asked for 0.36 × 2.5. Answers clustered at 0.09 (9 papers), 0.9 (11 papers), and 9 (4 papers).
Ms. Whitfield has one 40-minute block before the unit assessment and a cabinet of base-ten blocks, place-value charts, and hundredths grids that have not been used in this unit. She wants the review to be built on student reasoning and estimation rather than on a list of rules to memorize.
1. Explain what the six '153 r 3' papers reveal about place value, show how the required multiplication check exposes the error, and write one question Ms. Whitfield could ask those six students.
Read the worked answer
The six papers are correct up to 3,672 − 3,600 = 72, and 72 ÷ 24 = 3 is also correct. The error is where the 3 is recorded: it belongs in the ones place of the quotient, giving 153, but the students wrote it as a remainder. Their quotient has only two places (100 and 50) instead of three.
The multiplication check catches it: 153 × 24 = 3,672, which matches the dividend exactly, while 153 × 24 + 3 = 3,675, which does not. Students who believe they have a remainder of 3 will find their check off by 3, and that mismatch - not the teacher's correction - is what should drive the revision.
A productive question is: 'You found 3 groups of 24. Where does the 3 belong - with the groups you already counted, or off to the side?' A follow-up: 'How many groups of 24 fit in 3,672 altogether?' The 15.3 papers need a different question, about why a quotient of 15.3 would mean fewer than 16 groups when 24 × 16 is already less than the dividend.
2. Using 3,672 ÷ 24 = 153 as the anchor, find 36.72 ÷ 24 and 367.2 ÷ 2.4 without doing new long division. Justify each with place-value reasoning.
Read the worked answer
36.72 ÷ 24 = 1.53. The dividend is one hundredth of 3,672 (36.72 = 3,672 ÷ 100) while the divisor is unchanged, so the quotient is one hundredth of 153, namely 1.53.
367.2 ÷ 2.4 = 153. Multiply dividend and divisor by the same power of ten so the divisor becomes a whole number: 367.2 ÷ 2.4 = 3,672 ÷ 24. Multiplying both numbers by 10 does not change the quotient, because it is equivalent to multiplying the fraction 367.2/2.4 by 10/10.
Estimation confirms both. For the first, 24 × 1.5 = 36, so a quotient near 1.5 is sensible for a dividend near 36.7. For the second, 2.4 × 150 = 360, so 153 is sensible for a dividend near 367.
The teaching point is that the digits of the quotient do not change when both numbers are scaled by the same factor; only when the dividend alone is scaled does the decimal point shift.
3. Decide which of 0.09, 0.9, and 9 is the correct value of 0.36 × 2.5, using estimation and not the decimal-place-counting shortcut. Then explain what role the shortcut should play.
Read the worked answer
Reason about size before computing.
- Multiplying by 2.5, which is greater than 1, must make the product larger than 0.36. So 0.09 is impossible.
- Multiplying by 2.5, which is less than 3, must make the product less than 0.36 × 3 = 1.08. So 9 is impossible.
Only 0.9 survives both bounds, and the exact product is 0.900 = 0.9.
The 'count the decimal places' shortcut gives two places in 0.36 plus one in 2.5, so three places: 900 → 0.900 = 0.9. It agrees here, but it is bookkeeping, not reasoning - it tells a student where to put the point, never whether the answer is the right size. Ms. Whitfield can make the distinction visible by writing 0.36 × 2.5 and 36 × 25 side by side (both give 900 as digits) and asking which product is bigger and why, so that estimation becomes the check and the shortcut becomes the confirmation.