Addition and Subtraction to 1000
Warm-up
Estimate: 473 + 318. Students jot estimates in 10 seconds. Share. Most should be around 800. Now compute. How close was your estimate? What benchmark did you use?
Explore
Three-strategy challenge: solve 548 - 273 using (1) decompose by place value, (2) add up from 273 to 548 on a number line, (3) compensate (273 is close to 275; 548 - 275 = 273, add back 2). Compare all three strategies: do they give the same answer? Which is fastest?
Formalize
Regrouping demystified: 342 + 175. Ones: 2+5=7. Tens: 40+70=110. Hundreds: 300+100=400. Combine: 400+110+7=517. But 110 is 1 ten and 10 tens, i.e., 1 hundred and 1 ten. So: 400+100+10+7 = 517. This is regrouping: when the tens sum exceeds 99, the hundred carries over.
Addition and Subtraction to 1000
Inverse verification: I computed 637 - 284 = 353. Check: 353 + 284 = ? If it equals 637, the answer is correct. If not, find the error. Teaching students to self-check with the inverse operation is a metacognitive strategy that transfers to all of mathematics.
Practice
Students solve 6 three-digit addition and subtraction problems using at least two different strategies each, showing estimates. Exit ticket: estimate then compute 564 + 278.
Exit ticket
Students solve 6 three-digit addition and subtraction problems using at least two different strategies each, showing estimates. Exit ticket: estimate then compute 564 + 278.
Step 1: Build both numbers: 3 flats 5 rods 6 units, and 2 flats 8 rods 7 units.
Step 2: Combine units first: 6 + 7 = 13 units. Trade 10 of them for a rod: 1 rod moves to the rods pile, 3 units stay.
Step 3: Combine rods: 5 + 8 = 13, plus the traded rod = 14 rods. Trade 10 rods for a flat: 1 flat moves up, 4 rods stay.
Step 4: Combine flats: 3 + 2 = 5, plus the traded flat = 6 flats. Read the answer off the mat: 643.
Step 5: NOW show the column algorithm beside the materials and match every mark to a trade: the little "carried 1" above the tens column IS the traded rod; the 1 above the hundreds IS the traded flat. The algorithm is just the materials' story written in shorthand — never a separate piece of magic.
The trap: column subtraction on 600 − 234 hits 0 − 4 immediately, and borrowing across two zeros produces the classic tangled mess.
Way 1 — think money: 600 is 5 hundreds + 9 tens + 10 ones (break a hundred into ten tens, break one ten into ten ones — do it with materials once so it's real). Now subtract cleanly: 10 − 4 = 6, 9 − 3 = 6, 5 − 2 = 3 → 366.
Way 2 — add up instead (often smarter): from 234, climb to 600: 234 + 6 = 240, +60 = 300, +300 = 600. Total climb: 6 + 60 + 300 = 366. Same answer, zero borrowing.
Way 3 — shift both numbers (elegant): 600 − 234 has the same gap as 599 − 233 (slide both down 1). Now no borrowing anywhere: 599 − 233 = 366.
The lesson: the algorithm is ONE tool, and for across-zero problems it's often the clumsiest one. Strategy choice is part of the skill.
Step 1: Before computing, estimate by rounding to friendly numbers: 438 ≈ 440, 175 ≈ 180 → about 620. (Or rougher: 400 + 200 = 600.) Write the estimate down — it's the answer's bodyguard.
Step 2: A student computes and gets 5,113. Compare to the estimate: 5,113 vs ≈620. The bodyguard tackles it — something went badly wrong.
Step 3: Autopsy the error: the student added 8 + 5 = 13 and wrote BOTH digits in the ones column instead of trading the ten. Every column overflowed sideways, inflating the number.
Step 4: Recompute with trades: 8 + 5 = 13 → write 3, carry 1. Tens: 3 + 7 + 1 = 11 → write 1, carry 1. Hundreds: 4 + 1 + 1 = 6. Answer: 613 — comfortably inside the estimate's neighbourhood ✓.
The non-negotiable habit: estimate BEFORE, compare AFTER. An answer without an estimate is a stranger you let in without looking.