Long multiplication (age 10+)

Activities to try

1. 1

   Step-by-step long multiplication with an estimate check

   Why it works: Each partial-product row multiplies the whole of 2463 by a part of 37 (by 7 ones, then by 30), and the rows are added to recombine the parts. The placeholder zero is there because the second row is really ×30, not ×3. The estimate bounds the answer's size, so a lost zero shows up as a clearly wrong magnitude. A carry slip usually changes only one or two digits and can still look plausible, so it has to be caught by checking the working itself.

   1. Before calculating, estimate 2463×37 by rounding both factors: 2500×40 = 100000, and write this down as the expected ballpark.
   2. Work 2463×37 by long multiplication, writing the ×ones row (2463×7 = 17241) and the ×tens row underneath.
   3. When starting the ×tens row, write the placeholder zero in the ones column first, then multiply by 3 tens to get 73890.
   4. Add the two rows: 17241 + 73890 = 91131, keeping the columns aligned.
   5. Compare the final answer with your estimate of ~100000 — 91131 is in the right ballpark, so the magnitude is plausible.
2. 2

   Spot-the-error and fill-the-gap in a worked long multiplication

   Why it works: Being able to diagnose a planted error or complete a missing step shows you understand each step's role — especially why the ×tens row is shifted by a placeholder zero and why rows must be aligned before adding. Estimation is a fast first filter for implausible answers.

   1. Examine worked examples of 4-digit × 2-digit long multiplication that contain a single planted mistake.
   2. Identify the error type: a missing placeholder zero, a misaligned partial-product addition, a dropped or misplaced carry, or multiplying by only one digit.
   3. Complete a fill-in-the-blank worked example where one partial product (or one digit of a partial product) is left empty, computing the missing value and checking it against the rest of the working.
   4. Estimate the correct answer by rounding both factors, and use it to judge whether the worked total is plausible before pinpointing the error.
   5. Rewrite the step correctly and recompute the final total.
3. 3

   Annotate each partial product with its place value

   Why it works: Each row corresponds to multiplying the whole number by a part of the multiplier; the placeholder zero marks that the second part is tens. Connecting the compact algorithm to the area model keeps the procedure meaningful.

   1. Take a completed long multiplication of 2463×37 and label the first row '×7 (ones)' and the second row '×30 (tens)'.
   2. Explain in writing or aloud why there is a zero at the end of the second row.
   3. Match each partial-product row to the corresponding region of an area/grid model: the rectangle 2463×30 and the rectangle 2463×7.
   4. State why the two rows are added together to get the total.