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Fraction Classroom · 2.6.0 · Student pages

Find the place-value error in a product

Audit a whole-number multiplication algorithm by naming each partial product, repairing its place value and checking the result against a useful bound.

Anonymous group code: __________   Date: __________

The two rows count two packs and thirty packs. Aligning 7,380 by place value keeps thirty copies of 246 in the total.Multiplier partCalculationLabels2 packs246 × 249230 packs246 × 307,38032 packs492 + 7,3807,872
The two rows count two packs and thirty packs. Aligning 7,380 by place value keeps thirty copies of 246 in the total.

Start with a diagnostic

diagnostic. Calculate 46 × 20. Enter the product and explain what the digit 2 means in the multiplier 20.

Build the idea

model. There are 32 packs with 246 labels in every pack. A draft shows 246 × 2 = 492, 246 × 3 = 738, then adds 492 + 738 = 1,230 as the answer to 246 × 32. What is the correct total, and where did the draft lose place value?

  1. Predict a useful range before repairing the work. Thirty packs contain 246 × 30 = 7,380 labels, while forty packs contain 9,840. A total for 32 packs must lie between those values. The draft's 1,230 labels cannot account for even thirty packs.
  2. Expand the multiplier: 32 = 30 + 2. Draw two columns labelled multiplier part and number of labels. Retain the correct ones row, 246 × 2 = 492, and say that it counts two full packs.
  3. Calculate the tens row using the standard algorithm: 246 × 3 = 738, but three tens require 738 tens, or 7,380 ones. In place-value columns, align this row one position to the left of the ones row; a zero in the ones column records that place value.
  4. Add the aligned partial products to obtain 7,872. Reconcile the algorithm with 246 × (30 + 2), then confirm that 7,872 lies between 7,380 and 9,840. Identify the original error as treating three tens as three ones; the final addition of the faulty rows was itself correct.

Worked example: The correct total is 7,872 labels. The first row counts two packs. The digit 3 in 32 represents thirty packs, so the second partial product must be 246 × 30 = 7,380. Add 492 + 7,380 = 7,872.

Practise and explain

Try p1–p3 with a partner. Use a drawing and label your units. Try e1 independently before comparing responses.

p1. Support: a draft for 124 × 12 adds 248 and 124 to get 372. Enter the corrected product, and label which row must count ten groups.

p2. Core: audit 307 × 24. A draft adds 1,228 and 614 to get 1,842. Enter the correct product and show the two correct partial products, including the zero inside 307.

p3. Challenge: a draft for 409 × 36 gives 409 × 6 = 2,454 and 409 × 30 = 12,027, then adds them to get 14,481. By how much must that total increase? Identify the faulty row and calculate its correction.

e1. Independent check: a draft for 218 × 43 adds 654 and 872 to get 1,526. Enter the correct product, write both corrected partial products in columns, and give a bound that rejects the draft total.

Transfer the idea

transfer. A different draft for 246 × 32 has the correct rows 492 and 7,380 but gives their sum as 7,772. What should be repaired, and why would shifting a row again make matters worse?