L2 · Multiplying 3-digit by 2-digit: standard algorithm

Learning goal

“I can multiply a 3-digit number by a 2-digit number using the standard algorithm - and I can explain what each row of the algorithm means in partial products.”

Standard algorithm — the conventional written procedure: multiply by the ones digit, then by the tens digit (shifted), then add the rows.

Partial product — one row of the algorithm: the first factor multiplied by one digit of the second factor, at its true place value.

Regroup — carrying a digit to the next place when a step's product is 10 or more (e.g. 7 x 8 = 56: write 6, regroup 5).

Placeholder zero — the 0 written at the right of the second row, because that row is really multiplying by tens, not ones.

Worked example

Compute 246 x 37 with the standard algorithm. Step 1: Split 37 into 7 and 30 - the same split as L1. The algorithm just writes the two partial products in rows instead of in a rectangle. Step 2: Row 1 - multiply by the ones digit: 246 x 7 = 1,722. Step 3: Row 2 - multiply by the tens digit. This is really 246 x 30, so write a placeholder 0 first, then 246 x 3 = 738 in front of it: the row reads 7,380. Step 4: Add the rows: 1,722 + 7,380 = 9,102. Say it plainly: the placeholder zero is not decoration. Row 2 is worth ten times what it would be without it, because the 3 in 37 is 3 tens. Check by estimate: 250 x 40 = 10,000, and 9,102 is sensibly close.

Check yourself — explain it

Why does row 2 of the algorithm end in 0?

Paired problem — example still visible

Compute 183 x 25 with the algorithm. -> Row 1: 183 x 5 = 915; Row 2: 183 x 20 = 3,660; total 4,575.

Faded completion — last step is yours

"324 x 16: Row 1 is 324 x 6 = 1,944. Row 2 is 324 x 10 = ___. Total = ___" -> 3,240; 5,184

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