Week 12 · Two-digit multiplication and remainders · 3 lessons
2
The standard way to multiply big numbers
Discover Mathematics P4: Term 1, Unit 7, Lessons 6–10
GoalUse the standard algorithm to multiply 2-digit and 3-digit numbers by 2-digit numbers.
New words
place holder · the 0 we write in the ones place of the second row so the digits stay in the right placespartial product · the answer of one row of the written methodreasonable · close to the estimate
Explanation
The standard algorithm multiplies by one digit of the second factor at a time. For 47 × 36, first multiply 47 by the 6 ones: 47 × 6 = 282. Then multiply 47 by the 3 tens. Because 3 tens is 30, write a 0 in the ones place first, then 47 × 3 = 141, so the row is 1,410. Add the two rows: 282 + 1,410 = 1,692. The zero is a place holder: without it we would multiply by 3 instead of 30. The same steps work for a 3-digit number: 213 × 24 uses 213 × 4 = 852 and 213 × 20 = 4,260, and 852 + 4,260 = 5,112. Always estimate: 47 × 36 is about 50 × 40 = 2,000 and the exact answer 1,692 is a little below, as it should be.
Row
Work
Result
1
47 × 6
282
2
47 × 30
1,410
Sum
282 + 1,410
1,692
Worked examples
Row
Work
Result
1
47 × 6
282
2
47 × 30
1,410
Sum
282 + 1,410
1,692
Find 47 × 36. Row 1: 47 × 6 = 282. Row 2: 47 × 30 = 1,410. Sum: 1,692. Estimate 50 × 40 = 2,000, so it is reasonable.
Which way is faster? Choose 35 × 24. The child draws four boxes, the grown-up writes two rows. Compare the partial products: where do the numbers 140 and 700 appear in each method?
