Picture a primary school with around 1,250 pupils. Each pupil gets 24 worksheets through a term. How many sheets does the office have to print? That works out at 1,250 × 24. Take a quick guess at the answer before we calculate it. The shape of that sum, a four-digit number multiplied by a two-digit number, is exactly the maths we are tackling today.
Last lesson you set out 3-digit by 2-digit long multiplication, things like 247 × 34. Today you are taking the same method up one notch: a four-digit number multiplied by a two-digit number. The method is exactly the same. What changes is the room for slip-ups.
On the board: 3,847 × 26. Look at it for a few seconds. What could go wrong here that would not have gone wrong in 247 × 34? Two extra columns means two more places for a carry to land in the wrong column, two more places for a zero to disappear, two more places for the placeholder zero to be forgotten. By the end of this lesson you will know exactly where pupils most commonly slip, and how to catch yourself before the slip costs you the answer.
Two examples on the board, one at a time. Watch where the placeholder zero sits, watch where the carries go, watch the totals.
First row, multiply 3,847 by 6 (the units of 26):
First partial product: 23,082.
Second row, multiply 3,847 by 20 (the tens of 26). Write the placeholder zero in the units column first, then multiply by 2:
Second partial product: 76,940. Add: 23,082 + 76,940 = 100,022.
Before Example 2: on a count of three, everyone whisper the carry from 6 × 8 + 2. One, two, three. (The answer is 5, written above the thousands column. Good. Now we move on.)
The digits of 2,506 are 2 (thousands), 5 (hundreds), 0 (tens), 6 (units). The 0 in the tens column is the internal zero. Watch what happens to the carry when it arrives at that zero.
First row, multiply 2,506 by 5 (the units of 35):
First partial product: 12,530.
Second row, multiply 2,506 by 30 (the tens of 35). Placeholder zero first, then multiply by 3:
Second partial product: 75,180. Add: 12,530 + 75,180 = 87,710.
Two things to take away: the placeholder zero is non-negotiable, and an internal zero still needs any carry added to it. The multiplication being zero does not mean the column is zero. We will meet the third trap, the double carry, in your challenge problems shortly.
Three more 4-digit × 2-digit problems. The teacher walks the first; individual pupils come to the board for the next two while the class predicts each step and calls out any carries.
Problem A, teacher walks: 2,148 × 23
Problem B, pupil at the board, double-carry territory: 1,925 × 36
Problem C, pupil at the board, internal-zero trap: 3,604 × 27
For each problem, the class calls out: first partial product, the placeholder zero, second partial product, total. The pupil at the board writes; the class checks every carry. If the carry is wrong, the class catches it before the next column.
In your jotter, do 1,294 × 37. Set out every column exactly aligned and show each carry digit above the appropriate column.
Start by ruling four columns, Th, H, T, U, under the 1,294. Write 37 underneath with the 7 directly under the 4 and the 3 directly under the 9. Multiply by 7 on the first row. Then write the placeholder zero in the units column of the second row, and multiply by 3 on the rest of the second row. Then add. Carries go above the appropriate column, in pencil, small enough to cross out cleanly if you change your mind.
Six 4-digit × 2-digit problems. Work through them in your jotter, set out neatly with columns aligned. Estimate first, then compute, then tick if your estimate lands near the actual answer.
Two of the six have internal-zero traps. One is firmly in double-carry territory. Spot them as you work.
Fast finishers: try the extension bank with bigger digits and more carries.