Multiplication • Topic 3 of 4

Multiplication Algorithms

A multiplication algorithm is just a reliable step-by-step procedure for multiplying numbers, and CTET tests it less for the answer than for diagnosing where a child goes wrong. The foundation is one-digit multiplication, where a fact like 8 x 6 = 48 is recalled directly, or built up through repeated addition, skip counting, arrays or jumps on a number line. From there, multi-digit work rests on two ideas: place value and the distributive property. The simplest multi-digit case is multiplying by 10, 100 or 1000, where you multiply the non-zero digits and append the matching number of zeros, so 45 x 100 = 4500. For something like 32 x 3, the expanded form makes the logic clear: 32 = 30 + 2, so (30 x 3) + (2 x 3) = 90 + 6 = 96. When a place value product passes 9, you regroup, or carry. Take 48 x 6: 6 x 8 = 48, write 8 and carry 4; then 6 x 4 tens = 24 tens, add the carried 4 to get 28 tens, giving 288. The same routine, always starting from the ones place, scales up to 234 x 4 = 936 and beyond. For two-digit by two-digit, you multiply by each digit and shift for place value: 36 x 24 = (36 x 20) + (36 x 4) = 720 + 144 = 864, where the 720 line is written with a placeholder zero because you are really multiplying by 20, not 2. CTET's favourite errors live here: forgetting to shift the second partial product, dropping or mis-adding a carried digit, and mishandling a zero inside a number, as when a child treats 105 x 3 as 3015 instead of 315. The remedy in every case is to return to expanded form and the area model, moving from concrete to pictorial to the compact algorithm.

✅ Solved examples

1. Multiply 23 x 10 and 7 x 3000, explaining the rule.
To multiply by a multiple of 10, multiply the non-zero digits and append the zeros. 23 x 10 = 230 (one zero appended). For 7 x 3000, first 7 x 3 = 21, then append three zeros: 21000.
2. Use expanded form to find 32 x 3.
Break 32 into 30 + 2. Then (30 x 3) + (2 x 3) = 90 + 6 = 96. So 32 x 3 = 96. This is the distributive property behind the standard algorithm.
3. Find 48 x 6 using the standard algorithm with carrying.
6 x 8 ones = 48: write 8 in the ones place, carry 4 tens. 6 x 4 tens = 24 tens, plus the carried 4 tens = 28 tens. Write 8 in the tens place and carry 2 to the hundreds. The product is 288.
4. Multiply 36 x 24 using partial products.
36 x 24 = 36 x (20 + 4) = (36 x 20) + (36 x 4) = 720 + 144 = 864. The 720 comes from multiplying by 20, so it is written one place to the left (with a zero placeholder). Sum: 144 + 720 = 864.

✏️ Practice — try these, take hints as needed

1. A child writes 105 x 3 = 3015. Find the correct answer and identify the error.
Check the middle digit: 0 x 3 is not 3.
Use expanded form: 100 x 3, 0 x 3, 5 x 3.
315 (the child wrongly treated 0 x 3 as 3; correct is 300 + 0 + 15 = 315)
2. Compute 234 x 4 using the standard algorithm.
Start from the ones place and carry.
4 x 4 = 16, then 4 x 3 plus carry, then 4 x 2 plus carry.
936
3. Find 45 x 100.
Multiply the non-zero digits, then append zeros.
100 has two zeros.
4500
4. Multiply 256 x 32 using two partial products.
256 x 2 and 256 x 30, then add.
Remember to shift the second line for place value.
256 x 2 = 512 and 256 x 30 = 7680, so 512 + 7680 = 8192

📝 Topic test — 8 questions

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