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Matemáticas

Column Multiplication Practice Questions, Laid Out in Full

Twelve column multiplications from 34 × 6 to 486 × 37, each answer showing both partial products and the placeholder zero, so a misaligned digit can be traced.

La respuesta corta

Column multiplication splits one calculation into partial products, one for each digit of the second number. For 47 × 23, work out 47 × 3 = 141, then 47 × 20 = 940 — written with a zero holding the units column — and add them: 141 + 940 = 1,081.

Los ejercicios

  1. Ejercicio 1Básico2 puntos

    Work out 34 × 6 using the column method.

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    204

    1. 6 × 4 = 24 — write 4 in the units column, carry 2 into the tens.
    2. 6 × 3 tens = 18 tens = 180.
    3. 180 + the carried 20 = 200.
    4. 200 + 4 = 204.
  2. Ejercicio 2Básico2 puntos

    Work out 58 × 7 using the column method.

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    406

    1. 7 × 8 = 56 — write 6, carry 5 tens.
    2. 7 × 5 tens = 35 tens = 350.
    3. 350 + the carried 50 = 400.
    4. 400 + 6 = 406.
  3. Ejercicio 3Básico2 puntos

    Work out 246 × 3 using the column method.

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    738

    1. 3 × 6 = 18 — write 8, carry 1 ten.
    2. 3 × 4 tens = 120, plus the carried 10 = 130 — write 3 in the tens, carry 1 hundred.
    3. 3 × 2 hundreds = 600, plus the carried 100 = 700.
    4. 700 + 30 + 8 = 738.
  4. Ejercicio 4Básico3 puntos

    Work out 407 × 4 using the column method.

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    1,628

    1. 4 × 7 = 28 — write 8, carry 2 tens.
    2. 4 × 0 tens = 0, plus the carried 20 = 20 — write 2 in the tens column.
    3. 4 × 4 hundreds = 1,600.
    4. 1,600 + 20 + 8 = 1,628. The zero in 407 still owns a column, and skipping it turns the answer into 168.
  5. Ejercicio 5Estándar3 puntos

    Work out 47 × 23 using long multiplication, labelling both partial products.

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    1,081

    1. Row 1 — 47 × 3 = 141.
    2. Row 2 — 47 × 20: write the placeholder 0 in the units column, then 47 × 2 = 94, giving 940.
    3. 141 + 940 = 1,081.
    4. Check: 47 × 20 = 940 and 47 × 3 = 141, and 940 + 141 is a little over a thousand, as it should be.
  6. Ejercicio 6Estándar3 puntos

    Work out 86 × 45 using long multiplication.

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    3,870

    1. Row 1 — 86 × 5 = 430.
    2. Row 2 — 86 × 40: placeholder 0, then 86 × 4 = 344, giving 3,440.
    3. 430 + 3,440 = 3,870.
    4. Check: 86 × 45 sits between 86 × 40 = 3,440 and 86 × 50 = 4,300.
  7. Ejercicio 7Estándar3 puntos

    Work out 132 × 24 using long multiplication.

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    3,168

    1. Row 1 — 132 × 4 = 528.
    2. Row 2 — 132 × 20: placeholder 0, then 132 × 2 = 264, giving 2,640.
    3. 528 + 2,640 = 3,168.
    4. Check: 132 × 24 should be a little over 132 × 25 = 3,300 minus 132, which is 3,168.
  8. Ejercicio 8Estándar3 puntos

    Work out 305 × 27 using long multiplication.

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    8,235

    1. Row 1 — 305 × 7 = 2,135.
    2. Row 2 — 305 × 20: placeholder 0, then 305 × 2 = 610, giving 6,100.
    3. 2,135 + 6,100 = 8,235.
    4. The zero in 305 contributes nothing to the value but still holds the tens column in both rows.
  9. Ejercicio 9Avanzado4 puntos

    Work out 486 × 37 using long multiplication.

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    17,982

    1. Row 1 — 486 × 7 = 3,402.
    2. Row 2 — 486 × 30: placeholder 0, then 486 × 3 = 1,458, giving 14,580.
    3. 3,402 + 14,580 = 17,982.
    4. Check by a different route: 486 × 37 = 486 × 40 − 486 × 3 = 19,440 − 1,458 = 17,982.
  10. Ejercicio 10Avanzado4 puntos

    Work out 209 × 68 using long multiplication.

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    14,212

    1. Row 1 — 209 × 8 = 1,672.
    2. Row 2 — 209 × 60: placeholder 0, then 209 × 6 = 1,254, giving 12,540.
    3. 1,672 + 12,540 = 14,212.
    4. Check: 200 × 68 = 13,600 and 9 × 68 = 612, and 13,600 + 612 = 14,212.
  11. Ejercicio 11Avanzado4 puntos

    A pupil works out 154 × 36 and writes 1,386. Find the error and give the correct answer.

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    5,544 — the second row was written as 154 × 3, not 154 × 30

    1. The pupil's rows were 154 × 6 = 924 and 154 × 3 = 462, and 924 + 462 = 1,386.
    2. The 3 in 36 stands for thirty, so the second row must be 154 × 30 = 4,620 with a placeholder zero.
    3. Correct: 924 + 4,620 = 5,544.
    4. The wrong answer is close to a quarter of the right one, which is the signature of a missing placeholder zero.
  12. Ejercicio 12Avanzado3 puntos

    A school orders 24 boxes of exercise books, each holding 125 books. How many books is that? Check your answer by multiplying the other way round.

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    3,000 books

    1. Row 1 — 125 × 4 = 500.
    2. Row 2 — 125 × 20: placeholder 0, then 125 × 2 = 250, giving 2,500.
    3. 500 + 2,500 = 3,000.
    4. Check the other way: 24 × 100 = 2,400 and 24 × 25 = 600, and 2,400 + 600 = 3,000.

Dónde se falla

  • Omitting the placeholder zero, so 47 × 20 is added as 94 instead of 940 and the answer comes out roughly a quarter of the truth.
  • Adding the carried digit before multiplying rather than after — doing (5 + 5) × 7 instead of 7 × 5 then + 5.
  • Writing the carry above the wrong column, so it is added into hundreds instead of tens.
  • Skipping a zero inside the number being multiplied, turning 407 × 4 into 47 × 4 and the answer into 168.
  • Ending the second row under the units column instead of the tens, which misaligns every digit in the addition.

Every row is a complete multiplication on its own

The two rows under a long multiplication are not fragments. In 47 × 23 the first row is the whole of 47 × 3 and the second is the whole of 47 × 20, and each can be checked by itself. A pupil who can say out loud what a row is meant to be — "this row is forty-seven twenties" — will catch a wrong row immediately, because 940 is obviously about a thousand and 94 obviously is not.

That is why the answers below label both rows. When an answer comes out wrong it is nearly always one row that is wrong, and naming the rows turns a hunt through fourteen digits into a check of two numbers.

Why the second row starts with a zero

The digit 2 in 23 is not two, it is twenty. Writing 47 × 2 = 94 and then adding it as 94 is the single most common error in the whole method, and it makes the answer 846 short of the truth. The zero written in the units column is not decoration and it is not a rule to be memorised — it is the statement that this row is worth ten times what the digits say.

The same logic carries into a three-digit multiplier, where the third row takes two zeros. If a pupil is told the zeros are "just what you do", they will drop one under pressure. If they know the second row is tens and the third is hundreds, they will not.

Tracing a wrong answer back to the digit that caused it

Column multiplication has exactly three places to go wrong: a times-table fact, a carry, or an alignment. Checking them in that order finds the fault fast. Recompute each row against the tables first, then look at the small carried digits, then look at whether each row ends in the right column.

An answer that is roughly a tenth of what it should be is an alignment fault every time. An answer that is out by a small amount — twenty or thirty — is a lost carry. An answer that is wildly wrong in one digit is a table fact. The size of the error names the cause before you have found it.

In England, pupils are expected to multiply numbers up to four digits by a one- or two-digit number using a formal written method, including long multiplication for two-digit numbers, by the end of Year 5.

The size of long multiplication expected by the end of Year 5 in England
4 digits × 2 digitsThe size of long multiplication expected by the end of Year 5 in England[1]

Preguntas frecuentes

Why does my child need the column method when a calculator is faster?

Because the method is the place value. Setting out 47 × 20 as 940 is how a child learns that the 2 in 23 means twenty, and that idea is what later carries decimals, standard form and algebraic expansion. A child who has never seen where 940 comes from struggles with (x + 3)(x + 7) years afterwards.

Is the grid method wrong?

No. The grid method computes the same partial products in a different arrangement, and for many children it makes the place value more visible. The column method is what the English national curriculum names as the formal written method, so most children need to end up fluent in it, but arriving there via the grid is entirely reasonable.

How can I tell whether the mistake is multiplication or times tables?

Look at the size of the error. An answer roughly ten times too small is an alignment or placeholder problem. An answer out by twenty or thirty is a lost carry. An answer that is wrong by an odd amount in one place is usually a recalled fact — check 7 × 8 and 6 × 9 before blaming the method.

What size of numbers should a Year 5 pupil manage?

Numbers up to four digits multiplied by a one- or two-digit number, using long multiplication for the two-digit case. That is the statutory expectation in England for the end of Year 5, and it is why the harder questions here run to three digits by two digits rather than stopping at 47 × 23.

Fuentes

  1. National curriculum in England: mathematics programmes of study — Department for Education

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