How to do long multiplication

How to do long multiplication

While sitting in an examination hall confronted with a Mathematics exam, all of a sudden you hear the invigilator saying “Ten more minutes to go so kindly check your name and index numbers on the answer sheet”. If you are done with solving this call may not bother you all that much. However, if you have some long calculations and multiplications to make the call seems more like a death sentence. If it is in your control, you would love to glue the invigilator’s lips so that you do not hear anything that may possibly make you lose concentration. Nevertheless, if you know how to do long multiplications in quick time there may be no need to waste the glue for any such purpose.

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First of all you should know about long multiplication as it involves atleast two digits being multiplied with each other. If one number is for instance 4 digits and the other being only one digit then this is not long multiplication.

How to do long multiplication

Make sure that in long multiplication you have to multiply each digit of either of the two numbers to be separately in the way shown.

How to do long multiplication

After multiplying the first digit with the number, put a cross or multiplication sign as shown, before multiplying with the other digit with the same number.

How to do long multiplication

Once you have multiplied both the digits now add both the products that you have obtained separately. The total sum of the both the obtained products will give you the required answer.

How to do long multiplication

If it seems confusing for you then it is better to first separately multiply each digit with the number and then add the two products as shown.

What is long multiplication?

Long multiplication is a method used to solve multiplication problems with large numbers. One thing that can really help you in long multiplication is if you know the multiplication table by heart. This will speed up your work and make it more accurate.

The first step in long multiplication is to write down the numbers on top of each other. You align the numbers on the right. Don’t worry about the decimal points when lining up the numbers; just write them down and line up the right-most number.

Second Step

Now we are going to start multiplying. We’ll use the numbers from the first example above: 469 x 32. We begin with the ones place in the bottom number. This is the 2 in 32. We multiply 2×469 and write it down under the line.

How to do long multiplication

Adding a Zero for the Tens Space

Now we need to multiply by the next number to the left of the 2. This is the 3 in 32. Because the 3 is in the tens place we need to hold the tens place by placing a zero in the 1’s place before we start multiplying.

How to do long multiplication

Finish multiplying

Multiply the 3 by the top number (469) and write this number next to the zero.

How to do long multiplication

If there were more numbers we would add more rows and continue to add more zeros. For example, if there were a 4 in the hundreds spot (i.e. the number on the bottom was 432) we would add two zeros in the next row and then multiply 469 by 4.

After we have multiplied all the numbers on the bottom, we add up the rows of numbers to get the answer. In this case there are two rows, but there would be more if the number we were multiplying by on the bottom (the 32) had more digits.

How to do long multiplication

Another Long Multiplication Example

Below is an example long multiplication problem where the added zeros are shown in red and the carry numbers for each step are shown in blue.


Associate professor in mathematics, UNSW

Disclosure statement

David Harvey receives funding from the Australian Research Council.


UNSW provides funding as a member of The Conversation AU.

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Multiplication of two numbers is easy, right?

At primary school we learn how to do long multiplication like this:

How to do long multiplication

Methods similar to this go back thousands of years, at least to the ancient Sumerians and Egyptians.

But is this really the best way to multiply two big numbers together?

In long multiplication, we have to multiply every digit of the first number by every digit of the second number. If the two numbers each have N digits, that’s N 2 (or N x N) multiplications altogether. In the example above, N is 3, and we had to do 3 2 = 9 multiplications.

Around 1956, the famous Soviet mathematician Andrey Kolmogorov conjectured that this is the best possible way to multiply two numbers together.

In other words, no matter how you arrange your calculations, the amount of work you have to do will be proportional to at least N 2 . Twice as many digits means four times as much work.

Kolmogorov felt that if a short cut was possible, surely it would have already been discovered. After all, people have been multiplying numbers for thousands of years.

This is a superb example of the logical fallacy known as “the argument from ignorance”.

A quicker way

Just a few years later, Kolmogorov’s conjecture was shown to be spectacularly wrong.

In 1960, Anatoly Karatsuba, a 23-year-old mathematics student in Russia, discovered a sneaky algebraic trick that reduces the number of multiplications needed.

For example, to multiply four-digit numbers, instead of needing 4 2 = 16 multiplications, Karatsuba’s method gets away with only nine. When using his method, twice as many digits means only three times as much work.

This stacks up to an impressive advantage as the numbers get bigger. For numbers with a thousand digits, Karatsuba’s method needs about 17 times fewer multiplications than long multiplication.

But why on earth would anyone want to multiply such big numbers together?

In fact, there are a tremendous number of applications. One of the most visible and economically significant is in cryptography.

Big numbers in real life

Every time you engage in encrypted communication on the internet — for example, access your banking website or perform a web search — your device performs a head-spinning number of multiplications, involving numbers with hundreds or even thousands of digits.

Very likely your device uses Karatsuba’s trick for this arithmetic. This is all part of the amazing software ecosystem that keeps our web pages loading as snappily as possible.

For some more esoteric applications, mathematicians have to deal with even larger numbers, with millions, billions or even trillions of digits. For such enormous numbers, even Karatsuba’s algorithm is too slow.

A real breakthrough came in 1971 with the work of the German mathematicians Arnold Schönhage and Volker Strassen. They explained how to use the recently published fast Fourier transform (FFT) to multiply huge numbers efficiently. Their method is routinely used by mathematicians today to handle numbers in the billions of digits.

The FFT is one of the most important algorithms of the 20th century. One application familiar in daily life is digital audio: whenever you listen to MP3s, music streaming services or digital radio, FFTs handle the audio decoding behind the scenes.

An even quicker way?

In their 1971 paper, Schönhage and Strassen also made a striking conjecture. To explain, I’ll have to get a bit technical for a moment.

The first half of their conjecture is that it should be possible to multiply N-digit numbers using a number of basic operations that is proportional to at most N log (N) (that’s N times the natural logarithm of N).

Their own algorithm did not quite reach this target; they were too slow by a factor of log (log N) (the logarithm of the logarithm of N). Nevertheless, their intuition led them to suspect that they were missing something, and that N log (N) should be feasible.

In the decades since 1971, a few researchers have found improvements to Schönhage and Strassen’s algorithm. Notably, an algorithm designed by Martin Fürer in 2007 came agonisingly close to the elusive N log (N).

The second (and much more difficult) part of their conjecture is that N log (N) should be the fundamental speed limit — that no possible multiplication algorithm could do better than this.

Have we reached the limit?

A few weeks ago, Joris van der Hoeven and I posted a research paper describing a new multiplication algorithm that finally reaches the N log (N) holy grail, thus settling the “easy” part of the Schönhage–Strassen conjecture.

The paper has not yet been peer-reviewed, so some caution is warranted. It is standard practice in mathematics to disseminate research results before they have undergone peer review.

Instead of using one-dimensional FFTs — the staple of all work on this problem since 1971 — our algorithm relies on multidimensional FFTs. These gadgets are nothing new: the widely-used JPEG image format depends on 2-dimensional FFTs, and 3-dimensional FFTs have many applications in physics and engineering.

In our paper, we use FFTs with 1,729 dimensions. This is tricky to visualise, but mathematically no more troublesome than the 2-dimensional case.

Really, really big numbers

The new algorithm is not really practical in its current form, because the proof given in our paper only works for ludicrously large numbers. Even if each digit was written on a hydrogen atom, there would not be nearly enough room available in the observable universe to write them down.

On the other hand, we are hopeful that with further refinements, the algorithm might become practical for numbers with merely billions or trillions of digits. If so, it may well become an indispensable tool in the computational mathematician’s arsenal.

If the full Schönhage–Strassen conjecture is correct, then from a theoretical point of view, the new algorithm is the end of the road – it is not possible to do any better.

Personally, I would be very surprised if the conjecture turned out to be wrong. But we shouldn’t forget what happened to Kolmogorov. Mathematics can sometimes throw up surprises.

Basic Method — for Small Numbers

When calculating a multiplication where one of the numbers is small, such as 68435 × 18, it may be fastest to simply add together multiples of the smaller number:

  • 5 × 18 = 9 0 ⇒ ………… 0
  • 9 + 3 × 18 = 6 3 ⇒ ………. 3 0
  • 6 + 4 × 18 = 7 8 ⇒ …….. 8 30
  • 7 + 8 × 18 = 15 1 ⇒ …… 1 830
  • 15 + 6 × 18 = 123 ⇒ 123 1830

In fact when I coach beginners in Mental Math, one of the first things I work on is expanding the knowledge of times tables to other useful numbers — such as 18 — to make this easier to perform.

But when using this basic method for larger multiplications, such as 29136 × 5847, we don’t have enough working memory to calculate each multiple of e.g. 5847 without forgetting the numbers we’ve already calculated! So we need another method — one that’s more efficient in terms of memory.

Below I’ll show you the cross-multiplication method that most advanced mental calculators use for multiplications.

Cross-Multiplication Method

Some nice things about this method are:

  • You only need to know your times tables up to 9 × 9
  • However large the multiplications, you never have to remember many numbers at once
  • It’s very straightforward once you know the simple pattern

To see how this works, we’ll take our example of 29136 × 5847:

1st digit — units place:

To begin, we simply multiply 6 × 7 = 42. Then the rightmost digit of the answer is 2 and we can “carry” the 4 for the next step:

2nd digit — factors with 10:

Next we consider 40 × 6 and 7 × 30 , as these are the digit products ( 4 × 6 and 7 × 3 ) that come with a factor of 10, just like the 40 we remembered from the previous step.

The quickest way is to start with the 4 from the 40 that we carried, then add on the 4 × 6 and 7 × 3 :

  • 4 + 4 × 6 = 28
  • 28 + 7 × 3 = 49

These addition-multiplication pairs are quick to do with practice.

Again we can write down the “9” in the tens place of the final answer, and keep the 4 for the following step.

3rd digit — factors with 100:

We continue with 800 × 6 , 40 × 30 and 7 × 100 , as these are the digit products that come with a factor of 100.

Again — start with the 4 from the 400 that we carried, then add on the other products:

  • 4 + 8 × 6 = 52
  • 52 + 4 × 3 = 64
  • 64 + 7 × 1 = 71

Write down the “1” in the 100s place of the final answer, and keep the 7 for the following step:

Notice that in each step, the order of the colors on the top is the mirror image of the colors on the bottom, as each matching pair of digits must represent the same power of 10.

You can add the digits in any order you like, but I find it helpful to always start with the bottom-left and top-right product (here the 8 × 6 ) and systematically move simultaneously rightwards on the bottom number and leftwards on the top number.

4th digit — factors with 1000:

By now the pattern should be fairly clear, so I’ll continue with minimal commentary:

  • 7 + 5 × 6 = 37
  • 37 + 8 × 3 = 61
  • 61 + 4 × 1 = 65
  • 65 + 7 × 9 = 128

5th digit — factors with 10,000:

This time, notice that the “6” has already been multiplied by every digit of the bottom number, so will not be active for the rest of the calculation:

  • 12 + 5 × 3 = 27
  • 27 + 8 × 1 = 35
  • 35 + 4 × 9 = 71
  • 71 + 7 × 2 = 85

6th digit — factors with 100,000:

From now on the calculation gets simpler as there are fewer and fewer products of the same magnitude:

  • 8 + 5 × 1 = 13
  • 13 + 8 × 9 = 85
  • 85 + 4 × 2 = 93

7th digit — factors with 1,000,000:

  • 9 + 5 × 9 = 54
  • 54 + 8 × 2 = 70

8th digit — factors with 10,000,000:

  • 7 + 5 × 2 = 17

As this is the final stage, we don’t have to carry anything, and simply write down the remaining digits:


So that is the standard cross multiplication method used by amateur human calculators, as well as current and past multiplication world-record holders such as Freddis Reyes and Marc Jornet Sanz! (Jeonghee Lee prefers a right-to-left method instead).

To conclude — the method in general is:

  • Start with the rightmost digit of each number:
    • calculate their product
    • write down the units digit
    • carry the tens digit for the next stage
  • For every subsequent digit of the answer:
    • take the carried number from before
    • add all products of the same magnitude by working systematically
    • write down the units digit in the final answer
    • carry the rest for the next stage

In the Memoriad competition, and in the Mental Calculation World Cup, competitors must multiply 8-digit numbers, such as 12345678 × 98702468, and the fastest competitors can do these in 15-30 seconds!

To practice this you can use the Memoriad software — although I recommend to start with smaller products of 3- and 4-digit numbers before working your way up.

To get in touch with me (Daniel Timms) about mental calculation training or anything on this site, you can contact me here.

How to do long multiplication

Sophie Bartlett

Long multiplication is multiplying a number with two or more digits by a two-digit number, e.g. 34 x 27, 851 x 82 or 4,274 x 93. This article explains the process for doing long multiplication following the column multiplication method used in primary schools at KS2.

This blog is part of our series of blogs designed for teachers, schools and parents supporting home learning.

What is long multiplication?

Long multiplication is a method of multiplying larger numbers together. In the primary curriculum long multiplication is taught for multiplying two, three and four digit numbers by two digit numbers.

Long multiplication methods

The mathematics appendix in the National Curriculum demonstrates the formal method of long multiplication:

How to do long multiplication

The first and third multiplication demonstrate the most common long multiplication methods.

Example 1: 24 x 16

26 is partitioned to become 20 and 6. 124 is multiplied by 6 first, which equals 744; 126 is then multiplied by 20 to become 2480; 2480 and 744 are added together to make 3224.

In order to succeed at long multiplication, it is essential that children are fluent in their times tables and if your child is struggling to learn them, then we recommend trying out these times tables games as a great next step.

How to do long multiplication

FREE KS2 Long Multiplication Worksheets

A series of ready-to-use worksheets to help pupils practice their long multiplication skills.

Example 3: 124 x 26

Here, you have the steps to do long multiplication for the third example:

  1. Set the question in the formal method
  2. Remember to start the process of multiplication with the units
  3. Multiply 6 by 4
  4. Write the answer down correctly – including any carrying
  5. Multiply 6 by 2
  6. Add anything that you have carried from the previous multiplication.
  7. Multiply 6 by 2
  8. Write the answer down correctly
  9. Drop a zero as we are now multiplying with 10s
  10. Multiply 2 by 4
  11. Write the answer correctly
  12. Multiply 2 by 2
  13. Write the answer down
  14. Multiply 2 by 1
  15. Write the answer correctly
  16. Add the two answers up together correctly

Now, all you need to do is follow these steps for any other similar long multiplication question.

When will my child learn about long multiplication in primary school?

Children first learn about how to do long multiplication in Year 5, where they are expected to multiply numbers up to 4 digits by a one- or two-digit number using a formal written method, including long multiplication for two-digit numbers. This is continued in Year 6.

  • In year 5 the national curriculum objectives for multiplication and division include ‘pupils should be taught to multiply numbers up to 4 digits by a one- or two-digit number using a formal written method, including long multiplication for two-digit numbers.’
  • In year 6 the national curriculum objectives for multiplication and division include ‘pupils should be multiply multi-digit numbers up to 4 digits by a two-digit whole number using the formal written method of long multiplication.’

Long multiplication questions

1. 746 x 23 =

2. A box contains trays of melons. There are 15 melons in a tray. There are 3 trays in a box. A supermarket sells 40 boxes of melons. How many melons does the supermarket sell?

3. Write the two missing digits to make this long multiplication correct.

How to do long multiplication

4. A toy shop orders 11 boxes of marbles. Each box contains 6 bags of marbles. Each bag contains 45 marbles. How many marbles does the shop order in total?

(Answer: 11 x 6 x 45 = 2,970)

5. A shop sells sheets of sticky labels. On each sheet there are 36 rows and 18 columns of labels. How many labels are there altogether on 45 sheets?

(Answer: 36 x 18 x 45 = 29,160)

Read more
  • What Is A Square Number: Explained For Primary Parents And Kids
  • What Is The Lowest Common Multiple: Explained For Primary Parents And Kids
  • What Is A Cube Number: Explained For Primary Parents And Kids

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