Question

How do I do multi digit multiplication?

Answer 1

A web search will turn up numerous videos that are available to answer that question. Often, you may find them more satisfactory than the static explanation of a text answer.

The fundamental idea is that the product is the sum of the products of the parts of the number(s). That is, the distributive property applies.

Here is an example.

... 12 × 34

... = (10 +2)×(30 +4)

... = 10(30 +4) +2(30 +4)

... = 10·30 + 10·4 + 2·30 + 2·4

... = 300 + 40 + 60 + 8

... = 408

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Here's an interpretation of what we've just seen that is a little unconventional. The product is the following sum:

... (1·3)×100 + (1·4 + 2·3)×10 + (2·4)×1

If you look at the place values of the digits being multiplied, you see that the multiplier here (×100 or ×10 or ×1) is the product of the place values of the digits involved. That is, when a digit in the 10s place is multiplied by another in the 10s place, their product contributes to the 100s place (10×10) of the result.

One method of multidigit multiplication that is taught is to only write down the partial sums obtained by adding products with the same "place" contribution in the result. That is, the product of 1s place digits (2 and 4 in our example) will go in the 1×1=1s place of the result.

The sum of products of the 10s and 1s place digits (1·4 + 2·3) = 10 will go in the 10×1 = 10s place of the result.

The product of the 10s place digits (1·3) = 3 will go in the 10×10 = 100s place of the result.

If you're good at keeping numbers in your head (gets easier with practice), this method can be faster than other more conventional ways to do it.

For numbers of more digits and/or of different lengths (say a 3-digit by 5-digit number), there are more partial sums, but the idea stays the same. It can be helpful to add leading zeros to the short number to make the numbers the same length.

Here's an example with two 5-digit numbers. Digits are chosen to be different so you can see what is being multiplied at each stage. Pay attention to the pattern being used to select digits to play with.

$17986\\03524\\\\=(6\cdot 4)\times 1+(8\cdot 4+6\cdot 2)\times 10+(9\cdot 4+8\cdot 2+6\cdot 5)\times 100\\+(7\cdot 4+9\cdot 2+8\cdot 5+6\cdot 3)\times 1000\\+(1\cdot 4+7\cdot 2+9\cdot 5+8\cdot 3+6\cdot 0)\times 10000\\+(1\cdot 2+7\cdot 5+9\cdot 3+8\cdot 0)\times 10^5\\+(1\cdot 5+7\cdot 3+9\cdot 0)\times 10^6+(1\cdot 3+7\cdot 0)\times 10^7+(1\cdot 0)\times 10^8\\=63,382,664$

It can be convenient to write down partial sums vertically aligned with the numbers being multiplied. (Put the sum where its place value indicates it should go.) Here, we have proceeded from right to left, but you can also do it proceeding from left to right. (Of course, the product of anything with zero is zero, so can be skipped or ignored.)

Some find it convenient to write the higher-order digits of a sum on separate lines, vertically aligned according to place value. For example, the partial sums we got in the exercise above were 24, 44, 82, 104, 87, 64, 26, and 3. Those might be written like this ...

$\begin{array}{cccccccc}3&6&4&7&4&2&4&4\\2&6&8&0&8&4&2\\&&1\\6&3&3&8&2&6&6&4\end{array}$

where the last line in this array is the sum of the rows above, hence the result of the multiplication.

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When the numbers have decimal digits, the number of decimal places in the result will be the total of the numbers of decimal places in the numbers being multiplied. For example, 8.31×5.6 has 2+1=3 total decimal digits, so the result will have 3 decimal digits. (It is 46.536.) Sometimes such a multiplication will have a product that ends in zeros. Those zeros are counted when placing the decimal point. (1.2×1.5 = 1.80 with 2 decimal digits.)