How can I easily double any size number in my head?

Question

I'm a software engineer, and I often double numbers especially when doing binary to decimal conversions. When numbers get large, I have trouble doubling a number in my head without using paper. For example, I can double 128 in my head easily because it's common and I have it memorized, but numbers like 183 get more difficult.

Is there some clever trick I can use to mentally double any number? I'm probably being idealistic, but it would be nice to have 4 digit numbers be just as easy to double as 2 digit numbers.

Answer 1

I usually look for an easy calculation that is close by the original one.

For example:

$2 \times 183 = 2 \times 180 + 6=366$

or

$ 2\times 1481= 2 \times 1500 - 38 = 2960+2$

Answer 2

It may help to partition the number into smaller numbers that lead with a number less than 5. The idea being that you can double each of these smaller numbers independently then combine their answers without having to worry about "carrying the one".

Example: $$18397238$$ Partitions into $$\underline{18}\,\underline{397}\,\underline{2}\,\underline{38}.$$ Doubling each one yields $$\underline{36}\,\underline{794}\,\underline{4}\,\underline{76}$$

This is efficient if there are a lot of numbers less than $5$ in your number, but not so much otherwise.

Edit: It should be noted that this technique can work in tandem with many of the other methods mentioned, and can actually make them easier to use. For example, if you first think of the number grouped according to my strategy, then use the left-to-right method outlined in David K's answer on each group, every digit has a carried 1 except the last, so you don't even have to think about what the digit to the right is when you are doubling.

Answer 3

You can very well work from left to right (which is the way you spell and utter the number). Just double the current digit and anticipate a carry from the next one (if $5$ to $9$).

Examples:

$192$ yields $284$ without carries and $\bar384$ with.

$9999$ yields $08888/\bar1\bar9\bar9\bar98$.

$123456789$ yields $246802468/246\bar9\bar1\bar3\bar5\bar78$.

I don't mean that you should double all digits then process the carries. On the opposite, generate double+carry form left to right, in a single pass.

This essentially amounts to mechanically applying the regular table below: $$ 00,01,02,03,04\to0\\ 05,06,07,08,09\to1\\ 10,11,12,13,14\to2\\ 15,16,17,18,19\to3\\ 20,21,22,23,24\to4\\ 25,26,27,28,29\to5\\ 30,31,32,33,34\to6\\ 35,36,37,38,39\to7\\ 40,41,42,43,44\to8\\ 45,46,47,48,49\to9\\ 50,51,52,53,54\to0\\ 55,56,57,58,59\to1\\ 60,61,62,63,64\to2\\ 65,66,67,68,69\to3\\ 70,71,72,73,74\to4\\ 75,76,77,78,79\to5\\ 80,81,82,83,84\to6\\ 85,86,87,88,89\to7\\ 90,91,92,93,94\to8\\ 95,96,97,98,99\to9\\ $$

Answer 4

There is a mental-arithmetic trick for doubling a number that works from left to right.

The rule for most of the digits of the result is as follows. First, multiply the digit by two, ignoring the tens place of this intermediate result. For example, if the original digit is $7,$ since twice $7$ is $14$ the result of this step is $4.$ (One way to think of this is "ignore the carry digit," which would have been the leading $1$ in this example.) Next, if the next digit to the right in the original number is $5$ or greater, add one.

For the first digit, however, do not discard the carry digit; just double the first digit and (if the next digit is $5$ or greater) add one. For the last digit, of course, there is no "next digit" so there is no "add one" step.

So to double $678948,$ we initially get the digits $13$ (from $2\times6+1,$ since $7 \geq 5$), then $5$ (since $2\times7=14$ and $4+1=5$), then $7,$ then $8$ (since $2\times9=18$ and the next digit, $4,$ is less than $5$), then $9,$ then $6.$ The final result is $1357896.$

There are variations on this method. If a digit (other than the first one) in the original number is $5$ or greater, you can subtract $5$ before doubling it. Alternatively, you can take the digits in "blocks" of more than one at a time. For example if you see a pair of consecutive digits that form an easy-to-double two-digit number, double that number; if this is not the first two digits of the original number discard the "carry" (that is, $77$ doubled is $154,$ but you just use $54$); then add $1$ if the next digit is $5$ or greater. (Even if you are going to treat the next two or three digits as another "block," you only need to look at the first digit of the block to decide whether to add one to the previous block of the result.)

You can even combine the two variations (subtract $5$ and multiply "blocks" of digits) if you like; just remember that if you subtract $5,$ subtract it from the leftmost digit of the block of digits. If we take the previous example, $678948,$ in blocks of two digits each, we get initially $2\times67+1=135$, then from $89$ we get $2\times39=78,$ and finally $2\times48=96.$

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The peculiar feature of multiplication by two that this trick exploits is that it's extremely easy to tell what the "carry" digit is going to be at any stage of the multiplication without actually working out all the digits to the right. So each time you are working on a particular digit or block of digits, you don't have to remember what previously happened during multiplication of the digits to the right; you just have to look at the next digit to the right to decide whether the digit you're about to include in your result should be (for example) $8$ or $9.$

It is not so easy to predict the carry digits when multiplying by numbers other than $2,$ which is (I think) why most schools don't bother to teach this trick.

The thing I find most difficult in mental arithmetic of this sort is remembering the result long enough to use it for whatever purpose I needed it. That's one reason I try to keep a pencil and some scrap paper handy when I'm programming or debugging.

Answer 5

If you want to get the digits from left to right (which is not said in the question, but maybe implicit, since going right-to-left doubling is really easy for any length of number), just add a final digit$~0$ at the right and then mentally divide the resulting number by$~5$. This is not hard; each time just find the quotient and remainder for a number below $50$. I think the other left-to-right answers given here just describe what this method amounts to.

Answer 6

I find doubling is easy to do in my head if I work right to left as I would with usual multiplication or addition.

For 183 for example: double 3 to get 6, double 8 to get 16, remember the 1 to carry, double 1 to get 2, add the carried 1 and get 366.

It doesn't write down nicely since it's more of a mental process than a mathematical one. I literally just do out the steps for hand multiplication; since it's only ×2 it's usually easy to keep track of numbers. Since you're working right to left you can always write down the digit you just found out, that way you just have the carry any higher order digits over without remembering anything else. Come to think of it, when doubling it's really easy since the number you carry over is only ever 1.

Answer 7

Responding just to one part:

I'm probably being idealistic, but it would be nice to have 4 digit numbers be just as easy to double as 2 digit numbers.

If you have a four digit number, e.g., $7215$, then you can think of it as: $72(100) + 15$.

So, responding to the quotation-pull above, let us double the 2 two digit numbers:

$72$ doubled is $144$, and $15$ doubled is $30$. Together, you get $14400 + 30 = 14430$.

Similarly, you give the example of $183$, which is: $1(100) + 83$.

$1$ doubled is $2$, and $83$ doubled is $166$. Together, you get $200+166 = 366$.

I have written out the steps here, but I think this method can be carried out mentally and should at least help with the goal stated in the body of your post ("to have 4 digit numbers be just as easy to double as 2 digit numbers") if not the title of your post ("double any size number").

Answer 8

The way which works for me is :

• Take the number, make a pair of two digits starting from right to left.
• Now, multiply each pair by two, and do remember the carry to the next i.e. left pair to it.
• In case the carry from right pair is zero, then it's fine, otherwise add the carry after multiplying this current pair by 2.

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For example: here c stands for carry.
A simple example for a power of $2$.

$ 4096 \to \underline{40}\ \underline{96} $

$ (\underline{40}\ \underline{96}) *2 \to (\underline{40}\ *2+[c] , \underline{96} *2)$

$ (\underline{40}\ *2+[1] , \underline{92}) \to (\underline{81}\ \underline{92}) $

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A simple example for a number with odd(5) length.

$ 98765 \to \underline{9}\ \underline{87}\ \underline{65} $

$ (\underline{9}\ \underline{87}\ \underline{65} \ )*2 \to (\underline{9}\ *2+[c] , \underline{87}\ *2+ [c], \underline{65} *2 ) $

$ (\underline{9}\ *2+[c] , \underline{87}\ *2+ [1], \underline{30} \to (\underline{9}\ *2+[1] , \underline{75}, \underline{30} )$

$ (\underline{18}+[1] , \underline{75}, \underline{30} ) \to (\underline{19} \ \underline{75}\ \underline{30} ) $

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A (pseudo)number this time, using the following Python code

import random
n=random.randint(1,2**(random.randint(2,100)))
print(n)

266103712878030

length is $15$

$266103712878030 \to \underline{2}\ \underline{66}\ \underline{10}\ \underline{37}\ \underline{12}\ \underline{87}\ \underline{80}\ \underline{30})$

$(\underline{2}\ \underline{66}\ \underline{10}\ \underline{37}\ \underline{12}\ \underline{87}\ \underline{80}\ \underline{30}) *2 \to $ $ (\underline{2}\ *2+[c] , \underline{66}\ *2+[c] , \underline{10}\ *2+[c] , \underline{37}\ *2+[c] , \underline{12}\ *2+[c] , \underline{87}\ *2+[c] , \underline{80}\ *2+[c] , \underline{30}*2)$

$ (\underline{2}\ *2+[c] , \underline{66}\ *2+[c] , \underline{10}\ *2+[c] , \underline{37}\ *2+[c] , \underline{12}\ *2+[c] , \underline{87}\ *2+[c] , \underline{80}\ *2+[c] , \underline{30} *2) \to$ $ (\underline{2}\ *2+[c] , \underline{66}\ *2+[c] , \underline{10}\ *2+[c] , \underline{37}\ *2+[c] , \underline{12}\ *2+[c] , \underline{87}\ *2+[1] , \underline{60}\ *2+[0] , \underline{60})$

$ (\underline{2}\ *2+[c] , \underline{66}\ *2+[c] , \underline{10}\ *2+[c] , \underline{37}\ *2+[c] , \underline{12}\ *2+[1] , \underline{74}+[1] , \underline{60}\ *2 , \underline{60}) \to$ $ (\underline{2}\ *2+[c] , \underline{66}\ *2+[c] , \underline{10}\ *2+[c] , \underline{37}\ *2+[0] , \underline{24}\ +[1] , \underline{75}\ , \underline{60} , \underline{60})$

$ (\underline{2}\ *2+[c] , \underline{66}\ *2+[c] , \underline{10}\ *2+[0] , \underline{74}\ *2+[0] , \underline{25}\ , \underline{75}\ , \underline{60} , \underline{60}) \to$ $ (\underline{2}\ *2+[c] , \underline{66}\ *2+[0] , \underline{20}\ +[0] , \underline{74}\ , \underline{25}\ , \underline{75}\ , \underline{60} , \underline{60})$

$ (\underline{2}\ *2+[1] , \underline{32}\ +[0] , \underline{20}\ , \underline{74}\ , \underline{25}\ , \underline{75}\ , \underline{60} , \underline{60}) \to $ $ (\underline{2}\ *2+[1] , \underline{32}\ , \underline{20}\ , \underline{74}\ , \underline{25}\ , \underline{75}\ , \underline{60} , \underline{60}) $

$ (\underline{2}\ *2+[1] , \underline{32}\ , \underline{20}\ , \underline{74}\ , \underline{25}\ , \underline{75}\ , \underline{60} , \underline{60}) \to $ $ (\underline{4}\ +[1] , \underline{32}\ , \underline{20}\ , \underline{74}\ , \underline{25}\ , \underline{75}\ , \underline{60} , \underline{60}) $

$ (\underline{5} , \underline{32}\ , \underline{20}\ , \underline{74}\ , \underline{25}\ , \underline{75}\ , \underline{60} , \underline{60}) $

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The reason i use this algorithm is that

1. It's easier to multiply two digits at a time, also i happen to remember the multiplication easily for the numbers under 100.
2. It is faster than to multiply single digit at a time (but slower than 3 or 4 at a time). We can negotiate this for simplicity as we are doing calculation in mind.
3. Also, when you re-iterate and make the final number from left to right you can easily recall the numbers you have just calculated, if not then multiply upcoming pair again.

Answer 9

As a simple trick you can use $(a-b)(a+b)=a^2-b^2$ sometimes. For example

$$102*98=100^2-2^2=9996$$

But it needs to memorize some perfect squares