You likely all know the 10 rule (to multiply by 10, just attach a ‘0’ after the number) but do you know the 11 rule? It is as easy! You should be able to do this one in your head for any two digit number.
Some may try this way: multiplying the number by
10, and add the original number to the result.
For example: 17 × 11 =?
17 × 10 = 170 + 17 = 187
So, 17 × 11 = 187
And, 156
X 11 =?
Add a zero to the end: 1560
Add 156 to this: 1560 + 156 = 1716
Here is a very nifty way to multiply any number by
11.
Multiplying
a 2-digit number by 11
To multiply a two-digit number by 11, just add the two digits and place the sum between the two digits.
For
example, suppose you need to multiply 45
by 11. According to the rule, add 4 and 5 and place it between the 4 and 5
to get 495. It’s as simple as that.
Let’s try with 52. - Separate the two digits in your mind (5__2).
- Notice the hole between them!
- Add the 5 and the 2 together (5+2=7)
- Put the resulting 7 in the gap to get 572. That's it!
Similarly, 11 x 54=594.
The only thing tricky to remember is that if the sum
of the two digits is greater than 9, then you only place the ‘units’ digit
between the two digits of the number being multiplied by 11 and “carry” the
‘tens’ digit to be added to the left digit of the multiplicand.
For example 11
x 57 ... 5__7 ... 5+7=12 ... put the 2 in the hole and add the 1 from the
12 to the 5 in to get 6 for a result of 627 ... 11 x 57 = 627
Let’s try
with 78 × 11. Since, 7 + 8 = 15, we
place the 5 between the 7 and 8 and add the 1 to the 7, to get [7 + 1][5][8]
or 858.
Multiplying
a multi-digit number by 11
The
product for any larger non-zero integer can be found by a series of additions
to each of its digits from right to left; two at a time i.e. beginning
at the right-side digit adding pairs of digits next to each other going to
the left, except
for the numbers on the edges.
First place the ones digit of the multiplicand as it is. Next, starting with the ones digit, add each digit to the digit to its left and put the result. If the sum of any pair of digits is greater than 9; place the ‘units’ digit and carry the ‘tens’ digit over to the next addition. Finally copy the left-most digit of the multiplicand to the left of the result to get the final product.
Let's say, for example, you need to multiply 54321 by 11. First, let's look at the problem the traditional long way...
54321
x 11
54321
+ 543210
=
597531
|
Now let's look at the easy way...
54321 ×
11 = 5[5+4][4+3][3+2][2+1]1 = 597531
A step-by-step example of 759 × 11:
1. The ones digit of the multiplicand, 9, is copied to the temporary result.
=>
result: 9
2. Add 5 + 9
= 14; so 4 is placed on the left side of the result and carry the 1.
=>
result:
49
3. Similarly
add 7 + 5 = 12, and then add the carried 1 to get 13. Place 3 to the result and
carry the 1.
=>
result:
349
4. Add the
carried 1 to the highest valued digit in the multiplicand,
7 + 1 = 8, and copy to the result to finish.
=>
Final
product of 759 × 11: 8349
Do you see the pattern? In a way, you're simply adding the digit to whatever comes before it. But Because of the carries, it may be easier to work from right to left.
Further
example: We carry out the process step by step:
62473 × 11 =
6[6+2][2+4][4+7][7+3]3
= 6[6+2][2+4][4+7][10]3
= 6[6+2][2+4][4+7+1][0]3
= 6[6+2][2+4][12][0]3
= 6[6+2][2+4+1][2][0]3
= 6[6+2][7][2][0]3
= 6[8][7][2][0]3
= 687203
9527136 × 11 = 9[9+5][5+2][2+7][7+1][1+3][3+6]6 = 104798496
What is 11x978
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