Multiply with 11 fast

Is there a way to multiply a two digits number with 11 fast?



Multiplying a two digits number with 11 is actually very simple: you just add the two digits and put them in the middle.

23*11=?
We just add 2 and 3 and put the result in the middle:
2+3=5
Now we put 5 between 2 and 3:
23*11=253

n:

11 * n:

Other examples:
34*11=374
72*11=792
22*11=242
53*11=583
54*11=594

But what if the sum of the two digits is bigger then 9? Do we still put the result in the middle?

Yes, but only the last digit of the sum. And we add 1 two the first digit of the initial number.

57*11=?
5+7=12
We put the last digit of the sum in the middle and we add 1 to the first digit of the initial number:
57*11=627
because:
5+1=6
2 is the last digit of 12
7 is the last digit of the initial number

What if the sum of the two digits is bigger then 9 and the first digit is 9?
95*11=?
9+5=14
We put the last digit of the sum in the middle and add 1 to the first digit of the initial number:
95*11=1045
In this case we have 4 digits numbers that start with this two digits 10.

Tests:
Part 1
24*11=?
26*11=?
31*11=?
17*11=?
36*11=?
81*11=?
45*11=?

Part 2
68*11=?
49*11=?
87*11=?
29*11=?
65*11=?
79*11=?
89*11=?

Part 3
93*11=?
97*11=?
95*11=?

Combinated
33*11=?
59*11=?
98*11=?
84*11=?
94*11=?
34*11=?

How does this work?

Let's say we have the two digits number $ \overline{ab} $

$ \overline{ab}*11=(a*10+b)*11=(a*10+b)*(10+1)= $
$ =(a*10+b)*10+(a*10+b)*1=a*100+b*10+a*10+b= $
$ =a*100+(a+b)*10+b=a*100+b+(a+b)*10= $
$ =\overline{a0b}+(a+b)*10 $

Let $ r=a+b $
if $ r\leq9 $ then $ (a+b)*10 = \overline{r0} $
$ \overline{a0b}+\overline{r0}=\overline{arb} $

if $ r\g9 $ then $ (a+b)*10 = \overline{1e0} $
$ \overline{a0b}+\overline{1e0}=\overline{aeb}+100 $

Average: 4.3 (39 votes)

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