Anyone Into The Number 11 ? Warning: Pure math ahead.

[iGuest 3]

WARNING: MATH AHEAD

MATH HATERS, GO AWAY. THIS IS YOUR LAST WARNING.

 

Hey guys,

Just got a really nice tip from my class teacher, or should I say math teacher. This is a nice way to find multiples of 11.

 

Here is how you do it.

For example if you want to find a the product of 248 when multipied with the factor 11.

 

Then see this,

Question: 248x11

Solution: 2(4+8)8 ---> We are not multiping here, just put the numbers 4 and 8 in braces to show that they are being multiplied, 4 and 8 are the last digits that make up the number.

2(12)8 ---> Since we cant write 12 just like that, we carry the one (1) from the number twelve (12) to the digit on the left, i.e. two (2).

So it becomes 2+1=3

 

Now write down all the numbers you have got in the same order.

Final Answer = 328

 

See another example and you will under-stand better.

 

Question: 123x11

Soltion: 1(2+3)3 ---> 1 is the first digit and the 3 is the last digit that make the number so keep them the first and last number even when your finding the product of the factor 11.

 

1(5)3

1(1+2)53 ---> here 1 and 2 are the next two digits from the right.

 

Final Answer: 1353

 

If you still have'nt understood it completely make a post below and I'll edit this post until you understand.

Happy multipling.

[iGuest 3]

OMG, I actually understood it.Guess I am either lazy enough to want to know how to do that, or it is a well written posting. I vote for the latter.

[evion 0]

I actually saw this on a video once but i forgot where and how.... Though I'd suggest that you show the easier method of multiplying 11s first, that is, using 2 digits.Say you have the number 43 and you want to multiply it with 11. All you have to do is 4+3=7. Then add 7 between the 2 digits and you get your answer 473! Of course, this will only work as easy as this if you have 2 digits that add up to less than or equal to 9. Anymore than 10 you will have to carry over the extra 1.

[iGuest 3]

248 x 11 is 2728You work it out like so (hard to explain)The first and last number go outside, while we reserve two places e.g 2(?)(?)8Then you add the first 2 numbers together, which go in the first bracket, you then add the last 2 numbers which goes in the last bracket.You work it out, carry the 1's if any to the next number.2(2+4)(4+8)82(6)(12)8carry the 12(7)(2)8and that's your answer, 2728.123 x 111(1+2)(2+3)3= 1353Let's make it harder666 x 116(6+6)(6+6)66(12)(12)6=7326I've not tested this with all numbers but I believe this rule to work 100%, I explained the 11 times any double digits in the verdics maths discussion here at Xisto because I thought it may have derived from verdics.Cheers,MC

[miCRoSCoPiC^eaRthLinG 0]

This method has been shown in Vedic maths as well as the Trachtenberg Speed System of Basic Mathematics - which was devised by the late Professor Jakow Trachtenberg while he was captive in one of the concentration camps of Hilter in Germany.

 

In order to maintain his sanity inside the camp, he started fiddling around with digits and gradually this new concept of ultra-high-speed mental calculations took shape. This is a highly established technique and during public demonstrations, have been very effectively utilized by little kids to multiply massive figures of 10 digits by 11 digits just under a short few minutes.

 

I have his book named the same - and it covers all aspects of Arithmatics, i.e. Addition, Subtraction, Multiplication & Division..

 

There's a software out too to teach you the same..

 

More info: http://www.speed-math.com/

[green_magi0 0]

plese post a version of this solution in english because im totally lost and how can 123*11 be more than that 200*11 problem? something went wrong there lmao but yes. I would like to see a more simplified versio nof this if there IS one.

[iGuest 3]

plese post a version of this solution in english because im totally lost and how can 123*11 be more than that 200*11 problem? something went wrong there lmao but yes. I would like to see a more simplified versio nof this if there IS one.

123 * 11 is not more than 200 * 11.

See we have 200, from my rules you take the outer numbers which are 2 and 0 and you put them on the end 2(?)(?)0 already I can tell you it's more than 123 because 1(?)(?)3

The only number missing from 2(?)(?)0 is the middle number 0 which we use to add with the first number and the last number to get the center bits.

2(2+0)(0+0)0 the answer is 2200

1stNum * 1000 + (1stNum + 2ndNum) * 100 + (2ndNum + 3rdNum) * 10 + 3rdNum

If I were to use 205 then it'd look like this

2 * 1000 + (2 + 0) * 100 + (0 + 5) * 10 + 5 = 2255

You can basically eliminate the multiplication part of this, this is just determining their position, 1000s, 100s, 10s, 1s.

I don't know how you can simplify it, basically you're not really using complex maths (or more commonly the maths we're taught) but using simple patterns and equations to get the answer. Now let's say I asked you what is 1234 * 11 and you had to work it out in your head, it uses the same rules, but adds and extra number. e.g.

1234 * 11

1 * 10000 + (1 + 2) * 1000 + (2 + 3) * 100 + (3 + 4) * 10 + 4 = 13574

but my way is:

1(?)(?)(?)4 where I know the outside numbers and if the 2nd digit i s going to be greater than 10, then I know I must adjust that outter number, the last digit is always the same.

Then all I have left to do is
1(1+2)(2+3)(3+4)4 but this is not maths, it's just a quick method for getting the answer.

To show how we usually do it in maths

  1234+12340

So you could impress people with really large numbers using this rule, just making sure you can keep it all in your head would be good.

Cheers,

MC

[lonelym 0]

That is very cool. I did notice that every last and first number was always the same. It was the middle that kept changing. Anyways, I prefer to multiply it to ten then just add an extra.

 

Example:

110 * 11 = (?)

(110 * 10) + (110) = 1210

 

So... its like doing half multiplication, and just adding an extra.

[sparkx 0]

I remember this from 6th grade math. It was one of the tricks for remembering your multiplication tables from 1-15. I forgot about it though. It is one of the things that will help you for about 2 months untill you forget about it again (at least I know that is true for me). I had also learned the little thing with 111*111=12321. You can do that with any number of 1's so 11*11=121. Its funny how that works out.Thanks,Sparkx

[Lydubs 0]

Thanks so much for this!Once I actually fully understand it , I'll teach it to my friends.I don't get how the first one, 248x11 is only 328 and the second one 123x11 could possibly be more.Don't worry about me, I (might) figure it out.

Edited November 8, 2007 by RoDanie (see edit history)

[Jared 0]

This method has actually got quite a simple proof. Just consider that abcdef is a 6 digit number with a being the first, b being the second, c being the third, etc. What happens when we multiply it by 11 is:

 

abcdef * 11= abcdef * 10 + abcdef * 1= abcdef0 + abcdefSo abcdef0 is a 7-digit number with 0 being the final digit.Similarily, we can add a 0 in front of the other number since 0's in front of a number mean nothing--they are just placeholders.So the product becomes:  abcdef0+ 0abcdef---------This naturally can be translated as:(a+0)(a+b)(b+c)(c+d)(d+e)(e+f)(f+0)

= a(a+(b+c)(c+d)(d+e)(e+f)f

 

And we naturally see in the addition part of the multiplication problem that when we add e to f, we would naturally carry the 1 to join d + e. So it makes perfect sense!

 

- Jared