unconventional math

[Baby Astronaut]

Transform each number into its square using only the digits of each number. You can add, subtract, and/or multiply the digits in various combinations, but all numbers must use the same exact formula. Hint: uses an old-school calculation algorithm when necessary. I came up with this in 3rd grade.

 

10

 

11

 

12

 

13

 

14

 

15

 

16

 

17

 

18

 

19

 

20

 

If you got the correct formula, it'll also work on numbers 0-9.

 

But can any formula work beyond 20?

[mrburns2012]

Yes, there's a formula for all 2 digit numbers (and from 0 to 9):

 

(a*10+b)^2 = (a+a+a+a+a+a+a+a+a+a+b)*(a+a+a+a+a+a+a+a+a+a+b)

 

In fact, there's one for 3 digit numbers too.

Edited January 21, 2009 by mrburns2012

[Baby Astronaut]

That's not it. And just so I don't confuse anyone, let me clear up two points.

 

• By formula I meant what paths/steps got you to the answer.
  
• When I asked if any formula can work beyond 20, I asked because I don't know if one exists. Don't take it to mean your formula has to work for numbers over 20 to be correct.
  

 

Another hint: there aren't more than 5 steps. For example, n + n + n + n + n + n is already wrong.

 

I was nine years old at the time I created it. And only knew basic elementary math like subtraction, addition, multiplication, etc.

[mrburns2012]

That's not it. And just so I don't confuse anyone, let me clear up two points.

 

• By formula I meant what paths/steps got you to the answer.
  
• When I asked if any formula can work beyond 20, I asked because I don't know if one exists. Don't take it to mean your formula has to work for numbers over 20 to be correct.
  

 

Another hint: there aren't more than 5 steps. For example, n + n + n + n + n + n is already wrong.

 

I was nine years old at the time I created it. And only knew basic elementary math like subtraction, addition, multiplication, etc.

 

Well, the most difficult target is a moving one.... I'll let someone else pick up the challenge under the new rules.

[Baby Astronaut]

I apologize for that. Of course your formula works, nice one!

 

Here is the solution I had in mind.

 

1. Multiply the second digit by itself.

2. Carry over if applicable.

3. Add the second digit to itself.

4. Carry over if applicable.

5. Multiply the first digit by itself.

 

I haven't gotten it to work for numbers over 20 though. If anyone can find one that works, please let me know. I originally wanted to multiply numbers without having to create a multiplication table, and only got as far as two digit numbers below twenty. You can use different numbers as well, for example 15 x 17, but they each must be 20 or less.

 

Anyone want to try finding a method that works for any value or number of digits?

[mrburns2012]

Good job. The only square I knew of when I was nine, was a shape with 4 equal sides.

 

ha, that almost rhymed.

[sandy85]

I have solutions for this math :

Multiply the second digit - Carry over if applicable - Add the second digit - Carry over if applicable - Multiply the first digit.