20050310, 03:20  #1 
Jun 2003
The Computer
3×131 Posts 
Multiplication Tendency
This seems to work all the time.
Square an integer. i.e. 2^2=4. Now, take the integer below it (1) and above it (3) and multiply it together. You get 3 and it is one less than the previous integers. So in other words... x^2= [(x1)*(x+1)]1. Is this already known? It seems quite simple, yet true. 
20050310, 03:46  #2 
Aug 2004
2^{2}·5 Posts 
Your are right. Here is why:
Last fiddled with by amateur on 20050310 at 03:49 
20050310, 04:06  #3 
Jun 2003
3·17·101 Posts 
The basic identitiy (a+b)(ab) = a^2b^2 is at work here.
Set a=x, b=1, and rearrange the terms, you'll get your result. 
20050310, 04:57  #4 
Jul 2004
Potsdam, Germany
3×277 Posts 
Personally, I like the formula a*b = ((a+b)/2)²  ((ab)/2)²
It looks complicated at first glance, but it helps me doing quick multiplication (mentally) in certain situations. Example: 21*19 = 20²  1² = 399 22*18 = 20²  2² = 396 
20050310, 20:32  #5  
∂^{2}ω=0
Sep 2002
República de California
10110110001110_{2} Posts 
Quote:


20050311, 00:23  #6  
Jun 2003
The Computer
3×131 Posts 
Quote:


Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
A Fib expression with multiplication  MattcAnderson  Homework Help  5  20161101 08:16 
multiplication and logarithm  bhelmes  Math  4  20161006 13:33 
Possible better multiplication algorithm  Triber  Computer Science & Computational Number Theory  17  20151110 17:48 
5 digit multiplication  MattcAnderson  Puzzles  8  20141205 15:09 
Montgomery Multiplication  dave_dm  Math  2  20041224 11:00 