Squaring Two-Digit Numbers Ending in 5

Date: 09/10/2001 at 22:03:08
From: Mary
Subject: Proving a squared number shortcut

I want to understand how the shortcut to squaring two-digit numbers 
ending in 5 works. The shortcut I am referring to is when you take 
the first digit (of a 2-digit number ending in 5), multiply it by the 
next consecutive number, and place it in front of 25. 

For example, 25 x 25 = 
   step 1: (2x3)= 6 
   step 2:  625. 

Can you prove this for me?

Date: 09/10/2001 at 22:41:03
From: Doctor Paul
Subject: Re: Proving a squared number shortcut

This is a nice little shortcut and its proof is nice and short too.  
The proof is realized when you understand what's really going on.

Pick your favorite two-digit number that ends in 5.  Let's call it

a5, where a could be 1, 2, 3, ..., 8, or 9

Then a5 is really a shorthand notation for the integer represented by 

10*a + 5.

Notice what happens when we square a5:

(a5)^2 = (10*a + 5)^2 = 100*a^2 + 100*a + 25 =

100(a^2 + a) + 25 = 100 * a * (a+1) + 25

and that is exactly the product of a and the next consecutive number 
with 25 placed after it.

Notice that this trick works for squaring any integer that ends in 
five - not just two-digit numbers that end in five. But the proof that 
it works for any integer would have to be modified a bit (since all 
integers that end in five cannot be represented by 10*a + 5).

See if you can prove it for 3-digit numbers that end in five.

Pick an arbitrary 3-digit number that ends in 5:  ab5

now rewrite ab5 = 100*a + 10*b + 5

square it and see that what you end up with is in fact what you think 
it should be.

For example:

665^2 can be computed by computing 66 * 67 = 4422 and appending 25.

Thus 665^2 = 442,225

I hope this is clear.  Please write back if you have any questions 
about what has been outlined above.

If you are interested in more "tricks" that simplify computations, get 
a copy of either of these books by Edward H. Julius: 

Rapid Math Tricks and Tips
More Rapid Math Tricks and Tips

If you've got a bit more money to spend, try The Trachtenberg Speed 
System of Basic Mathematics by Ann Cutler (translator).  You can get a 
bit of a preview about the Trachtenberg Speed System here:

   http://mathforum.org/dr.math/faq/faq.trachten.html   

Or visit the BEATCALC area of the Math Forum:

   http://mathforum.org/k12/mathtips/beatcalc.html   

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/