Squaring Two-Digit Numbers Ending in 5Date: 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/ |
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