Is 14798678562 or 15763530163289 a Perfect Square?Date: 12/08/2002 at 21:42:14 From: Christine Subject: Perfect square Hi, I am having problems figuring out this question that my teacher has asked the class. Is 14798678562 a perfect square? Is 15763530163289 a perfect square? Examine both the units digits and the digital roots of perfect squares to help determine whether or not a given number is a perfect square. Please explain your answers. Thanks for your help. Date: 12/08/2002 at 22:45:48 From: Doctor Ian Subject: Re: Perfect square Hi Christine, Suppose we have something that ends in a one, multiplied by itself: a 1 * a 1 ----- a 1 a 1 ----- ... 1 Whatever is going on in the other places, this is going to end in a 1. Now, consider the other digits: a 2 * a 2 ----- ... 4 ... ----- ... 4 a 3 * a 3 ----- ... 9 ... ----- ... 9 We can make a little table: (...0)^2 = ...0 (...1)^2 = ...1 (...2)^2 = ...4 (...3)^2 = ...9 (...4)^2 = ...6 (...5)^2 = ...5 (...6)^2 = ...6 (...7)^2 = ...9 (...8)^2 = ...4 (...9)^2 = ...1 So a perfect square can _only_ end in a 0, 1, 4, 5, 6, or 9. This should allow you to determine whether the first of your numbers is a perfect square. However, it isn't sufficient to draw a conclusion about the second number. But note that if we square numbers with more than two digits, a b c * a b c ------------- ac bc c^2 ab b^2 bc aa ab ac --------------------- the final two digits of the square will depend only on the final two digits of the number. So we could make a similar table for all the possibilities for the final two digits. Alas, even if you do this, it won't rule out numbers ending in 89, because '...89' is a possible square. However, according to Eric Weisstein's World of Mathematics: Square Number http://mathworld.wolfram.com/SquareNumber.html no number can be a perfect square unless its digital root is 1, 4, 7, or 9. You might already be familiar with computing digital roots if you've used the test for divisibility by 3. For example, to determine whether 2134576932400323 is divisible by 3, we add the digits: 2 + 1 + 3 + 4 + 5 + 7 + 6 + 9 + 3 + 2 + 4 + 0 + 0 + 3 + 2 + 3 = 54 then we do it again: 5 + 4 = 9 And we keep doing it until we end up with a single digit. If the digit is divisible by 3, then the entire number is also divisible by 3. So, find the digital root of your second number. If it's 1, 4, 7, or 9, the number _might_ be a perfect square, or it might not. But if the digital root isn't in this list, then the number is definitely _not_ a perfect square. Or you might consider this: Since the tests your teacher asked you to use are only conclusive in ruling out perfect squares, you might argue that the question only makes sense if neither of the numbers is a perfect square. Otherwise, you could run out of tests without determining the answer. I hope this helps! - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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