Odd Digits of Square Numbers
Date: 02/07/99 at 12:42:55 From: Kevin Bryceland Subject: Odd square numbers Dear Dr. Math, I cannot work out why there are no square numbers other than 1 and 9 that consist entirely of odd digits. Please help me. Yours thankfully, Kevin Bryceland
Date: 02/08/99 at 17:30:48 From: Doctor Nick Subject: Re: Odd square numbers Hi Kevin - The reason that these are the only two squares that have only odd digits is that a square number with more than 1 digit will have at least one even digit in the first two (that is, the right two) places. If n is even, then the right-most digit of n^2 is even. If n is odd, then the right-most digit of n^2 is odd, but the next digit is always even! A simple way to prove this is to use the fact that the right two digits of n^2 depend entirely on the right two digits of n. That means that we only have to check the squares of 1, 3, 5, 7, 9, 11, ..., 99 to see that they have an even number in the second place. You can cut the effort in half by noting that 1^2 and 99^2 end in the same two digits, 3^2 and 97^2 end in the same two digits, and so on, so you really only need to check the squares of 1, 3, 5, 7, ..., 49, which isn't too bad. In fact, you can cut it down even further. Notice that (50-m)^2 = 50^2 - 100m + m^2 so if 0 < m < 50, then m^2 and (50-m)^2 end in the same two digits. That means you only have to check 1, 3, 5, 7, ..., 25. Notice that the fact that we're using base 10 is very important here. In other bases there could be squares that have only odd digits. Have fun, - Doctor Nick, The Math Forum http://mathforum.org/dr.math/
Date: 04/01/01 at 13:17:04 From: Doctor Schwa Subject: Re: Odd square numbers To continue Dr. Nick's excellent response, you can cut down the checking even more by looking at (10a + b)^2 = 100a^2 + 20ab + b^2 The first term, 100a^2, doesn't affect the tens place at all; 20ab only changes the tens place by an even number, so you only need to check the squares of the one-digit numbers to verify that the tens place or the ones place is always even. As Dr. Nick already pointed out, if the ones place (b) is even, then b^2 will end with an even digit as well. So now there are only five numbers that need to be checked. --Dr. Schwa http://mathforum.org/dr.math/
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