Square Numbers, All Digits the SameDate: 07/10/2003 at 04:47:07 From: Elena Subject: About square numbers Is there any square number with all the same digits? Like 111, 222, etc. Date: 07/11/2003 at 09:56:22 From: Doctor Rob Subject: Re: About square numbers Thanks for writing to Ask Dr. Math, Elena. Aside from the one-digit squares 0, 1, 4, and 9, there are none. The proof goes like this. Any square number must end in 0, 1, 4, 5, 6, or 9. 0, 1, 4, and 9 are squares, but 5 and 6 are not. Now assume that the number has at least two digits. The possible two-digit endings are 00, X1, X4, 25, Y6, or X9, where X is an even digit and Y is an odd digit. That eliminates all but 00 and 44 as same-digit two-digit endings (since X cannot be 1 or 9 and Y cannot be 6). 00 and 44 are not squares. Now assume that the number has at least three digits. Then the possible three-digit endings are 000 and Y44, where Y is an odd digit. That leaves only 00...000 as a possibility, which is also impossible. I leave it to you to figure out why what I said above is true. Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 06/17/2003 at 02:28:26 From: Lucy Subject: Square Numbers A positive integer N is a square number and all its digits are the same. Find all such numbers N. The most difficult part for me is how to prove my answers. I'm pretty sure the only possible answers are 1, 4 and 9. But these answers only contain 1 digit and therefore every digit is of course the same. If these answers are right, but how do I prove there are no more answers? Date: 06/17/2003 at 03:58:51 From: Doctor Jacques Subject: Re: Square Numbers Hi Lucy, There are indeed no other possible answers. To see this, note first that the last digit of the square of a number only depends on the last digit of the number itself. For example: 7^2 = 49 17^2 = 289 37^2 = 1369 and so on. The square of a number that ends in 7 always ends in 9. This is because, if n^2 = A then (10+n)^2 = 100 + 20n + n^2 = 10(10+2n) + A i.e. when you add 10 to a number, you add a multiple of 10 to its square, which means that you don't change the last digit. In the same way, the last two digits of the square of a number only depend on the last two digits of the number, for example: 14^2 = 196 314^2 = 98596 and so on.. the square of a number that ends in 14 always ends in 96. You can read more details about this in the Dr. Math archives: Perfect Squares with Congruences http://mathforum.org/library/drmath/view/56081.html Let us now assume that there is a square number, N = a^2, with more than one digit, and that all its digits are the same. In particular, the last two digits will be the same. If you look at the last two digits of the squares of all the numbers up to 100 (actually, you only need to check up to 50), you will find that the only possibilities for the last two digits are those given above. In particular, the only case when the last two digits are the same is when these digits are 00 or 44. If all the digits of N are the same, it means that N consists only of 4's: N = a^2 = 44...44 As this is an even number, "a" itself must be even (if "a" were odd, its square would be odd also). So we may write a = 2b, and we have: N = (2b)^2 = 44...44 4*b^2 = 44...44 and, if we divide by 4, we get: b^2 = 11...11 which means that b^2 is a square number that ends in 11, and we have seen that this is not possible. This shows that there cannot be a square number with at least two digits where all digits are the same. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 06/18/2003 at 01:37:54 From: Lucy Subject: Thank you (Square Numbers) Thanks a lot for your time and effort! It has helped me greatly. |
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