Perfect Square?Date: 02/18/2002 at 14:34:15 From: charles Subject: Advanced algebra If we use the digits 1,2,3,4,5,6,7 each only once to form a 7-digit number, can the resulting number be a perfect square? Date: 02/18/2002 at 20:47:51 From: Doctor Paul Subject: Re: Advanced algebra No. There are 7! = 5040 different ways to write a number that contains each of these digits exactly once. I had Maple (a math programming language) do this for me: > restart; > with(combinat, permute); [permute] > > K:=permute(7,7): > nops(K); 5040 > K[1]; [1, 2, 3, 4, 5, 6, 7] I created an array and called it K. K contains all 5040 permutions of the elements 1 through 7. But they aren't in integer format - they're in arrays. Just so you get the idea, the first couple of lines of K look like this: [[1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 7, 6], [1, 2, 3, 4, 6, 5, 7], [1, 2, 3, 4, 6, 7, 5], [1, 2, 3, 4, 7, 5, 6], [1, 2, 3, 4, 7, 6, 5], [1, 2, 3, 5, 4, 6, 7], [1, 2, 3, 5, 4, 7, 6], [1, 2, 3, 5, 6, 4, 7], [1, 2, 3, 5, 6, 7, 4], [1, 2, 3, 5, 7, 4, 6], [1, 2, 3, 5, 7, 6, 4], so what we need to do is convert each of these to integers and then check to see if it's a perfect square. To convert to an integer, just take each one and do something like the following: 7 + 6*10 + 5*10^2 + 4*10^3 + 3*10^4 + 2*10^5 + 1*10^6 = 1,234,567 I did this in Maple as well. I converted each array to an integer and then checked to see if it was a perfect square by seeing if the fractional part of its square root was zero: > for i from 1 to 5040 do > m:=0: > for j from 1 to 7 do > m:=m + K[i][j] * 10^(7-j); > if j = 7 and frac(sqrt(m)) = 0 then print(m) fi; > od: > od: Inasmuch as the output was blank, none of the 5040 combinations yields a perfect square. If you'd like to talk some more about my answer, feel free to write back. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/