If a certain six digit number is split into two parts, one constituting the first three digits and the other the last three digits, and the two parts are added and the resulting sum squared, it is found that the product is the original six-digit number. What is the number?
2007-12-31 05:26:22 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Let the 6 digit number be represented by
x + 1000y
where y and x are the first and last 3 digits respectively.
So x + 1000y = (x+y)^2
We need to find 3 digit integral solutions for x and y satisfying the above equation.
x + 1000y = x^2 + 2xy + y^2
x^2 + (2y-1)x + (y^2 - 1000y) = 0
x = {(1-2y) ± √[(2y-1)^2 - 4*(y^2 - 1000y)] }/2
= {(1-2y) ± √(3996y + 1) }/2
So 3996y + 1 must be a perfect square, k^2
3996y = k^2 - 1 = (k+1)(k-1)
Using an excel spreadsheet to test the possible values of y (I'm sure there is a more classical approach), only y = 494 and 998 result in perfect squares. 494 results in x = 209, but 998 results in x = 1.
answer: 494 209
*EDIT*
Some of the answerers are saying 998001, which works mathematically, but then you have to decide if 001 qualifies as a 3 digit number.
2007-12-31 05:35:54 · answer #1 · answered by Dr D 7 · 3⤊ 0⤋
494209 works.
So does 998001.
2007-12-31 13:57:26 · answer #2 · answered by fcas80 7 · 0⤊ 0⤋
Dr. D did all the heavy lifting - the answer is 998001
2007-12-31 13:53:19 · answer #3 · answered by Kenneth L 1 · 0⤊ 0⤋
An imaginary number??!?!?!?
Sorry Im dont know..this is hard..it is 666666
2007-12-31 13:35:13 · answer #4 · answered by saconners1 6 · 0⤊ 1⤋
omg thats a hard question good luck and sorry you prolly hate people that give answers like this
2007-12-31 13:34:25 · answer #5 · answered by trackhunni08 2 · 0⤊ 1⤋