A positive integer is expressible as 3X1 in base B and 1X3 in base B+6, where X is a base B (and also base B+6) digit.
Find all values of B and X.
Prove that no further solution is possible.
*** Of course, B is a positive integer.
3*b^2 + x*b + 1 = (b+6)^2 + x*(b+6) + 3
= b^2 + 12*b + 36 + x*b + 6*x + 3
2*b^2 - 12*b - 6*x - 38 = 0
b^2 - 6*b - 3*x - 19 = 0
b = (6 + sqrt(36 + 12*x + 76)) / 2
b = 3 + sqrt(9 + 3*x +19) = 3 + sqrt(28 + 3*x)
x b
7 10
12 11
24 13
...
All except the first are invalid as the digit x is larger than the base.
(Table function of graphing calculator app equivalent used; only integral b shown.)
Edited on May 2, 2016, 2:36 pm
|
Posted by Charlie
on 2016-05-02 14:35:06 |