Find all triplets (x, y, z) of positive integers that satisfy this equation:
2x+4y+8z= 328
In binary 328 = 101001000. Significantly there are only three 1's.
At first glance it looks like there might be 3!=6 solutions where each of x,y,z can cover a 1.
However, the 1's are in the 2^3, 2^6, 2^8 positions.
8^z can only cover 2^3 and 2^6 (powers multiples of 3)
4^y can only cover 2^6 and 2^8 (powers multiples of 2)
2^x can cover any left over.
If z=1, y can be 3 or 4
If z=2, y is 4
solutions
(8,3,1)
(6,4,1)
(3,4,2)
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Posted by Jer
on 2023-12-10 10:39:24 |
pure analytic solution
