Determine all possible triplets (a, b, c) of real numbers satisfying the following system of equations, where [n] denotes the greatest integer ≤n, and {n}=n–[n].
| {a} |
+ |
b |
+ |
[c] |
= |
5.5 |
| a |
+ |
[b] |
+ |
{c} |
= |
6.7 |
| [a] |
+ |
{b} |
+ |
c |
= |
8.4 |
Letting A={a}, B={b} and C={c}, it must be that A+B=0.5 or 1.5, A+C=0.7 or 1.7, and B+C=0.4 or 1.4, with 0<=A,B,C<1.
If we solve the generic system A+B=X, A+C=Y, B+C=Z, we get A=(X+Y-Z)/2, B=(X-Y+Z)/2 and C=(-X+Y+Z)/2.
Trying all eight possible combinations of values for A, B and C, we get valid solutions only for X=0.5, Y=0.7 and Z=0.4 (when A=0.4, B=0.1 and C=0.3) or for X=1.5, Y=1.7 and Z=1.4 (when A=0.9, B=0.6 and C=0.8).
In the first case we must solve
0.4 + [b]+0.1 + [c] = 5.5, so [b]+[c]=5
[a]+0.4 + [b] + 0.3 = 6.7, so [a]+[b]=6
[a] + 0.1 + [c]+0.3 = 8.4, so [a]+[c]=8
but this produces non-integer values for some of the variables.
In the second case we must solve
0.9 + [b]+0.6 + [c] = 5.5, so [b]+[c]=4
[a]+0.9 + [b] + 0.8 = 6.7, so [a]+[b]=5
[a] + 0.6 + [c]+0.8 = 8.4, so [a]+[c]=7
which produces [a]=4, [b]=1, [c]=3.
The original problem has one solution only:
a=4.9,
b=1.6 and
c=3.8.
Solutions