Let $a,b,c,d,e,f$ be six real numbers with sum 10, such that \[ (a-1)^2+(b-1)^2+(c-1)^2+(d-1)^2+(e-1)^2+(f-1)^2 = 6. \] Find the maximum possible value of $f$. 

[i]Cyprus[/i]
1. Let \( A = a-1 \), \( B = b-1 \), \( C = c-1 \), \( D = d-1 \), \( E = e-1 \), and \( F = f-1 \). Then we have:
   \[
   A^2 + B^2 + C^2 + D^2 + E^2 + F^2 = 6
   \]
   and
   \[
   A + B + C + D + E + F = 4
   \]

2. We need to find the maximum possible value of \( f \). Since \( f = F + 1 \), we need to maximize \( F \).

3. From the given conditions, we can express the sum of \( A, B, C, D, \) and \( E \) in terms of \( F \):
   \[
   A + B + C + D + E = 4 - F
   \]
   and
   \[
   A^2 + B^2 + C^2 + D^2 + E^2 = 6 - F^2
   \]

4. By the Cauchy-Schwarz inequality, we have:
   \[
   5(A^2 + B^2 + C^2 + D^2 + E^2) \geq (A + B + C + D + E)^2
   \]
   Substituting the known values, we get:
   \[
   5(6 - F^2) \geq (4 - F)^2
   \]

5. Expanding and simplifying the inequality:
   \[
   30 - 5F^2 \geq 16 - 8F + F^2
   \]
   \[
   30 - 5F^2 \geq 16 - 8F + F^2
   \]
   \[
   30 - 16 \geq 5F^2 + F^2 - 8F
   \]
   \[
   14 \geq 6F^2 - 8F
   \]
   \[
   6F^2 + 8F - 14 \leq 0
   \]

6. Solving the quadratic inequality \( 6F^2 + 8F - 14 \leq 0 \):
   \[
   3F^2 + 4F - 7 \leq 0
   \]
   The roots of the quadratic equation \( 3F^2 + 4F - 7 = 0 \) are found using the quadratic formula:
   \[
   F = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \( a = 3 \), \( b = 4 \), and \( c = -7 \):
   \[
   F = \frac{-4 \pm \sqrt{16 + 84}}{6}
   \]
   \[
   F = \frac{-4 \pm \sqrt{100}}{6}
   \]
   \[
   F = \frac{-4 \pm 10}{6}
   \]
   \[
   F = 1 \quad \text{or} \quad F = -\frac{7}{3}
   \]

7. The inequality \( 3F^2 + 4F - 7 \leq 0 \) holds for:
   \[
   -\frac{7}{3} \leq F \leq 1
   \]

8. Therefore, the maximum possible value of \( F \) is \( 1 \). Since \( f = F + 1 \), the maximum possible value of \( f \) is:
   \[
   f = 1 + 1 = 2
   \]

The final answer is \(\boxed{2}\).