Consider the set of all ordered $6$-tuples of nonnegative integers $(a,b,c,d,e,f)$ such that \[a+2b+6c+30d+210e+2310f=2^{15}.\] In the tuple with the property that $a+b+c+d+e+f$ is minimized, what is the value of $c$?


[i]2021 CCA Math Bonanza Tiebreaker Round #1[/i]
1. We start with the equation given in the problem:
   \[
   a + 2b + 6c + 30d + 210e + 2310f = 2^{15}
   \]
   We need to find the tuple \((a, b, c, d, e, f)\) such that \(a + b + c + d + e + f\) is minimized.

2. First, we calculate \(2^{15}\):
   \[
   2^{15} = 32768
   \]

3. We use the Greedy Algorithm to minimize the sum \(a + b + c + d + e + f\). We start with the largest coefficient, which is 2310.

4. Determine the maximum value of \(f\) such that \(2310f \leq 32768\):
   \[
   f = \left\lfloor \frac{32768}{2310} \right\rfloor = 14
   \]
   This gives:
   \[
   2310 \times 14 = 32340
   \]
   Subtract this from 32768:
   \[
   32768 - 32340 = 428
   \]

5. Next, we consider the coefficient 210 for \(e\):
   \[
   e = \left\lfloor \frac{428}{210} \right\rfloor = 2
   \]
   This gives:
   \[
   210 \times 2 = 420
   \]
   Subtract this from 428:
   \[
   428 - 420 = 8
   \]

6. Now, we have:
   \[
   a + 2b + 6c + 30d = 8
   \]
   We need to minimize \(a + b + c + d\).

7. Since 30 is larger than 8, \(d = 0\).

8. Next, we consider the coefficient 6 for \(c\):
   \[
   c = \left\lfloor \frac{8}{6} \right\rfloor = 1
   \]
   This gives:
   \[
   6 \times 1 = 6
   \]
   Subtract this from 8:
   \[
   8 - 6 = 2
   \]

9. Now, we have:
   \[
   a + 2b = 2
   \]

10. To minimize \(a + b\), we set \(b = 1\) and \(a = 0\):
    \[
    2 \times 1 = 2
    \]

11. Therefore, the tuple \((a, b, c, d, e, f)\) that minimizes \(a + b + c + d + e + f\) is:
    \[
    (0, 1, 1, 0, 2, 14)
    \]

12. The value of \(c\) in this tuple is:
    \[
    c = 1
    \]

The final answer is \(\boxed{1}\)