1. Interpret the sum as a geometric series

The given sum is $49^{125}+49^{124}+\cdots+49^2+49+1.$

This is a geometric progression with first term $1$, common ratio $49$, and total number of terms $126$. Hence, $S=1+49+49^2+\cdots+49^{125}=\frac{49^{126}-1}{49-1}.$ So, $S=\frac{49^{126}-1}{48}.$

2. Condition for divisibility by $49^k+1$

We want the greatest positive integer $k$ such that $49^k+1 \mid S.$

Since $S=\frac{49^{126}-1}{48},$ it is enough to study when $49^k+1$ divides $49^{126}-1$.

Let $a=49.$ Then we need the greatest $k$ such that $a^k+1 \mid a^{126}-1.$

3. Use the standard divisibility result

A standard result is: $a^m+1 \mid a^n-1 \quad \text{if } \frac{n}{m} \text{ is an odd multiple of }2,$ but more precisely, $a^m+1 \mid a^n-1 \iff n \text{ is an odd multiple of } m \text{?}$ Let us derive carefully.

If $a^k+1$ divides $a^{126}-1$, then modulo $a^k+1$, $a^k\equiv -1.$ Therefore, $a^{126}=(a^k)^{126/k} \equiv (-1)^{126/k}.$ For this to be congruent to $1$, we need:

• $k$ must divide $126$, and
• $\dfrac{126}{k}$ must be even.

Thus, $\frac{126}{k} \text{ is even} \iff 126=2k\cdot t \iff k\mid 63.$ So the greatest such $k$ is the greatest divisor of $63$, namely $k=63.$

4. Check directly

For $k=63$, $49^{63}+1 \mid 49^{126}-1=(49^{63}-1)(49^{63}+1).$ Hence $49^{63}+1$ divides $49^{126}-1$.

Also, since $49^{63}+1$ is odd and $48=2^4\cdot 3,$ we have $\gcd(49^{63}+1,48)=2.$ But the original expression is $S=\frac{49^{126}-1}{48},$ and from the geometric factorization, $S=\frac{(49^{63}-1)(49^{63}+1)}{48}.$ This indeed remains divisible by $49^{63}+1$ because $49^{63}-1=(49-1)(1+49+\cdots+49^{62})=48(1+49+\cdots+49^{62}),$ so $S=(49^{63}+1)(1+49+\cdots+49^{62}).$ Thus $49^{63}+1$ is definitely a factor.

5. Greatest possible value

We need the greatest divisor $k$ of $63$. That is clearly $k=63.$

6. Evaluate options

• A: 32 — not greatest.
• B: 60 — $60\nmid 63$, so not possible.
• C: 63 — correct.
• D: 65 — not possible.

Therefore, the correct option is $\boxed{63}.$