Question

The exponent of 7 in \( ^{100}C_{50} \) is :

• A. \( 4 \)
• B. \( 2 \)
• C. \( 1 \)
• D. \( 0 \) Correct Answer: (D) \( 0 \) Solution:

Hint:

Kummer's Theorem: The exponent of a prime \( p \) in \( ^nC_r \) is the number of "carries" when adding \( r \) and \( n-r \) in base \( p \).
For \( 50 + 50 \) in base 7:
\( 50 = (1 \cdot 7^2) + (0 \cdot 7^1) + (1 \cdot 7^0) = 101_7 \).
Adding \( 101_7 + 101_7 = 202_7 \).
Since there were 0 carries during the addition, the exponent is 0.

Answer 1

The Correct Option is D

Step 1: Set up what we need.
We want how many times $7$ divides $\,^{100}C_{50} = \dfrac{100!}{50!\,50!}$. Count the $7$s on top, then subtract the $7$s on the bottom.

Step 2: Count factors of $7$ in $100!$.
Add the floor terms.
\[ \left\lfloor\frac{100}{7}\right\rfloor + \left\lfloor\frac{100}{49}\right\rfloor = 14 + 2 = 16 \]

Step 3: Count factors of $7$ in $50!$.
\[ \left\lfloor\frac{50}{7}\right\rfloor + \left\lfloor\frac{50}{49}\right\rfloor = 7 + 1 = 8 \]
The bottom has two copies of $50!$, so that is $8 + 8 = 16$.

Step 4: Subtract.
\[ 16 - 16 = 0 \]
So $7$ does not divide the number at all. That is option 4.
\[ \boxed{0} \]