Question

50C36 is divisible by 19,25,53,192

• A. 19
• B. 25
• C. 53
• D. 192

Answer 1

Answer: (A), (B)

19, 25

Answer 2

To determine which of the given numbers (19, 25, 53, 192) divide 50C36, we first simplify the binomial coefficient and then analyze its prime factorization.

1. Simplify the binomial coefficient: Using the property $\binom{n}{k} = \binom{n}{n-k}$, we have: $\binom{50}{36} = \binom{50}{50-36} = \binom{50}{14}$

   The expression for $\binom{50}{14}$ is: $\binom{50}{14} = \frac{50 \times 49 \times 48 \times 47 \times 46 \times 45 \times 44 \times 43 \times 42 \times 41 \times 40 \times 39 \times 38 \times 37}{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$

2. Check divisibility by 19: Observe the numerator of $\binom{50}{14}$. It contains the term $38$. $38 = 2 \times 19$. The denominator is $14! = 14 \times 13 \times \dots \times 1$. None of the numbers from 1 to 14 are multiples of 19. Since 19 is a prime number and it is a factor in the numerator but not in the denominator, $\binom{50}{14}$ is divisible by 19.

3. Check divisibility by 25: We need to find the exponent of the prime factor 5 in $\binom{50}{14}$. We use Legendre's formula, which states that the exponent of a prime $p$ in $n!$ is $E_p(n!) = \sum_{i=1}^{\infty} \lfloor \frac{n}{p^i} \rfloor$. The exponent of $p$ in $\binom{n}{k}$ is $E_p(n!) - E_p(k!) - E_p((n-k)!)$. For $p=5$, $n=50$, $k=14$: $E_5(50!) = \lfloor \frac{50}{5} \rfloor + \lfloor \frac{50}{25} \rfloor = 10 + 2 = 12$. $E_5(14!) = \lfloor \frac{14}{5} \rfloor = 2$. $E_5(36!) = \lfloor \frac{36}{5} \rfloor + \lfloor \frac{36}{25} \rfloor = 7 + 1 = 8$. The exponent of 5 in $\binom{50}{14}$ is $E_5(\binom{50}{14}) = E_5(50!) - E_5(14!) - E_5(36!) = 12 - 2 - 8 = 2$. Since the power of 5 in $\binom{50}{14}$ is 2, it is divisible by $5^2 = 25$.

4. Check divisibility by 53: 53 is a prime number. For a binomial coefficient $\binom{n}{k}$ to be divisible by a prime $p$, the prime $p$ must be less than or equal to $n$. In this case, $n=50$ and $p=53$. Since $53 > 50$, 53 cannot be a factor of $50!$. Therefore, $\binom{50}{14}$ cannot be divisible by 53.

5. Check divisibility by 192: First, find the prime factorization of 192: $192 = 2 \times 96 = 2^2 \times 48 = 2^3 \times 24 = 2^4 \times 12 = 2^5 \times 6 = 2^6 \times 3$. For $\binom{50}{14}$ to be divisible by 192, it must be divisible by $2^6$ and $3$.

   Let's find the exponent of the prime factor 2 in $\binom{50}{14}$: For $p=2$, $n=50$, $k=14$: $E_2(50!) = \lfloor \frac{50}{2} \rfloor + \lfloor \frac{50}{4} \rfloor + \lfloor \frac{50}{8} \rfloor + \lfloor \frac{50}{16} \rfloor + \lfloor \frac{50}{32} \rfloor = 25 + 12 + 6 + 3 + 1 = 47$. $E_2(14!) = \lfloor \frac{14}{2} \rfloor + \lfloor \frac{14}{4} \rfloor + \lfloor \frac{14}{8} \rfloor = 7 + 3 + 1 = 11$. $E_2(36!) = \lfloor \frac{36}{2} \rfloor + \lfloor \frac{36}{4} \rfloor + \lfloor \frac{36}{8} \rfloor + \lfloor \frac{36}{16} \rfloor + \lfloor \frac{36}{32} \rfloor = 18 + 9 + 4 + 2 + 1 = 34$. The exponent of 2 in $\binom{50}{14}$ is $E_2(\binom{50}{14}) = E_2(50!) - E_2(14!) - E_2(36!) = 47 - 11 - 34 = 2$. The power of 2 in $\binom{50}{14}$ is $2^2 = 4$. Since 192 requires a factor of $2^6 = 64$, and $\binom{50}{14}$ only has $2^2$ as a factor, it is not divisible by $2^6$. Therefore, $\binom{50}{14}$ is not divisible by 192.

Based on the analysis, 50C36 is divisible by 19 and 25.