We are given the task to find the largest value of \(n\) for which \(40^n\) divides \(60!\).
First, let's express 40 in terms of its prime factors:
\(40 = 2^3 \times 5^1\)
Therefore, \(40^n = (2^3 \times 5^1)^n = 2^{3n} \times 5^n\).
To determine the maximum value of \(n\) such that \(2^{3n} \times 5^n\) divides \(60!\), we first need to find the number of times 2 and 5 appear in the prime factorization of \(60!\).
The formula to find the exponent of a prime \(p\) in \(n!\) is:
\(\text{Exponent of } p \text{ in } n! = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots\)
Let's apply this to primes 2 and 5:
Calculating the power of 2 in \(60!\):
- \(\left\lfloor \frac{60}{2} \right\rfloor = 30\)
- \(\left\lfloor \frac{60}{4} \right\rfloor = 15\)
- \(\left\lfloor \frac{60}{8} \right\rfloor = 7\)
- \(\left\lfloor \frac{60}{16} \right\rfloor = 3\)
- \(\left\lfloor \frac{60}{32} \right\rfloor = 1\)
Total power of 2 = \(30 + 15 + 7 + 3 + 1 = 56\)
Calculating the power of 5 in \(60!\):
- \(\left\lfloor \frac{60}{5} \right\rfloor = 12\)
- \(\left\lfloor \frac{60}{25} \right\rfloor = 2\)
Total power of 5 = \(12 + 2 = 14\)
Now, set up the equations for division:
- \(3n \leq 56\) (for power of 2)
- \(n \leq 14\) (for power of 5)
The limiting factor is the power of 5, as \(n \leq 14\) while \(3n \leq 56\) allows for \(n \leq 18.67\).
Thus, the largest integer \(n\) such that \(40^n\) divides \(60!\) is \(14\).
Therefore, the correct answer is \(14\).
Similar Questions
- Let $S$ be a set of 5 elements and $\text{P}(S)$ denote the power set of $S$. Let $\text{E}$ be an event of choosing an ordered pair $(A, B)$ from the set $\text{P}(S) \times \text{P}(S)$ such that $A \cap B = \emptyset$. If the probability of the event $\text{E}$ is $\frac{3^p}{2^q}$, where $p, q \in \mathbb{N}$, then $p + q$ is equal to _______
- Let the mean and variance of 7 observations 2, 4, 10, x, 12, 14, y, $x > y$, be 8 and 16 respectively. Two numbers are chosen from $\{1, 2, 3, x-4, y, 5\}$ one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is :
- A building construction work can be completed by two masons A and B together in 22.5 days. Mason A alone can complete the construction work in 24 days less than mason B alone. Then mason A alone will complete the construction work in :
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Important Questions from Arithmetic
- Let $S$ be a set of 5 elements and $\text{P}(S)$ denote the power set of $S$. Let $\text{E}$ be an event of choosing an ordered pair $(A, B)$ from the set $\text{P}(S) \times \text{P}(S)$ such that $A \cap B = \emptyset$. If the probability of the event $\text{E}$ is $\frac{3^p}{2^q}$, where $p, q \in \mathbb{N}$, then $p + q$ is equal to _______
- Let the mean and variance of 7 observations 2, 4, 10, x, 12, 14, y, $x > y$, be 8 and 16 respectively. Two numbers are chosen from $\{1, 2, 3, x-4, y, 5\}$ one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is :
- A building construction work can be completed by two masons A and B together in 22.5 days. Mason A alone can complete the construction work in 24 days less than mason B alone. Then mason A alone will complete the construction work in :
- Let $m$ and $n$, ($m < n$), be two 2-digit numbers. Then the total numbers of pairs $(m, n)$, such that $gcd (m, n) = 6$, is
- Two distinct numbers $a$ and $b$ are selected at random from $1, 2, 3,..., 50$. The probability, that their product $ab$ is divisible by 3, is
