The largest non-negative integer k such that | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To find the largest $k$ such that $24^k$ divides $13!$,we first find the prime factorization of $13!$.$13! = 2^{10} 	imes 3^5 	imes 5^2 	imes 7

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) To find the largest $k$ such that $24^k$ divides $13!$,we first find the prime factorization of $13!$.$13! = 2^{10} 	imes 3^5 	imes 5^2 	imes 7 	imes 11 	imes 13$.We have $24^k = (2^3 	imes 3)^k = 2^{3k} 	imes 3^k$.For $24^k$ to divide $13!$,we must have $3k \le 10$ and $k \le 5$.From $3k \le 10$,we get $k \le \frac{10}{3} \approx 3.33$.From $k \le 5$,we get $k \le 5$.The largest integer $k$ satisfying both conditions is $k = 3$.

The largest non-negative integer $k$ such that $24^k$ divides $13!$ is

Similar Questions

The number of divisors of $7!$ is

The number of positive divisors of $252$ is

The sum of all positive divisors of $960$ is

The arithmetic mean of five natural numbers is $40$. The largest exceeds the smallest number by $10$. If $\alpha$ is the maximum possible value for the largest of these $5$ numbers,then the number of positive integral divisors of $\alpha$ is

Let $d(n)$ denote the number of divisors of $n$ including $1$ and itself. Then, $d(225)$, $d(1125)$, and $d(640)$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module