Number of ways in which the number 831600 can | vedclass

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

Question

(D) First,find the prime factorization of $831600$: $831600 = 2^4 	imes 3^3 	imes 5^2 	imes 7^1 	imes 11^1$. For two factors to be relatively prim

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(D) First,find the prime factorization of $831600$: $831600 = 2^4 	imes 3^3 	imes 5^2 	imes 7^1 	imes 11^1$. For two factors to be relatively prime,each prime power factor (e.g.,$2^4, 3^3, 5^2, 7^1, 11^1$) must belong entirely to one of the two factors. There are $5$ distinct prime power factors. Each of these $5$ factors can be placed in either the first factor or the second factor,giving $2^5 = 32$ ways. Since the order of the two factors does not matter (splitting into $A$ and $B$ is the same as splitting into $B$ and $A$),we divide by $2!$. Number of ways $= \frac{2^5}{2} = \frac{32}{2} = 16$.

Number of ways in which the number $831600$ can be split into two factors which are relatively prime is

Similar Questions

If $(24, 92) = 24m + 92n$,then $(m, n)$ is

The exponent of $3$ in $100!$ is

If the constant term in the expansion of $\left(3 x^{3}-2 x^{2}+\frac{5}{x^{5}}\right)^{10}$ is $2^{k} \cdot l$,where $l$ is an odd integer,then the value of $k$ is equal to

It is given that the number $43361$ can be written as a product of $two$ distinct prime numbers $p_1$ and $p_2$. Further,assume that there are $42900$ numbers which are less than $43361$ and are coprime to it. Then,the value of $p_1+p_2$ is

Total number of divisors of $480$ that are of the form $4n + 2$,where $n \geq 0$,is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module