Advertisements
Advertisements
प्रश्न
If 49392 = a4b2c3, find the values of a, b and c, where a, b and c are different positive primes.
Advertisements
उत्तर
First find out the prime factorisation of 49392.
It can be observed that 49392 can be written as `2^4xx3^2xx7^3,` where 2, 3 and 7 are positive primes.
`therefore49392=2^4 3^2 7^3=a^4b^2c^3`
⇒ a = 2, b = 3, c = 7
APPEARS IN
संबंधित प्रश्न
Prove that:
`(a^-1+b^-1)^-1=(ab)/(a+b)`
Assuming that x, y, z are positive real numbers, simplify the following:
`sqrt(x^3y^-2)`
Assuming that x, y, z are positive real numbers, simplify the following:
`(sqrtx)^((-2)/3)sqrt(y^4)divsqrt(xy^((-1)/2))`
Prove that:
`sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)=3/2`
Prove that:
`(2^n+2^(n-1))/(2^(n+1)-2^n)=3/2`
Show that:
`(x^(1/(a-b)))^(1/(a-c))(x^(1/(b-c)))^(1/(b-a))(x^(1/(c-a)))^(1/(c-b))=1`
The value of m for which \[\left[ \left\{ \left( \frac{1}{7^2} \right)^{- 2} \right\}^{- 1/3} \right]^{1/4} = 7^m ,\] is
If 9x+2 = 240 + 9x, then x =
The value of \[\sqrt{3 - 2\sqrt{2}}\] is
If `a = 2 + sqrt(3)`, then find the value of `a - 1/a`.
