Question

what is the least natural number by which 2560 should be multiplied so that the product is a perfect cube​

Answer 1

Step-by-step explanation:

If you break down 2560 into product of prime factors then 2560 = 2×2×2×2×2×2×2×2×2×5 i.e (2^9)×5.

Now , any perfect cube number is of the form , say if *N *is a perfect cube number then *N* = (a^3z)×(b^3y)×(c^3x)…. , *where a,b,c represent prime factors of N.*

In the above case the power to which 2 is raised is a multiple of 3 but the power of 5 isn't. So the smallest number by which 2560 should be multiplied so that the product is a perfect cube is 5*5.

The resultant number is 2^9 × 5^3 = 64000 which is the cube of 40