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Find the smallest number by which 2560 must be multiplied so that the product is a perfect cube Find an answer to your question find the smallest number by which 2560 must be multiplied so that the product is a perfect cube pls give step by step Find the smallest number by which 8788 be divided so that the quotient is a perfect cube

Smallest number multiplied to get perfect square - Teachoo

Write true or false for the following statements Hence, the smallest number by which it must be multiplied to make it a perfect cube is 25. (i) 650 is not a perfect cube

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(ii) perfect cubes may end with two zeroes

(iii) perfect cubes of odd. To find the smallest number by which 2560 must be multiplied so that the product is a perfect cube, we will follow these steps Prime factorization of 2560 we start by finding the prime factorization of 2560. To find the smallest number by which 2560 should be multiplied to make it a perfect cube, we first need to factorize 2560 into its prime factors

The prime factorization of 2560 is @$\begin {align*}2^9 \times 5\end {align*}@$. Find the smallest number by which 2560 must be multiplied so that the product is perfect cube See answers ayanaval ayanaval answer To make it a multiple of 3, we need to multiply by 5² = 25

Therefore, the smallest number by which 2560 must be multiplied to make it a perfect cube is 25.

But the number 5 is occurring one time only Hence, we need to multiply the same number twice to get the perfect cube Which is 5 × 5 2 will make the perfect cube Hence the smallest number by 2560 number must be multiplied so that the product is a perfect cube is 25.

Find the smallest number by which 2560 must be multiplied so that the product is a perfect cube. Therefore, 2560 is a not perfect cube However, if the number is multiplied by 5 × 5 = 2 5 the factors can be grouped into triples of equal factors such that no factor is left over Thus, 2560 should be multiplied by 25 to make it a perfect cube.

To determine the smallest number by which 2560 must be multiplied to make it a perfect cube, we first need to perform the prime factorization of 2560

Factor 2560 into its prime components. To solve these problems, we need to determine the smallest number by which 2560 must be multiplied to become a perfect cube and the smallest number by which 8788 must be divided to become a perfect cube We also need to evaluate the given statements as true or false. Solution for find the smallest number by which 2560 must be multiplied so that the product is a perfect cube.

Text solution verified explanation option [b] is correct The factors of 2560 is given by 2560=5×8×8×8 in this factors there are three 8 and one 5 So in order to make the number 2560 perfect cube we have to multiply it by 25 Therefore, 25 is the least number by which it must be multiplied so that it becomes a perfect cube.

Here, number of 2's is 9 and number of 5's is 1

So we need to multiply another 52 in the factorization to make 2560 a perfect cube Hence, the smallest number by which 2560 must be multiplied to obtain a perfect cube is 52 = 25 Therefore, option b is correct Let's find out the prime factors of the given number, ∴ 2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 as we can see, to make the pair of 4 triplets two 5 are required, which is 5×5

So, 25 will be the number multiplied to 2560, to get the perfect cube. Prime factorising 2560, we get, 2560=2 9 ×5 We know, a perfect cube has multiples of 3 as powers of prime factors So we need to multiply another 5 2 in the factorization to make 2560 a perfect cube

Hence, the smallest number by which 2560 must be multiplied to obtain a perfect cube is 5 2 =25.

No, 2560 is not a perfect cube because it cannot be expressed as the cube of a whole number In order to find the smallest natural number by which 2560 must be multiplied so that the product is a perfect cube, we need to factorize 2560. Long answer type questions find the smallest number by which 2560 must be multiplied so that the product is a perfect cube three number are to one another 2 4 the sum of their cubes is 0.334125

Find the number find the cube roots of the following integers (i) −2744000 (ii) −474552 (iii) −5832 views Factorising 2560 into prime factors 2560 =2×2×2×2×2×2×2×2×2×5 making them in groups of 3 equal factors, we are left with 5 ∴ to make it into a group of 3, we have to multiply it by 5×5 i.e