What is the smallest number by which 2916 should be divided so that the quotient is perfect cube?

Solution:

[i]81

Prime factors of 81 = 3\times3\times3\times3

Here one factor 3 is not grouped in triplets.

Therefore 81 must be divided by 3 to make it a perfect cube.

[ii] 128

Prime factors of 128 = 2\times2\times2\times2\times2\times2

Here one factor 2 does not appear in a 3’s group.

Therefore, 128 must be divided by 2 to make it a perfect cube.

[iii] 135

Prime factors of 135 = 3\times3\times3\times5

Here one factor 5 does not appear in a triplet.

Therefore, 135 must be divided by 5 to make it a perfect cube.

[iv] 192

Prime factors of 192 = 2\times2\times2\times2\times2\times3

Here one factor 3 does not appear in a triplet.

Therefore, 192 must be divided by 3 to make it a perfect cube.

[v] 704

Prime factors of 704 = 2\times2\times2\times2\times2\times2\times11

Here one factor 11 does not appear in a triplet.

Therefore, 704 must be divided by 11 to make it a perfect cube.

Step-by-step explanation:

Prime factorising 2916, we get,

2916=2×2×3×3×3×3×3×3

=2

2

×3

6

.

We know, a perfect cube has multiples of 3 as powers of prime factors.

Here, number of 2’s is 2 and number of 3’s is 6.

So we need to multiply another 2 in the factorization to make 2916 a perfect cube.

Hence, the smallest number by which 2916 must be multiplied to obtain a perfect cube is 2.

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1. Perfect Square:

A natural number x is a perfect square if there exists a natural number y such that x=y2. In other words, a natural number x is a perfect square, if it is equal to the product of a number with itself.

2. Properties of Squares Numbers:

[i] A number ending in 2, 3, 7, or 8 is never a perfect square.

[ii] The number of zeroes in the end of a perfect square is never odd. So, a number ending in an odd number of zeroes is never a perfect square.

[iii] Squares of even numbers are always even.

[iv] Squares of odd numbers are always odd.

3. General Properties of Perfect Squares:

[i] For any natural number n, we have n2=  [Sum of first n odd natural numbers]

[ii] The square of a natural number, other than 1, is either a multiple of 3 or exceeds a multiple of 3 by 1 .

[iii] The square of a natural number, other than 1, is either a multiple of 4 or exceeds a multiple of 4 by 1.

[iv] There are no natural numbers p  and q such that p2=2q2

4. Pythagorean Triplets:

For any natural number n greater than 1,  [2n, n2−1, n2+1], is a Pythagorean triplet.

5. Square roots:

The square root of a given natural number n is that natural number which when multiplied by itself gives n as the product and we denote the square root of n by n. Thus, n=m⇔n=m2.

6. Finding Square Roots:

[i] In order to find the square root of a perfect square, resolve it into prime factors; make pairs of similar factors and take the product of prime factors, choosing one out of every pair.

[ii] For finding the square root of a decimal fraction, make the even number of decimal places by affixing a zero, if necessary; mark off periods and extract the square root; putting the decimal point in the square root as soon as the integral part is exhausted.

7. Properties of Square Roots:

For positive numbers a and b, we have

[i] ab=a×b

[ii] ab =ab

Is 2916 a perfect cube?

2916 doesn't have a perfect cube.

What is the smallest number by which we divide 6912 so that the quotient becomes a perfect cube find the cube root of the quotient?

Given: A number 6912 . To do: To find the smallest number by which 6912 must be divided so that the number formed is a perfect cube. Therefore, we should divide 6912 by 22=4 2 2 = 4 , the smallest number to get 1728 which is a cube of 12 .

What is the smallest number by which 1458 should be divided so that the quotient is a perfect cube?

The smallest number by which 1458 must be divided to make a perfect cube is 2.

What is the smallest number by which 1715 should be divided so that the quotient is a perfect?

In other words, 1715 should be divided by 5 [the smallest number] so that the quotient is a perfect cube [343].

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