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

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