The smallest number by which 18522 must be divided to obtain a 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.

Hint: The prime factorization is a method which is used to express a given number in form of multiplication of prime numbers. On the other hand, in factorization if a prime number occurs more than once, then factorization is expressed in exponential form.
To solve such type of questions i.e quotient will be a perfect cube or perfect square, we always do the prime factorisation of the given number and make the groups in triplet (to calculate perfect cube) and doublet (for perfect square). Here triplet means group of three same numbers and doublet means group of two same numbers.

Complete step by step solution: We will start here with prime factorisation of our given number:

\[2\]\[18522\]\[3\]\[9261\]\[3\]\[3087\]\[3\]\[1029\]\[7\]\[343\]\[7\]\[49\]\[7\]\[7\]\[1\]

So Prime factorisation of \[18522{\text{ }} = {\text{ }}2 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7\]
Since we have to make quotient a perfect cube, hence we make group of three same number here.
i.e. \[18522{\text{ }} = {\text{ }}2 \times \]\[3 \times 3 \times 3\]\[ \times \]\[7 \times 7 \times 7\]
Clearly it can be seen that 2 is the only factor which is extra here. If we divide our number by 2 here then its quotient can become a perfect cube here.
i.e. \[\dfrac{{18522}}{2} = 9261\] which is a perfect cube of \[21\].

Hence, our desired answer is 2.

Note: Students always remember in prime factorisation you have to use prime numbers. Prime numbers are those which can be divided by one or the number itself. Example \[2,3,5,7 \ldots .\]
Also one is not a prime number. Many students use numbers other than prime and get wrong answers. Secondly, always remember you have to choose the same number to make a triplet because \[{2^3} = 2 \times 2 \times 2\], the cube is always a group of three numbers. We divide only with that number which is not forming a triplet.

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