Find the smallest number by which 46305 must be multiplied so that the product is a perfect cube

today question is find the smallest number by which given number must be multiplied so that the product is a perfect square also first 23805 second part is 12150 third party 7688 let this question taking the first part first part but we are given in first part is given to 38052 let's find the prime factors for two what will be the prime factors of this it will be what do you live this so it will divided with I will get 761 then with free will get 15879 with three again we will get 529 then with

again if will divided with 23 we will that we will get 23 and again you will divide B 23 will get 100 so prime factor of 23805 are 5 x 3 x 3 x 23 x 23 so as you can see that she is in a 23 is in a but 5 is not enough there Jaipur if X with 5 it will become in a pair it will will get a pair of 5 and this whole number with the cover but I swear therefore there 45 is the required number Elite sofa first part 5 is the required number that move to the second part

what is s Park 12150 212412 15012 150 led to the prime factorization so what we will get on after doing prime factorization will / with two will get 6075 then again with 3 will get 2025 just again we will get 20 25 then again with three we will get 675 and again with 3 we will get 225 then again with three we will get 75 then again with three I will get to 5 then we will get 5 and again with 5

will get one so what are the prime factorization of 12150 2 x 3 x 3 x 3 x 3 10 x free x 5 x 5 which AC and DC that 5 is in a train is there is a pair of 3 and another pair of tree but 12 and 13 is there which are not in a show 2 x 3 is 66 is the number if you X with it will become a perfect square 26 is the required required number 6 is required number 3rd part third part but we have given you have given

7688 so what 7688 if will do the prime factorization prime factorization but will get will get to feel divided Bittu will get 3844 and again Bittu will get 1922 then again Bittu will get 961 then again with 31 will get 31 then again with 31 will get 12 everything the prime factorization of 7688 3 X 2 x 2 X 31 X 31 everything there is a factor of 2 and there is a factor of 2 and there is a factor of 31 there is a pair of 2 and there is a pair of 31 but to is

not in the pair therefore to is the near airport to is required number I hope you understood the solution thank you

Ex 7.1, 2 (i) - Chapter 7 Class 8 Cubes and Cube Roots

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Ex 7.1, 2 Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube. (i) 243 We see that 243 = 3 × 3 × 3 × 3 × 3 Since 3 does not occur in triplets, ∴ 243 is not a perfect cube. So, we multiply by 3 to make triplet So, our number becomes 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 Now, it becomes a perfect cube. So, we multiply 243 by 3 to make it a perfect cube

Class VIII Math
NCERT Solution for Cubes and Cube Roots

Q1.   Which of the following numbers are not perfect cubes?

(i) 216

(ii) 128

(iii) 1000

(iv) 100

(v) 46656

Sol.     (i) We have 216 = 2 × 2 × 2 × 3 × 3 × 3

Grouping the prime factors of 216 into triples, no factor is left over.

∴ 216 is a perfect cube.

(ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Grouping the prime factors of 128 into triples, we are left over with 2 as ungrouped factor.

∴ 128 is not a perfect cube.

(iii) We have 1000 = 2 × 2 × 2 × 5 × 5 × 5

Grouping the prime factors of 1000 into triples, we are not left over with any factor.

∴ 1000 is a perfect cube.

(iv) We have 100 = 2 × 2 × 5 × 5

Grouping the prime factors into triples, we do not get any triples. Factors 2 × 2 and 5 × 5 are not in triples.

∴ 100 is not a perfect cube.

(v) We have 46656 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

Grouping the prime factors of 46656 in triples we are not left over with any prime factor.

∴ 46656 is a perfect cube.

Q2.   Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.

(i) 243

(ii) 256

(iii) 72

(iv) 675

(v) 100

Sol.     (i) We have 243 = 3 × 3 × 3 × 3 × 3

The prime factor 3 is not a group of three.

∴ 243 is not a perfect cube.

Now, [243] × 3 = [3 × 3 × 3 × 3 × 3] × 3

or 729 =3 × 3 × 3 × 3 × 3 × 3

Now, 729 becomes a perfect cube.

Thus, the smallest required number to multiply 243 to make it a perfect cube is 3.

(ii) We have 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Grouping the prime factors of 256 in triples, we are left over with 2 × 2.

∴ 256 is not a perfect cube.

Now, [256] × 2 = [2 × 2 × 2 × 2 × 2 × 2 × 2 × 2] × 2

or 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

i.e. 512 is a perfect cube.

Thus, the required smallest number is 2.

(iii) We have 72 = 2 × 2 × 2 × 3 × 3

Grouping the prime factors of 72 in triples, we are left over with 3 × 3.

∴ 72 is not a perfect cube.

Now, [72] × 3 =[2 × 2 × 2 × 3 × 3] × 3

or 216 = 2 × 2 × 2 × 3 × 3 × 3

i.e. 216 is a perfect cube.

∴ The smallest number required to multiply 72 to make it a perfect cube is 3.

(iv) We have 675 = 3 × 3 × 3 × 5 × 5

Grouping the prime factors of 675 to triples, we are left over with 5 × 5.

∴ 675 is not a perfect cube.

Now, [675] × 5 = [3 × 3 × 3 × 5 × 5] × 5

or 3375 = 3 × 3 × 3 × 5 × 5 × 5

Now, 3375 is a perfect cube.

Thus, the smallest required number to multiply 675 such that the new number perfect cube is 5.

(v) We have 100 = 2 × 2 × 5 × 5

The prime factor are not in the groups of triples.

∴100 is not a perfect cube.

Now [100] × 2 × 5 = [2 × 2 × 5 × 5] × 2 × 5

or [100] × 10 = 2 × 2 × 2 × 5 × 5 × 5

1000 = 2 × 2 × 2 × 5 × 5 × 5

Now, 1000 is a perfect cube.

Thus, the required smallest number is 10.

Q3.   Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.

(i) 81

(ii) 128

(iii) 135

(iv) 192

(v) 704

Sol.     (i) We have 81 = 3 × 3 × 3 × 3

Grouping the prime factors of 81 into triples, we are left with 3.

∴ 81 is not a perfect cube.

Now, [81] 3 = [3 × 3 × 3 × 3] + 3

or 27 = 3 × 3 × 3

i.e. 27 is a prefect cube

Thus, the required smallest number is 3.

(ii) We have 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Grouping the prime factors of 128 into triples, we are left with 2.

∴ 128 is not a perfect cube

Now, [128] 2 = [2 × 2 × 2 × 2 × 2 × 2

or 64 = 2 × 2 × 2 × 2 × 2 × 2

i.e. 64 is a perfect cube.

∴ The smallest required number is 2.

(iii) We have 135 = 3 × 3 × 3 × 5

Grouping the prime factors of 135 into triples, we are left over with 5.

∴ 135 is not a perfect cube

Now, [l35] 5 = [3 × 3 × 3 × 5] 5

or 27 = 3 × 3 × 3

i.e. 27 is a perfect cube.

Thus, the required smallest number is 5.

(iv) We have 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

Grouping the prime factors of 192 into triples, 3 is left over.

∴ 192 is not a perfect cube.

Now, [192] 3 =[2 × 2 × 2 × 2 × 2 × 2

or 64 = 2 × 2 × 2 × 2 × 2 × 2

i.e. 64 is a perfect cube.

Thus, the required smallest number is 3.

(v) We have 704 = 2 × 2 × 2 × 2 × 2 × 2 × 11

Grouping the prime factors of 704 into triples, 11 is left over.

∴ [704] 11 =[2 × 2 × 2 × 2 × 2 × 2

or 64 = 2 × 2 × 2 × 2 × 2 × 2

i.e. 64 is a perfect cube.

Thus, the required smallest number is 11.

Q4.   Parikshit makes a cuboid of plasticine of sidec 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?

Sol:    Sides of the cuboid arc: 5 cm, 2 cm, 5 cm

∴ Volume of the cuboid = 5 cm × 2 cm × 5 cm

To form it as a cube its dimension should be in the group of triples.

∴ Volume of the required cube = [5 cm × 5 cm × 2 cm] × 5 cm × 2 cm × 2 cm

=[5 × 5 × 2 cm3] = 20 cm3

Thus, the required number of cuboids = 20.

1.   Find the cube root of each of the following numbers by prime factorisation method.

(i) 64

(ii) 512

(iii) 10648

(iv) 27000

(v) 15625

(vi) 13824

(vii) 110592

(viii) 46656

(ix) 175616

(x) 91125

Sol.  (i) By prime factorisation, we have

(ii) By prime factorisation, we have

(iii) By prime factorisation, we have

Thus, cube root of 10648 is 22.

(iv) By prime factorisation, we have

Thus, cube root of 27000 is 30.

(v) By prime factorisation, we have

Thus, cube root of 15625 is 25.

(vi) By prime factorisation, we have

Thus, cube root of 13824 is 24.

(vii) By prime factorisation, we have

Thus, the cube root of 110592 is 48

(viii) By the prime factorisation, we have

Thus, the cube root of 46656 is 32

(ix) By prime factorisation, we have

175616 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 7 × 7 × 7

= 2 × 2 × 2 × 7 = 56

Thus the cube root of 175616 is 56.

(x) By prime factorisation, we have:

91125 = 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5

= 3 × 3 × 5 = 45

Thus, the cube root of 91125 is 45.

Q2.   State true or false.

(i) Cube of any odd number is even.

(ii) A perfect cube does not end with two zeros.

(iii) If square of a number ends with 5, then its cube ends with 25.

(iv) There is no perfect cube which ends with 8.

(v) The cube of a two digit number may he a three digit number.

(vi) The cube of a two digit number may have seven or more digits.

(vii) The cube of a single digit number may be a single digit number.

Sol.     (i) False     (ii) True             (iii) False           (iv) False

(v) False     (vi) False          (vii) True

Q3.   You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.

Sol.     (i) Separating the given number (1331) into two groups:

1331 → 1 and 331

∴ 331 end in 1.

∴ Unit's digit of the cube root = 1

∴ Ten's digit of the cube root = I

(ii) Separating the given number (4913) in two groups:

4913 → 4 and 913

Unit's digits:

∵ Unit's digit in 913 is 3.

∴ Unit's digit of the cube root = 7

[73 = 343; which ends in 3]

Ten's digit:

13 = 1, 23 = 8

and 1 < 4 < 8

i.e. 13 < 4 < 23

∴ The ten's digit of the cube root is 1.

(iii) Separating 12,167 in two groups:

12167 → 12 and 167

Unit's digit:

∵ 167 is ending in 7 and cube of a number ending in 3 ends in 7.

∴ The unit's digit of the cube root = 3

Ten's digit:

∵ 23 = 8 and 33 = 27

Also, 8 < 12 < 27

or 23 < 12 < 32

∴ The tens digit of the cube root can be 2.

(iv) Separating 32768 in two groups:

32768 → 32 and 786

Unit's digit:

768 will guess the unit's digit in the cube root.

∵ 768 ends in 8.

Unit's digit in the cube root = 2

Ten's digit:

∵ 33 = 27 and 43 – 64

Also, 27 < 32 < 64

or 33 < 32 < 43

The ten's digit of the cube root = 3.

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