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Class 10 Class 12

Is 1372 a perfect cube? If not, find the smallest natural number by which 1372 must be multipled so that the product is a perfect cube.


We have 1372 = 2 x 2 x 7 x 7 x 7

Since, the prime factor 2 does not appear in a group of triples.  


We have 1372 = 2 x 2 x 7 x 7 x 7Since, the prime factor 2 does not ap
∴ 1372  is  not a  perfect cube.

Obviously, to make it a perfect cube we need one more 2 as its factor.

i.e.            [1372] x 2 = [2 x 2 x 7 x 7 x 7] x 2

or                2744 = 2 x 2 x 2 x 7 x 7 x 7

which is a perfect cube.

Thus, the required smallest number = 2.



775 Views

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


(i) We have 81 = 3 x 3 x 3 x 3


(i) We have 81 = 3 x 3 x 3 x 3Grouping the prime factors of 81 into t
Grouping the prime factors of 81 into triples, we are left with 3.

∴ 81 is  not a perfect cube
Now,   [81] divided by3= [3 x 3 x 3 x 3] divided by3
or    27 = 3 x 3 x 3
i.e. 27 is a perfect cube
Thus, the required smallest number is 3

(ii) we have 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2

(i) We have 81 = 3 x 3 x 3 x 3Grouping the prime factors of 81 into t
Grouping the prime factors of 128 into triples, we are left with 2
∴  128 is  not a perfect cube
Now,  [128] divided by2 = [2 x 2 x 2 x 2 x 2 x 2 x 2]divided by2
or         64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube
∴  the smallest required number is 2.

(iii) we have 135 =  3 x 3 x 3 x 5

(i) We have 81 = 3 x 3 x 3 x 3Grouping the prime factors of 81 into t
Grouping the prime factors of 135 into triples, we are left over with 5.
∴  135 is not a perfect cube
Now, [135]divided by5 = [ 3 x 3 x 3 x 5] divided by5
or      27 = 3 x 3 x 3
i.e. 27 is a perfect cube.
Thus, the required smallest number is 5
(iv) We have 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3

(i) We have 81 = 3 x 3 x 3 x 3Grouping the prime factors of 81 into t
Grouping the prime factors of 192 into triples, 3 is left over.
∴  192 is not a perfect cube.
Now,     [192] divided by3= [2 x 2 x 2 x 2 x 2 x 2 x 3]divided by3
or            64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube.
Thus,  the required smallest number is 3.
(v) We have 704 = 2 x 2 x 2 x 2 x 2 x 2 x 11

(i) We have 81 = 3 x 3 x 3 x 3Grouping the prime factors of 81 into t
Grouping the prime factors of 704 into triples, 11 is left over
∴  [704]divided by11 = [2 x 2 x 2 x 2 x 2 x 2 x 11]divided by11
or    64 = 2 x 2 x 2 x 2 x 2 x 2
i.e. 64 is a perfect cube
Thus, the required smallest number is 11.


 

1360 Views

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


(i) We have 243 = 3 x 3 x 3 x 3 x 3


(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
The prime factor 3 is not a group of three.
∴ 243 is  not a perfect cube.
Now, [243] x 3 = [3 x 3 x 3 x 3 x 3] x 3
or, 729, = 3 x 3 x 3 x 3 x 3 x 3  
Now, 729 becomes a [perfect cube
Thus, the smallest required number to multipkly 243 to make it a perfect cube is 3.

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

(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
Grouping the prime factors of 256 in triples, we are left over with 2 x 2.
∴ 256 is  not a perfect cube.
Now, [256] x 2 = [2 x 2 x 2 x 2 x 2 x 2 x 2 x 2] x 2
or, 512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
i.e. 512 is a perfect cube.
thus, the required smallest number is 2.

(iii) we  have 72  = 2 x 2 x 2 x 3 x 3

(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
Grouping the prime factors of 72 in triples, we are left  over with 3 x 3
∴ 72 is  not a perfect cube.
Now, [72] x 3 = [2 x 2 x 2 x 3 x 3] x 3
or,     216 = 2 x 2 x 2 x 3 x 3 x 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 x 3 x 3 x 5 x 5
Grouping the prime factors of 675 to triples, we are left over with 5 x 5

(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
∴  675 is not a perfect cube.
Now, [675] x 5 = [3 x 3 x 3 x 5 x 5] x 5
Now, 3375  is a  perfect cube
Thus, the smallest required number to multiply 675 such that the new number is a perfect cube is 5.

(v) We have 100 = 2 x 2 x 5 x 5
The prime factor are not in the groups of triples.


(i) We have 243 = 3 x 3 x 3 x 3 x 3The prime factor 3 is not a group
∴  100 is not a perfect cube.
Now, [100] x 2 x 5 = [2 x 2 x 5 x 5] x 2 x 5
or,   [100] x 10 = 2 x 2 x 2 x 5 x 5 x 5

1000 = 2 x 2 x 2 x 5 x 5 x 5

Now, 1000 is a perfect cube
Thus, the required smallest number is 10


1158 Views

Which of the following are perfect cubes?

1. 400 2. 3375 3. 8000 4. 15625

5. 9000 6. 6859 7. 2025 8. 10648


                                   

                                   

                                   

1677 Views

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.


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

1331 rightwards arrow 1 and 331
∵ 331 end in 1
∴ Unit's digit of the cube root = 1

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∴ Ten's digit of the cube root = 1

∴ cube root of 1331 equals space 11

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

4913 rightwards arrow4 and 913
Unit's digit:
∵ Unit's digit in 913  is 3
∴ Unit's digit of the cube root = 7

[7= 343 : which ends in 3]

Ten's digit:
∵ 1= 1, 2 = 8
and 1 < 4 < 8
i.e. 1 <  4 < 2
∴ Then ten's digit of the cube root is 1.

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(iii) Separatibng 12167 in two groups:
     12167rightwards arrow12 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
∵ 2 = 8 adn 3= 27
Also,  8 < 12 < 27
or,  2< 12 < 32  

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

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(iv) separating 32768 in two groups:

32768 rightwards arrow32 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:
∵ 3= 27 and 4- 64
Also, 27 < 32 < 64
or, 3< 32 < 4

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

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799 Views

Check which of the following are perfect cubes.

(i) 2700 (ii) 16000 (iii) 64000 (iv) 900

(v) 125000 (vi) 36000 (vii) 21600 (viii) 10000

(ix) 27000000 (x) 1000
What pattern do you observe in these perfect cubes?


(i) We have 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5
We do  not get complete triples of prime factors, i.e. 2 x 2 and 5 x 5 are left over.

(i) We have 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5We do  not get complete
∴ 2700 is not a perfect cube.
(ii) we  have 1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5
After grouping id 3's we get 5 x 5 which is ungrouped in triples.

(i) We have 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5We do  not get complete
∴ 1600 is not a perfect cube.
(iii) We have 64000 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5
Since, we get groups of triples.

(i) We have 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5We do  not get complete
∴ 64000 is a perfect cube.
(iv) We have 900 = 2 x 2 x 3 x 3 x 5 x 5
which are ungrouped in triples.


(i) We have 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5We do  not get complete
∴ 900 is not a perfect cube.
(v) We have 125000 = 2  x 2 x 2 x 5 x 5 x 5 x 5 x 5 x 5
As we get all the prime factors in the group of triples.

(i) We have 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5We do  not get complete
∴ 125000 is a perfect cube
(vi) we have 36000 = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 5 x 5 x 5
While grouping the prime factors of 36000 in triples, we are left over with 2 x 2 and 3 x 3

(i) We have 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5We do  not get complete
∴ 36000 is not a perfect cube
(vii) We have 21600 = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5
While grouping the prime factors of 21600 in triples, we are left with 2 x 2 and 5 x 5.

(i) We have 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5We do  not get complete
∴ 21600 is not a perfect cube
(viii) we have 10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5
While grouping the prime factors into triples, we are left over with 2 and 5.


(i) We have 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5We do  not get complete
∴ 10000 is not a perfect cube
(ix) we have 27000000 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5 x 5 x 5 x 5 x 5


(i) We have 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5We do  not get complete
Since, all the prime factors of 27000000 appear in groups of triples.
∴ 27000000 is a perfect cube.
(x) We have 1000 = 2 x 2 x 2 x 5 x 5 x 5


(i) We have 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5We do  not get complete
∴ 1000 is a perfect cube.



1205 Views