Question

5. By what number should each of the
following numbers be multiplied to get a
perfect square in each case? Also find the
number whose square is the new number.
(1) 7776
(ii) 3468
(iii) 4056
(iv) 2880 (v) 605
(vi) 3675
(vii) 8820​

Answer 1

Answer:

Step-by-step explanation:

(i)7776 × 6 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

= (2 × 2) × (2 × 2) × (2 × 2) (3 × 3) × (3 × 3) × (3 × 3)

= (2 × 2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 3 × 3 × 3)

= 216 × 216

= (216)2

So, the product is the square of 216

(ii)3468 = (2 × 2) × 3 × (17 × 17)

In the above factors only 3 are unpaired

So, mulityply the number with 3 to make it paired

3468 × 3 = (2 × 2) × (3 × 3) × (17 × 17)

= (2 × 3 × 17) × (2 × 3 × 17)

= 102 × 102

= (102)2

= 210 × 210

= (210)2

So, the product is the square of 210

3468 = (2 × 2) × 3 × (17 × 17)

In the above factors only 3 are unpaired

So, mulityply the number with 3 to make it paired

3468 × 3 = (2 × 2) × (3 × 3) × (17 × 17)

= (2 × 3 × 17) × (2 × 3 × 17)

= 102 × 102

= (102)2

So, the product is the square of 102

So, the product is the square of 105

(ii) 2880

2880 = 5 × (3 × 3) × (2 × 2) × (2 × 2) × (2 × 2)

In the above factors only 5 is unpaired

So, multiply the number with 5 to make it paired

Again,

2880 × 5 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

= (2 × 2) × (2 × 2) × (2 × 2) (3 × 3) × (5 × 5)

= (2 × 2 × 2 × 3 × 5) × (2 × 2 × 2 × 3 × 5)

= 120 × 120

= (120)2

So, the product is the square of 120

(v) 4056

4056 = (2 × 2) × (13 × 13) × 2 × 3

In the above factors only 2 and 3 are unpaired

So, multiply the number with 6 to make it paired

Again,

4056 × 6 = 2 × 2 × 13 × 13 × 2 × 2 × 3 × 3

= (2 × 2) × (13 × 13) × (2 × 2) (3 × 3)

= (2 × 2 × 3 × 13) × (2 × 2 × 3 × 13)

= 156 × 156

= (156)2

So, the product is the square of 156

(vi) 3468

3468 = (2 × 2) × 3 × (17 × 17)

In the above factors only 3 are unpaired

So, mulityply the number with 3 to make it paired

3468 × 3 = (2 × 2) × (3 × 3) × (17 × 17)

= (2 × 3 × 17) × (2 × 3 × 17)

= 102 × 102

= (102)2

So, the product is the square of 102

(v) 4056

4056 = (2 × 2) × (13 × 13) × 2 × 3

In the above factors only 2 and 3 are unpaired

So, multiply the number with 6 to make it paired

Again,

4056 × 6 = 2 × 2 × 13 × 13 × 2 × 2 × 3 × 3

= (2 × 2) × (13 × 13) × (2 × 2) (3 × 3)

= (2 × 2 × 3 × 13) × (2 × 2 × 3 × 13)

= 156 × 156

= (156)2

So, the product is the square of 156

(iv) 2880

2880 = 5 × (3 × 3) × (2 × 2) × (2 × 2) × (2 × 2)

In the above factors only 5 is unpaired

So, multiply the number with 5 to make it paired

Again,

2880 × 5 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

= (2 × 2) × (2 × 2) × (2 × 2) (3 × 3) × (5 × 5)

= (2 × 2 × 2 × 3 × 5) × (2 × 2 × 2 × 3 × 5)

= 120 × 120

= (120)2

So, the product is the square of 120

(vi) 3675

3675 = (5 × 5) × (7 × 7) × 3

In the above factors only 3 is unpaired

So, multiply the number with 3 to make it paired

Again,

(v) 605

605 = 5 × (11 × 11)

In the above factors only 5 is unpaired

So, multiply the number with 5 to make it paired

Again,

605 × 5 = 5 × 5 × 11 × 11

= (5 × 5) × (11 × 11)

= (5 × 11) × (5 × 11)

= 55 × 55

= (55)2

So, the product is the square of 55

(vi)3675 × 3 = 5 × 5 × 7 × 7 × 3 × 3

= (5 × 5) × (7 × 7) × (3 × 3)

= (3 × 5 × 7) × (3 × 5 × 7)

= 105 × 105

= (105)2

(vii) 8820

8820 = (2 × 2) × (3 × 3) × (7 × 7) × 5

In the above factors only 5 is unpaired

So, multiply the number with 5 to make it paired

8820 × 5 = 2 × 2 × 3 × 3 × 7 × 7 × 5 × 5

= (2 × 2) × (3 × 3) × (7 × 7) (5 × 5)

= (2 × 3 × 7 × 5) × (2 × 3 × 7 × 5)

= 210 × 210

= (210)2

So, the product is the square of 210

i hope this helps