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By what number should each of the following numbers by multiplied to get a perfect square in each case? Also find the number whose square is the new number.(i) 8820(ii) 3675(iii) 605(iv) 2880(v) 4056(vi) 3468(vii) 7776 |
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Answer» (i) 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 Again, 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 (ii) 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, 3675 × 3 = 5 × 5 × 7 × 7 × 3 × 3 = (5 × 5) × (7 × 7) × (3 × 3) = (3 × 5 × 7) × (3 × 5 × 7) = 105 × 105 = (105)2 So, the product is the square of 105 (iii) 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 (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 (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, multiply 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 (vii) 7776 7776 = (2 × 2) × (2 × 2) × (3 × 3) × (3 × 3) × 2 × 3 In the above factors only 2 and 3 are unpaired So, multiply the number with 6 to make it paired Again, 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. |
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