question_answer
1)
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
Answer:
(i) 243 3 | 243 |
3 | 81 |
3 | 27 |
3 | 9 |
3 | 3 |
| 1 |
By prime factorisation, \[243=\underline{3\times 3\times 3}\times 3\times 3\] | grouping the factors in triplets The prime factor 3 does not appear in a group of three. Therefore, 243 is not a perfect cube. To make it a cube, we need one more 3. In that case \[243\times 3=\underline{3\times 3\times 3}\times \underline{3\times 3\times 3}\] = 729, which is a perfect cube. Hence, the smallest number by which 243 should be multiplied to make a perfect cube is 3. (ii) 256 2 | 256 |
2 | 128 |
2 | 64 |
2 | 32 |
2 | 16 |
2 | 8 |
2 | 4 |
2 | 2 |
| 1 |
By prime factorisation, \[256=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times 2\times 2\] | grouping the factors in triplets In the above factorisation 2 remains after grouping 2's in triplets. Therefore, 128 is not a perfect cube. To make it a perfect cube, we need one 2's more. In that case, \[256\times 2=\underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\times \underline{2\times 2\times 2}\] \[={{2}^{3}}\times {{2}^{3}}\times {{2}^{3}}\] | by laws of exponents \[={{(2\times 2\times 2)}^{3}}\] | by laws of exponents \[={{8}^{3}}=512,\] which is a perfect cube. Hence, the smallest number by which 256 must be multiplied to obtain a perfect cube is 2. The resulting perfect cube is \[512=({{8}^{3}})\]. (iii) 72 By prime factorisation, \[72=\underline{2\times 2\times 2}\times 3\times 3\] | grouping the factors in triplets The prime factors 3 does not appear in a group of three. Therefore, 72 is not a perfect cube. To make it a cube, we need one more 3. In that case, \[72\times 3=\underline{2\times 2\times 2}\times \underline{3\times 3\times 3}\] \[={{2}^{3}}\times {{3}^{3}}\] | by laws of exponents \[={{(2\ \times 3)}^{3}}\] | by laws of exponents \[={{6}^{3}},\] which is a perfect cube. Hence, the smallest number by which 72 must be multiplied to obtain a perfect cube is 3. (iv) 675 3 | 675 |
3 | 225 |
3 | 75 |
5 | 25 |
5 | 5 |
| 1 |
By prime factorisation, \[675=\underline{3\times 3\times 3}\times 5\times 5\] | grouping the factors in triplets The prime factor 5 does not appear in a group of three. Therefore, 675 is not a perfect cube. To make it a cube, we need one more 5. In that case, \[675\times 5=\underline{3\times 3\times 3}\times \underline{5\times 5\times 5}\] \[={{3}^{3}}\times {{5}^{3}}\] | by laws of exponents \[={{(3\times 5)}^{3}}\] | by laws of exponents \[={{15}^{3}},\] which is a perfect cub Hence, the smallest number by which 675 must be multiplied to obtain a perfect cube is 5. The resulting perfect cube is 3375 \[(={{15}^{3}})\] (v) 100 By prime factorisation, \[100=2\times 2\times 5\times 5\] | grouping the factors in triplets The prime factors 2 and 5 do not appear in a group of three. Therefore, 100 is not a perfect cube. To make it a perfect cube, we need one 2 and one 5 more. In that case, \[100\,\times 2\times 5=\underline{2\times 2\times 2}\times \underline{5\times 5\times 5}\] \[={{2}^{3}}\times {{5}^{3}}\] | by laws of exponents \[={{(2\times 5)}^{3}}\] | by laws of exponents \[={{10}^{3}},\] which is a perfect cube. Hence, the smallest number by which 100 must be multiplied to obtain a perfect cube is \[2\times 5=10\]. The resulting perfect cube is \[1000(={{10}^{3}})\].