Foundation June 2017 Paper 3 Q15
15 Show that there are exactly five 3-digit cube numbers. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 of 4 | ||
| Identifies any 3-digit cube number | M1 | 125 or 216 or 343 or 512 or 729 |
| 125 and 216 and 343 and 512 and 729 | M1dep | |
| 125 and 216 and 343 and 512 and 729 and 64 and 1000 | A1 | |
| Alternative method 2 of 4 | ||
| Identifies any 3-digit cube number | M1 | 125 or 216 or 343 or 512 or 729 |
| \(5^3 = 125\) and \(9^3 = 729\) and 5, 6, 7, 8, 9 or \(9 - 4 = 5\) | M1dep | |
| \(5^3 = 125\) and \(9^3 = 729\) and 5, 6, 7, 8, 9 or \(9 - 4 = 5\) and \((4^3 =)\) 64 and \((10^3 =)\) 1000 | A1 | |
| Alternative method 3 of 4 | ||
| \(\sqrt[3]{100} = 4.6\ldots\) | M1 | |
| \(\sqrt[3]{999} = 9.9\ldots\) or \(\sqrt[3]{1000} = 10\) | M1 | |
| \(\sqrt[3]{100} = 4.6\ldots\) and \(\sqrt[3]{999} = 9.9\ldots\) or \(\sqrt[3]{1000} = 10\) and 5, 6, 7, 8, 9 or \(9 - 4 = 5\) | A1 | |
| Alternative method 4 of 4 | ||
| \(5^3 = 125\) | M1 | |
| \(10^3 = 1000\) or \(\sqrt[3]{1000} = 10\) | M1 | |
| \(4^3 = 64\) and \(5^3 = 125\) and \(10^3 = 1000\) or \(\sqrt[3]{1000} = 10\) and 5, 6, 7, 8, 9 or \(9 - 4 = 5\) | A1 | |