Now that you know several mathematical operators, try this. Select a number from 1 to 10. Using the number selected thrice and only thrice, with the help of any number of mathematical operators can you get the result as 6? Likewise, compute for all numbers from 1 to 10.
3 + 3 = 6 is not the right solution, because 3 has been used only twice.
| 1. | ![]() |
= | 6 |
| 2. | 2 + 2 + 2 | = | 6 or 22 + 2 = 6 |
| 3. | 3 × 3 – 3 | = | 6 |
| 4. | + + ![]() |
= | 6 |
| 5. | 5 ÷ 5 + 5 | = | 6 |
| 6. | 6 + 6 – 6 | = | 6 |
| 7. | 7 – (7 ÷ 7) | = | 6 |
| 8. | + +
![]() |
= | 6 |
| 9. | ( ×
) –
|
= | 6 |
| 10. | (log 10 + log 10 + log 10)! | = | 6 |