3388
Problem
Using some or all of the operations of addition, subtraction, multiplication and division and using the digits 3, 3, 8 and 8 each once and only once make an expression equal to 24.
Printable NRICH Roadshow resource.
Student Solutions
Well done to Alex of Waingels Copse School, Reading, who cracked this Tough Nut. He ended up checking through every combination of the numbers and operations until he found the one that worked, using a spreadsheet to save a bit of time.
$$ \frac{8}{3 - 8\div3} = 24 $$
Before Alex found the solution above, lots of people tried bending the rules:
Callum (Madras College, St Andrew's, Scotland) and Bei(Riccarton High School, Christchurch, New Zealand) both noticed that the problem can be done with a square root sign:
$ \sqrt{8\times8}\times\sqrt{3\times3}$ or $\sqrt{(3\times8)\times(3\times8)} $
David (Alcester Grammar School) added in factorial notation:
$$ (3! \times 8) - (3 \times 8) $$
Ben(Madras College) used a bit of rotation:
$$ 8 \times 3 + \frac{3}{\infty} $$
Sarah(Madras College) decided that working in another base might help. Since there is an 8 in the problem, she decided to try base 9, and came up with:
$$ 38_{nine} - 3_{nine} - 8_{nine} = 26_{nine} = 24 $$
Tim Whitmore (Madras College) used recurring decimals:
$$ (8 + 8) \div (.\dot{3} + .\dot{3}) $$
Finally, Ravi, (St Xavier's College, Calcutta) used the greatest
integer function, [x].
[x] is the greatest integer which is less than or equal to x.
$$ [(8\times8)\div3] + 3 = 24 $$