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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.
$$ [(8\times8)\div3] + 3 = 24 $$
Two brothers were left some money, amounting to an exact number of pounds, to divide between them. DEE undertook the division. "But your heap is larger than mine!" cried DUM...
The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 � 1 [1/3]. What other numbers have the sum equal to the product and can this be so for any whole numbers?
Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.