From the people behind the Alan Turing Cryptography Competition.
You are reading the website of the 2020 edition of the MathsBombe Competition, which ended on Sunday 26 April at 11:59 pm

# Puzzle 1

Amy the algebraist has an abacus with five rows, each of which contains nine beads. Used conventionally, this can represent any whole number between 0 and 99999. In order to represent the number $a_1+10 a_2 +100 a_3 + 1000 a_4 + 10000 a_5$ one should put the abacus in position $$(a_1,a_2,a_3,a_4,a_5)$$, meaning that on the right hand side there are $$a_1$$ beads from the top row, $$a_2$$ beads from the second row, $$a_3$$ beads from the third row, $$a_4$$ beads from the 4th row, and $$a_5$$ beads from the fifth row, with all other beads being on the left hand side. Here $$a_1, a_2, a_3, a_4$$ and $$a_5$$ are each whole numbers between $$0$$ and $$9$$.

Being an algebraist, Amy has to do lots of calculations involving the golden mean $$\phi=\frac{1+\sqrt{5}}{2}$$. To help, she uses her abacus but now puts it in position $$(a_1,a_2,a_3,a_4,a_5)$$ to denote the number $a_1+(\phi\times a_2)+(\phi^2\times a_3)+(\phi^3\times a_4) +(\phi^4\times a_5).$ Some numbers now have multiple representations, because $$\phi^2=\phi+1$$. For example, positions $$(2,1,0,0,0)$$ and $$(1,0,1,0,0)$$ both represent the number $$2+\phi$$.

How many ways are there of representing $$4+7\phi$$ on Amy's abacus?