Congratulations to the following for solving this problem!
Hamsika Maddali – The Netherhall School
Premalatha Balan
This problem revolves around Pascal’s Triangle: an arrangement of numbers that starts with a 1 at the top, and each number below is the sum of the two numbers above it.
One method to approach this problem is to consider a generic sequence of increasing length, building up to 8.
Let us start with length 3, which requires 2 iterations to reach a single element.
The powers should start to look familiar. For length 4, we require 3 iterations.
You should notice that for a length 
sequence, the powers follow the 
th row of Pascal’s triangle. So for a sequence of length 8, the powers will follow the 7th row of Pascal’s triangle.
Since we generalised the sequence at the start of the solution, replacing each variable with their original value yields a closed expression for the solution. Let us call this 
.
We were also asked to find the number of trailing zeros in , which can be achieved by considering factors of 10 and therefore 2 and 5. There are more factors of 2 than 5, so we only need to find the number of factors of 5 in . There are 35, so the number of trailing zeros is 35.
Some of you may question the “rigour” of this solution. A better method would be to prove that the powers of the variables follow the th row of Pascal’s triangle by induction, but that is beyond what is expected.
Notes from the team
1) Apologies for omitting the word “trailing” in the question. We gave credit to anyone who found an expression for or identified that the number of digits was 81.
2) Although we accept all solutions, a mathematical one is preferred over brute force.
3) In later problems asking for proofs, please note that proof by exhaustion will not be sufficient for credit.
4) We wish all year 11 students the best of luck in their GCSE exams! Remember to revise for all subjects, not just the best one.
