Are we nearly there?

Thank you to everyone who submitted solutions to this problem. There were lots of good solutions; well done!

Ollie and Ziggy, from Priestlands School, made this observation:

The line goes: up 1 step, diagonally 1 step, across 1 step, diagonally 2 steps, and so on.

Each new diagonal line is 1 step longer than the last diagonal line.

Matt, Tom and Ollie, also from Priestlands School, were able to tell us how to work out the number of points in the previous rows.

The lengths of the diagonals can be found by adding $1+2+3+4+5+...$, ending at the length of the final diagonal.

These sums are also known as the triangle numbers.

Sunny, from Carmel Pak U Secondary School, Hong Kong, and Phil, from Our Lady of Lourdes School, Canada were able to use this to find the coordinates of 100. Here is Sunny's solution:



Looking at the 15th line, this has $1+2+3+4+5+6+7+8+9+10+11+12+13+14 = 105$, so we are ahead by $5$ dots.

Since line three goes down, line 4 goes up and in general the even lines go up and the odd lines go down, the line is going down.

Therefore: $105$ is at $(14,1)$.

$104$ is at $(13,2)$

$103$ is at $(12,3)$

$102$ is at $(11,4)$

$101$ is at $(10,5)$

$100$ is at $(9,6)$

Sunny then went on to describe how to find the number at $(60,40)$.

Since $60+40=100$, $(60,40)$ is on line $100$, so the final number on this diagonal is $4950$, as this is $1+2+3+4+...+98+99$.

Since $100$ is even, this line is going up, so we need to go back $59$ steps to get back to $(60,40)$.

This means the number is $4950 - 59 = 4891$.

Ziggy and Ollie found an alternative route that visited all the numbers:

Hannah sent in this solution which explained very clearly how she was able to work out how far along the route each point was: