2010 AMC 10A Problems/Problem 25
We can find the answer by working backwards. We begin with on the bottom row, then the
goes to the right of the equal's sign in the row above. We find the smallest value
for which
(and that does not violate the fact that
must be the greatest perfect square less than
), which is
.
We continue with the same way with the next row, but at the fourth row, we see that solving yields
, but in that case we would have to make the
a
since
. So we make it a
and solve
. We continue on using this same method where we move the perfect square up by
each time until we get the
to be a value greater than the RHS plus the perfect square number. When we repeat this until we have
rows, we get:
Hence the solution is the last digit of , which is
.