Reducing Combinations Now we are into a phase of
reducing combinations. Look at the last cell of the block for the 7
in Figure 4. It can only be 1 or 4, but we also know that we need a
total of 10 to complete the 15 block that intersects with the last
cell of the 7 block in two. Making 10 with either a 1 or 4 in it
would give us either 1 and 9 or 4 and 6. The 1 and 9 combination
doesn't work, because the only other square available in that 15
block intersects with the 16 block where only 1, 4 or 6 are
available as candidates. So, it must be 4 and 6 and the 4 must be
the last digit in the 7 block. The remaining 6 can nosw complete the
15 block. We can also complete the 7 block with the remaining 1 to
complete the sum.
Locked Values Tackling the 16 block in Figure 5, where
just two cells remain, we must make 5 from a 1 and a 4 - the only
digits remaining in that known combination. We can see that the
second cell cannot be a 1, because there is already a 1 in the 15
block - it's a locked value - so it must be 4. And we have solved
the 16 block.
All that remains in Figure 6 is the last number in the 15 block,
which must be a 7 to make up the sum, and then the 5 in the 12
block, which is proved by the sum of the 6 block. Job
done!
