Can U Dig It?
Puzzle
(Apologies, but we’ve drawn a blank with this puzzle’s instructions. One thing we do know is that the answer is a positive integer…)
Solution
We are presented with what appears to be a word search, but with two twists: one letter has been replaced by a dash, and the puzzle asks for a positive integer. My immediate instinct was to look for words. find, trash, and fire were all easy to spot, but it was not clear to me where to go from here. At the same time, I noticed that the northeast corner had what appeared to be a jumbled version of find the start, while the south portion seemed to resemble twenty-six. The latter somewhat reminded me of January’s puzzle.
With no leads, the obvious next step was to find a lot more words. I used a tool to extract the text from the puzzle.1 Using a very long list of English words, I was able to extract terms in two categories: standard word search (straight lines) and snaking (any adjacent letters allowed).
Several words stuck out significantly: hexadecimal, binary, integer, base ten. Since I had already noticed that the “dig it?” in the title could be interpreted as “digit,” I was able to figure out most of the words once I noticed thumb and pinky among the bank. Identifying the can-related words was a bit harder, but words like aluminum and opener helped me confirm I was on the right track. After staring at the grid for a while, I realized that the digit terms were in the shape of an arch—or a U—constituting the third and final titular clue. With this information, I was able to finish most of the grid, including the elusive soda-hallux pair.
To be honest, I never really finished finding all the words,2 but I was able to squint and make out “find the twentieth smallest number with digits totaling twenty-six.” This task is trivial with some code; here is a Julia script for reference.
function digit_sum(n)
total = 0
while n > 0
total += n % 10
n ÷= 10
end
return total
end
counter = 1
numbers = []
while length(numbers) < 20
if digit_sum(counter) == 26
push!(numbers, counter)
end
counter += 1
end
println(numbers[end])
Of course, this gives us 3599.
Looking back, I found this puzzle to be extremely difficult—perhaps too difficult for me, as I needed a fair amount of luck—but that made it all the more rewarding when I finally solved it.
Enjoy Reading This Article?
Here are some more articles you might like to read next: