On the screen is shown a “00/81”. A number pad with number keys 1 to 9 is on the right. When a number key is entered, a knob is lit in red and the value of the number is added to the numerator of the displayed ratio. There are 10 unlit knobs.
The puzzle should be obvious. Match the numerator and the denominator with 10 entries.
So how would I solve it? My hint as stated, was:
the last entry is the most important
The first 9 entries are to get you as close to the required number as possible. This is actually similar to blackjack, a card game where you try to get as close to 21 points as possible without going over. The difference is that you don’t get to choose your next number (or card value) in blackjack. You do here.
[image by pavlen]
So my first instinctual thought was to finish up the first 9 entries to get to the last entry. My second instinctual thought was to make these 9 entries all the same number value. It’s a timed effort. I don’t have time to go figure out different combinations of sums.
Based on those thoughts, my next (instinctual) step was to divide the required number by 9, rounding down to nearest integer. Don’t ask me why at this point, because I’ll know the reason only after thinking it through, and I’m not thinking it through at this point.
So we have 81 divide by 9, and rounded down, we get 9. We have a problem. It’s not just 9. It’s exactly 9. There’s no number value 0 for the final entry. So we use 8 (1 less than 9).
Then we multiply 8 by 9 (entries) to get 72. And then 81 (required number) – 72 is 9, the final entry. So the answer is 8,8,8,8,8,8,8,8,8,9.
This method gives the simplest of combinations (only 2 distinct numbers used) and is the fastest to get you to the end point, the final entry. It’s the final entry that “corrects” the sum to the required number.
I have to say, the changing of 9 to 8 is (I’m starting to use the word generously) instinctive. I don’t know how changing 9 to 8 would solve the problem. I just know.
Let’s use this method on the second part of the puzzle: 67.
67 divide by 9 rounded down gives 7. 7 multiplied by 9 is 63. 67 – 63 is 4. So the answer is 7,7,7,7,7,7,7,7,7,4.
There you have it, the algorithm to solving the math puzzle. The second puzzle uses the “normal” mode of the algorithm. The first puzzle tested the edge case of the algorithm.
Hmm… can we write a function that spits out the combination using the above algorithm? Can we write a function that spits out all possible combinations?