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Fill in the nine X spots in the following equation using each of 1, 2, ..., 9 exactly once, so that the equation is satisfied. The XX in the denominator denotes concatenation of two digits.

$$ \frac{X}{XX} + \frac{X}{XX} + \frac{X}{XX} = 1 $$


Source: the problem for January 1 on Mathematics Calendar 2021, published by the Korean Mathematical Society. The calendar has 365 small mathematical amusements, one for each day. A sample image of the calendar can be found on P. S. Park's blog post (Korean, contains spoiler to the above problem).

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  • $\begingroup$ I could reach 0.75 , not 1 . Maybe a Program is going to be very useful. $\endgroup$
    – Anonymous
    Commented Jan 20, 2021 at 2:34

2 Answers 2

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The following works:

9/12 + 5/34 + 7/68.

Confession:

I found this after seeing Bubbler's remark that Joe Kerr's non-solution is pretty close to being right. I was already trying things along similar lines, with e.g. 8/12 + things making 1/3 or 9/12 + things making 1/4, but I hadn't yet tried 17s in the denominator :-).

Someone asked in comments whether there was a logical path to the solution. Not really, especially as part of the actual path to the solution was getting a nudge from OP's comment on someone else's answer! But I can say a few things:

First of all, note that if three numbers add up to 1 then their average is 1/3. So maybe all three are pretty large, or maybe one of them is large on its own and the others fill the gap. A promising approach is to begin with 8/12 (= 2/3) or 9/12 (= 3/4) and look for two fractions adding up to the 1/3 or 1/4 left over. E.g., write 1/4 = 1/6 + 1/12 and replace those two with other equivalent fractions. Another is to pick some denominator (let's consider 13 as an example), find some fractions with that denominator that obviously add up to 1, and then replace two of the fractions with equivalent ones. So e.g. you might start with 1/13 + 3/13 + 9/13 = 1 and then try to replace the first two fractions with ones having denominators 26, 39, 52, 65, 78. Unfortunately this doesn't work out because 13 and 39 share a 3, 26 and 52 share a 2, putting 65 in the denominator will give you a 5 in the numerator, and for 78 in the denominator there are only a few things that could conceivably work and they all turn out not to. ... Beyond that it's basically trial and error, learning along the way some things that aren't likely to work (e.g., 13 is generally a bad denominator for the reasons above along with the fact that you can't use 2/26 or 2/52, or 1/13 or 2/26 or 3/39 or 5/65; this latter problem is shared even more badly by 12) and therefore trying things that use progressively larger denominators. I hadn't yet tried many things using 17 at the point when I saw Bubbler's comment on Joe's answer :-).

Of course, if you wanted to be really methodical you could just ask a computer. There are only 9! ways to arrange those digits; some of them are equivalent and some of them are ridiculous, but 9! is pretty small so easier just to brute-force it. I did this (after posting the answer) and it turns out that the only solutions are the 6 permutations of this one.

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  • $\begingroup$ Correct, it's the intended answer. And yeah, rot13(ab bar jbhyq rnfvyl rkcrpg fhpu n ynetr cevzr snpgbe va gur qrabzvangbe) :) $\endgroup$
    – Bubbler
    Commented Jan 20, 2021 at 4:00
  • $\begingroup$ Clever solution, is there any specific logical path or deduction which led to the solution? (Of course, what you did is to play with the problem) . $\endgroup$
    – Anonymous
    Commented Jan 20, 2021 at 6:38
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    $\begingroup$ There isn't anything I would call a logical path. I'll add a few notes on the thought processes involved to my answer. $\endgroup$
    – Gareth McCaughan
    Commented Jan 20, 2021 at 10:30
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My answer:

9/12, 5/34, and 8/76.

Reasoning:

(9/12 = .75), (5/34= .14705882352) and (8/76= .10526315789) When these are added together, they add up to 1.00232198142. This rounds down to 1.

Although this is not a (fully correct) answer, as it does not add exactly to 1, it was the closest I could find.

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  • $\begingroup$ Of course rounding is not allowed, but your fractions are pretty close to the actual answer. $\endgroup$
    – Bubbler
    Commented Jan 20, 2021 at 3:38

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