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Seven tiles need to be placed so that one is placed in each of the curved "regions" formed by the three circles.

One tile has been placed for you.

Each circle has a corresponding statement. Your placement should ensure that these statements hold true.
What is the unique placement? (and why?) enter image description here

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2 Answers 2

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A solution by hand:

First, the minimum sum in the upper-left loop is 2+3+4+5 = 14, so X >= 14.
The maximum product in the bottom is $3*6*7*8$ = 1008
The product includes at least 1 power of 3, at most 2 powers of 3, and at most 1 power of 5, and no primes greater than 7.
$14*15$ = $2*3*5*7$, maybe
$15*16$ = $2^4 *3*5$, maybe
$16*17$ includes 17, no
$17*18$ includes 17, no
$18*19$ includes 19, no
$19*20$ includes 19, no
$20*21$ = $2^2 * 3 * 5 * 7$, maybe
$21*22$ includes 11, no
$22*23$ includes 11 and 23, no
$23*24$ includes 23, no
$24*25$ includes two powers of 5, no
$25*26$ includes 13 and 2 powers of 5, no
$26*27$ includes 13 and 3 powers of 3, no
$27*28$ includes 3 powers of 3, no $28*29$ includes 29, no
$29*30$ includes 29, no
$30*31$ includes 31, no
$31*32$ includes 31, no
$32*33$ > 1008, too big.

So X is 14, 15, or 20. If X=14 or X=15, then the upper-left loop requires 2,3,4,5 or 2,3,4,6, and the upper right loop will be too big. So X=20, and the bottom loop has 3,4,5,7.
The right loop must have 3, one of (4,5,7), and two of (2,6,8) that sums to 21. To have an odd sum, it must use the 4. The other two digits sum to 14, so must be 6,8.
The left loop must have 3, one of (5,7), and two of (2,6,8) that sums to 20. The only way to do that is 3,7,2,8.

Now that it's known which digits fall in each loop, the exact placement is straightforward.

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  • $\begingroup$ Nice approach. Thx for posting $\endgroup$
    – JLee
    May 22 at 11:56
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The unique placement is

enter image description here

where x =

20

Here is my thinking:

I looked at the equations A+B+C+3+1=B+E+F+3, which reduces to A+B+C+1=B+E+F, and CxDxFxG = (A+B+C+3)(B+E+F+4) and my eyes started to glaze over because of all the variables.

Then I had the idea to brute force it, so I

went to this website and generated a list for all 720 permutations of 2,4,5,6,7,8. I put them in an Excel spreadsheet, inserting a 3 in the middle of each permutation. Then I broke out each of the numbers into its own column. At that point, the first 7 lines of the 720 looked like this: enter image description here

Then, I named each region in the diagram a letter from A-G and added a few columns to calculate each of the three relationships.

enter image description here

Then I added a couple columns that I could filter on based on the relationship calculations.

enter image description here

Then I filtered down the rows to where the last 2 columns were true, leaving me with just 1 out of the 720 records.

enter image description here

I assume there is a way to use mathematics to arrive at this same result, or to at least help one trim down the 720 possibilities, but I do not think that I am the guy for that, at least not right now. However, if a math-based solution is posted, I will read it and most likely upvote it because it is interesting.

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  • $\begingroup$ Whilst this is the right answer, you are correct in that there is a less brutish way $\endgroup$ May 22 at 10:31

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