Behind the Question Mark 2
In the following puzzle, dashed arrows denote addition while solid arrows denote multiplication:
If all circles contain distinct positive integers from to , how many different possible integers replace the question mark?
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1 solution
The 3 multiplicative circles can only be 2,3,6 or 2,4,8 which comply with xy = z < 10 for x ≠ y ≠ z. For 2,3,6, the complementary summative components are {1,4,5,7,8 & 9} with joint collaboration of 2 or 3 for a total sum of 36 or 37 respectively. On the other hand, for 2,4,8, the complementary summative components are {1,3,5,6,7 & 9} with joint collaboration of 2 or 4 for a total sum of 33 or 35 respectively.
In short, we have :
1) sum of {1,4,5,7,8 & 9} + 2 = 36
2) sum of {1,4,5,7,8 & 9} + 3 = 37
3) sum of {1,3,5,6,7 & 9} + 2 = 33
4) sum of {1,3,5,6,7 & 9} + 4 = 35
Looking at 1), obviously 9 should be in one of the lower sum circles but not at the middle one. Why? Because the total is even and the 4 upper summands will have the same total as those of the left and right lower circles as their 2 sums.
36 - (double the sum of upper circles) = (even number at middle lower circle) = {4 or 8}
Since 3 is being placed at the far right upper circle as factors of 6, then 4 by the summation of 1+3 is impossible to be placed at middle lower, leaving 8 to be there. This 8 must be the the sum of 1 and 7 as the only possible summands available within {1,4,5,7,8 & 9}. 9 must have either 1 or 7 as its summand, but 8 is already taken (midlow) so 1+8 is a no go and 7+2 is now confirmed. From the above, 1 must be at the ? circle and be paired with 4 for a sum of 5.
Solution 1 :
4-5-(? = 1)-8-7-9-2-6-3
Looking at 2), using the same arguments as above, midlow must be an odd number with its summands both available among {1,4,5,7,8 & 9}. That could be one of these three possibilities : 1+4=5 or 1+8=9 or 4+5=9. But 8 as a MIDDLE summand is impossible since it's already too large by itself and any other summands paired to it will exceed a sum of 9. And keeping in mind that a 1 must be in an upper circle (it can only be summand and never a sum as the smallest of the bunch), for 4+5=9 a 1 must be paired with either of the summands, but 5 is unavailable being another upper circle and 6 is the 'leader' of the multiplication group. Knowing 5 as the midlow, it's easy to see that ? is again 1 for a Solution 2 as below :
8-9-(? = 1)-5-4-7-3-6-2
Looking at 3), using the same arguments as above, midlow must be an odd number with its summands both available among {1,3,5,6,7 & 9}. That could be one of these two possibilities : 1+6=7 or 3+6=9. Both have a mutual summand of 6 which cannot be paired with the collaborator 2 since 8 is part of another group. Even without solving it fully we already reached the conclusion of ? = 6. Anyway, the Solution 3 is :
1-7-(? = 6)-9-3-5-2-8-4
Looking at 4), using the same arguments as above, midlow must be an odd number with its summands both available among {1,3,5,6,7 & 9}. That could be one of these two possibilities : 1+6=7 or 3+6=9. Both have a mutual summand of 6 which cannot be paired with the collaborator 4 since 10 is a two digits number. Even without solving it fully we already reached the conclusion of ? = 6. Anyway, the Solution 4 is :
3-9-(? = 6)-7-1-5-4-8-2
Conclusion : ? = 1 or ? = 6.
Answer = 2 different positive integers.