Numberrangements
Logic puzzles require you to think. You will have to be logical in your reasoning.
In Numberrangements, you are given an arrangement of letters. The letters represent all of the whole numbers from 1 to the total number of letters used. Each letter represents a different number. Using the clues given, find which number each letter represents.
ABC
DEF
GHI
1. The sum of the top row is greater than the sum of the middle row, which is greater than the sum of the bottom row.
2. E is a prime factor of G.
3. F is greater than A.
4. The sum of B and G is equal to H.
5. I is not 1.
Answer
A=7, B=1, C=9
D=6, E=2, F=8
G=4, H=5, I=3
From Clue 2, E must be 2, 3, 5, or 7, but it cannot be 5 or 7 because that would force G to be a greater multiple of 5 or 7, and only the numbers 1 to 9 are used. E must be 2 or 3 and G must be 4, 6, 8, or 9.
Let's try I=2. In that case, E=3 and G=6. H must be greater than 6 because of Clue 4. The most G+H+I can be is 14 because of Clue 1, since the sum of all nine numbers is only 45. So, I cannot be 2, and it therefore must be at least 3 because of Clue 5.
If I=4, then G cannot be 4, and must be 6, 8, or 9. Again, since H must be greater than G, the bottom row's sum will be too high. If I = 5, then the minimums for G and H would be 4 and 6, respectively  a sum of 15. If I = 6, then the minimum sum again is 15.
This means I MUST be 3, which forces E=2 and G=4, 6, or 8. If G is 6 or 8, the sum of the bottom row will be too great, so G=4. The greatest that H can be is 7, which also means that B can only be 1, 2, or 3, from Clue 4. But 2 and 3 are already used, so B=1 and H=5.
This leaves A, C, D, and F, to be paired with 6, 7, 8, and 9, in some order. We know from Clue 1 that A+B+C > D+E+F, but E is already 1 greater than B, and from Clue 3, we know that F > A. This means that C must be at least 3 greater than D to make up for the top row trailing by at least 2. So, C must be 9, and D must be 6. This leaves 7 and 8 to be A and F, so of course, F=8 and A=7.
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Comments
charlottesodd
Dec 23, 2013
 Another difficult one (for me anyway)! Good though I like it 
eighsse
Dec 23, 2013
 Thanks glad you liked it and yes it's quite difficult. 
castiel
Feb 20, 2014
 Very fun! Would love to see more like this 
eighsse
Feb 20, 2014
 Thank you casteil! I will try to come up with more 
eighsse
Feb 20, 2014
 castile* sorry! 
eighsse
Feb 20, 2014
 LOLOL. Let's try again. castiel. 
EnderofGames
May 11, 2015
 I came up with 854/619/273, forgetting that people don't consider 1 to be prime. Silly me. 
eighsse
May 12, 2015
 Good try anyway EoG; I'm sure you'd have gotten it had you remembered the weirdness of 1 not being prime. It's debatably prime, really, but convention is to consider it nonprime  mostly because it keeps the fundamental theorem of arithmetic simple to state. 
asertus0
Jul 11, 2015
 I think I found alternative solution.
A=5; B=7; C=6; D=4; E=2; F=9; G=1; H=8; I=3 
eighsse
Jul 11, 2015
 The only problem is, 2 is not a prime factor of 1. 
Babe
Aug 06, 2015
 I was smart enough to not even try this one. Besides that, I was never a math person. Got by in school with a passing grade and that was it for math. I did excellent in English and other subjects. Just not math! UGH! 
Snowdog
Oct 12, 2016
 Terrific puzzle!
My initial reaction was that there is not enough info to sort it out. It is an interesting mix of limiting possibilities by deduction, then switching to a bit of trial and error. I don't see a direct deductive solution.
This is one of my favorites. 
eighsse
Oct 12, 2016
 Thanks, Snowdog! Very good to hear that you enjoyed it. 
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