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|Difficulty:|| (3.12) |
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.
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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