[Guest]

Rules: Assign a unique value from 1 to 9 for each rectangle A-I, so that the regions identified add up to the

displayed value.

Have fun!

Edited January 20, 2011 by savidbor

[Guest]

Possible values are...

A - 8

B - 5

C - 3

D - 1

E - 4

F - 6

H - 2

I - 9

Edited January 20, 2011 by zbaseball

[Guest]

Rules: Assign a unique value from 1 to 9 for each rectangle A-I, so that the regions identified add up to the

displayed value.

Have fun!

A=8

B=5

C=3

D=1

E=4

F=6

G=7

h=2

i=9

The trick:

1) Determine how many numbers are used for each value:

16 = 3 numbers

30 = 4 numbers

18 = 3 numbers

7 = 2 numbers

2) enumerate the legal permutations:

16 = 961 952 943 871 862 853 754

30 = 9876

18 = 981 972 963 954 873 864 765

7 = 61 52 43

3) Determine commonalities

16 & 30 share A only

16 & 18 share B only

16 & 7 share nothing

30 & 18 share I only

30 & 7 share F only

18 & 7 share nothing

4) Since 30 & 18 share I, and the only commonality between all permutations is 9, I = 9

etc., etc., etc.

IR SMART!

[Guest]

More clarification, since 30 & 18 can have only one value in common the only permutation of 18 that is legal is 954

Thus I the only common letter must be 9.

16 and 30 share one and only one thing in common. permutations of 16 involving 9 are eliminated. There can only be one value in common amongst the remaining set, thus 853 is the only choice. This means that A = 8

30 and 7 can only share one thing in common and that must be 6 so, F must be 6.

16 and 18 can only share one thing in common and thats a B = 5.

The remaining values are fill-ins.

H was never used in any permutation it gets the remaining value 2.

A=8

B=5

C=3

D=1

E=4

F=6

G=7

h=2

i=9

The trick:

1) Determine how many numbers are used for each value:

16 = 3 numbers

30 = 4 numbers

18 = 3 numbers

7 = 2 numbers

2) enumerate the legal permutations:

16 = 961 952 943 871 862 853 754

30 = 9876

18 = 981 972 963 954 873 864 765

7 = 61 52 43

3) Determine commonalities

16 & 30 share A only

16 & 18 share B only

16 & 7 share nothing

30 & 18 share I only

30 & 7 share F only

18 & 7 share nothing

4) Since 30 & 18 share I, and the only commonality between all permutations is 9, I = 9

etc., etc., etc.

IR SMART!

[Guest]

Thanks for the puzzle.

a-8

b-5

c-3

d-1

e-4

f-6

g-7

h-2

i-9

@zbaseball - you forgot to show 'G' in your answer though you obviously got it right.

[Guest]

First, write the equations that are illustrated:

A+F+G+I = 30

A+B+C = 16

B+E+I = 18

D+F = 7

Next, start whittling away. First, notice that the only way four numbers between 1 and 9 can equal thirty is if those numbers are 9,8,7, and 6. So A,F,G,I = 6,7,8,9. Thus, B,C,D,E,H = 1,2,3,4,5.

Since D+F = 7, the only combination from the groups above that work are D=1 and F=6. Now A,G,I = 7,8,9 and B,C,E,H = 2,3,4,5.

Since B+E+I = 18, if I<9, B+E>9, which is impossible. So I=9 and B+E = 9, which can only be made with 5+4. Now A,G = 7,8; B,E = 5,4; and C,H = 2,3.

Since A+B+C = 16, if A=7, B+C=9, which is impossible. So A=8, leaving G=7.

Since A+B+C = 16 and A=8, B+C=8. So B=5, C=3, and the remaining values are E=4 and H=2.

[Guest]

A + B + C + D + E + F + G + H + I = 45

D + F = 7 => D,F max 6

A + F + G + I = 30 => A,F,G,I = 6,7,8,9 => F = 6 => D = 1

B + E + I = 18

A + B + C = 16

F = 6

D = 1

A + B + C + E + G + H + I = 38

A + G + I = 24

B + E + I = 18

A + B + C = 16

=> 2A + 2B + C + E + G + 2I = 58

=> A + B + I - H = 20

=> C + H = I - 4 ( C,H min 2,3 => C+H min 5) => I = 9

I = 9

A + G = 15 => A,G = 7,8

B + E = 9

A + B + C = 16 ( A max 8, B max 5, C max 3) => A = 8 & B = 5 & C = 3

A + B - H = 11

C + H = 5 => C,H = 2,3

A = 8

B = 5

C = 3

G = 7

E = 4

H = 2

CONCLUSION:

A = 8

B = 5

C = 3

D = 1

E = 4

F = 6

G = 7

H = 2

I = 9

[Guest]

Thanks for the puzzle.

a-8

b-5

c-3

d-1

e-4

f-6

g-7

h-2

i-9

@zbaseball - you forgot to show 'G' in your answer though you obviously got it right.

HAH @spamwolf, thanks. I can figure out the puzzle but can't seem to manage my ABC's.

[Guest]

A=8, B=5, C=3, D=1, E=4, F=6, G=7, H=2, I=9

Method:

D+F=7 so D,F = {1,2,3,4,5,6}

A+F+G+I=30 so A,F,G,I = {6,7,8,9}

F therefore must be 6, making D=1

A+B+C=16, Since A={7,8,9}, 7<(B+C)<9 so B,C={2,3,4,5}

E+B+I=18, Since I={7,8,9} and B={2,3,4,5} 9<(B+I)<14 so 4<E<9, but D=6 and A,G,I={7,8,9} so E={4,5}

This now makes 13<(B+I)<14 Limiting I={8,9} and B={4,5} and there for C={2,3}

with E,B={4,5}, E+B=9 making I=9

B{4,5} + C{2,3} is now a maximum of 8 so A+(Bmax+Cmax)=A+(8)=16 so Amin=8 and since I=9 A=8

Since A=Amin, C=Cmax and B=Bmax so B=5 and C=3

This leaves E{4,5} and since B=5 the E=4

and G{7,8,9} with I=9 and A=8 then G=7

Though H's number is irrelevant the only number left is 2 so H=2

[Guest]

Rules: Assign a unique value from 1 to 9 for each rectangle A-I, so that the regions identified add up to the

displayed value.

Have fun!

My work was done on paper but my original equations were:

A+B+C=16

A+F+G+I=30

B+E+I=18

F+D=7

now...A,F,G,I all must represent integers 6-9 since those add up to 30. Since F+D=7, F must be 6 and D is therefore 1.

So now:

A+B+C=16

A+G+I=24

B+E+I=18

We know that A,G,I must represent 7-9, which leaves 2-5. The only way to make B+E+I=18 now is for I=9 and B,E to represent 5 and 4.

So now:

A+B+C=16

A+G=15

B+E=9

Of each letters' possible set of numbers, A,B,C must be their respective maximums so A=8, B=5, and C=3, which leaves G=7, E=4, and lonely H=2.

TLDR:

A=8

B=5

C=3

D=1

E=4

F=6

G=7

H=2

I=9