SOLUTION: Each letter 'a' through 'i' stands for a different digit 1 through 9. a+b+c=9 d+e+f+i=26 i+f+g+h=21 h+g+c=11 b+c+e+f+g=30 What are the values of a,b,c,d,e,f,g,h,i ? Than

Algebra.Com
Question 823348: Each letter 'a' through 'i' stands for a different digit 1 through 9.
a+b+c=9
d+e+f+i=26
i+f+g+h=21
h+g+c=11
b+c+e+f+g=30
What are the values of a,b,c,d,e,f,g,h,i ?
Thank you
Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
(1) a+b+c=9
(2) d+e+f+i=26
(3) i+f+g+h=21
(4) h+g+c=11
(5) b+c+e+f+g=30 

We know that

     1+2+3+4+5+6+7+8+9 = 45

So therefore we know that

(6)  a+b+c+d+e+f+g+h+i = 45

Add equations (1) and (2) 
(1) a+b+c         =  9
(2)      +d+e+f+i = 26           
    ------------------
(7) a+b+c+d+e+f+i = 35

Subtract equation (7) from equation (6)

(6)  a+b+c+d+e+f+g+h+i = 45
(7)  a+b+c+d+e+f    +i = 35
     ----------------------
(8)              g+h   = 10

Subtract (8) from (4), swapping
h and g in (8) 

(4)              h+g+c = 11
(8)              h+g   = 10
                 ----------
                     c =  1
Substitute in (5)

(5) b+c+e+f+g = 30

    b+1+e+f+g = 30

(9)   b+e+f+g = 29   

From (9), those four digits b,e,f, and g have to be large to add to 29, so if
they were as large as they could be, 9,8,7,6, their sum would be 9+8+7+6=30,
just 1 more than 29. And the only way we can make that sum 1 smaller is to
replace the 6 by a 5, and have 9+8+7+5=29.  But we don't yet know which letter
equals which number. However we know that these two sets are equal:

(10)   {b,e,f,g} = {9,8,7,5}

We already have c=1, so the set of the remaining four letters has to equal the
set of the remaining 4 numbers.  So

(11)   {a,d,h,i} = {2,3,4,6}

Put those together so we can look at them:

(11)   {a,d,h,i} = {2,3,4,6}
(10)   {b,e,f,g} = {9,8,7,5}

Now look at (1)

(1)    a+b+c=9

We already have c=1, so substituting,

(1)    a+b+1=9
(1)    a+b=8

So a and b must have sum 8. Look back at (11) and (10). 'a' must come from (11)
and 'b' must come from (10).  So the only way that can be is for 'a' to equal 3
and 'b' to equal 5, for no other pair adds to 8.

So far we have a=3, b=5, c=1

Now the sets in (10) and (11) become

(11)   {d,h,i} = {2,4,6}
(10)   {e,f,g} = {9,8,7}

Now we look at (8)

(8) g+h   = 10

'g' must come from (10) and 'h' must come from (11). So the only way that can
be is for 'g' to equal 8 and 'h' to equal 2, for no other pair adds to 10.

Now so far we have 

a=3, b=5, c=1, g=8, h=2,

Now (11) and (10) become:

(11)   {d,i} = {4,6}
(10)   {e,f} = {9,7}

We look at (3)

(3) i+f+g+h=21.  Substituting g=8 and h=2,
    i+f+8+2=21
     i+f+10=21
        i+f=11

'i' must come from (11) and 'f' must come from (10). So the only way that can
be is for 'i' to equal 4 and 'f' to equal 7, for no other pair adds to 11.

So now we have

a=3, b=5, c=1, g=8, h=2, i=4, f=7

Now (11) and (10) become:

(11)   {d} = {6}
(10)   {e} = {9}

So d=6 and e=9.

So the solution is:

a=3, b=5, c=1, d=6, e=9, f=7, g=8, h=2, i=4

Checking:

a+b+c = 9
3+5+1 = 9

d+e+f+i = 26
6+9+7+4 = 26

i+f+g+h = 21
4+7+8+2 = 21

h+g+c = 11
2+8+1 = 11

b+c+e+f+g = 30
5+1+9+7+8 = 30

Edwin
Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
Question 823348
(1) a+b+c=9
(2) d+e+f+i=26
(3) i+f+g+h=21
(4) h+g+c=11
(5) b+c+e+f+g=30 

We know that

     1+2+3+4+5+6+7+8+9 = 45

So therefore we know that

(6)  a+b+c+d+e+f+g+h+i = 45

Add equations (1) and (2) 
(1) a+b+c         =  9
(2)      +d+e+f+i = 26           
    ------------------
(7) a+b+c+d+e+f+i = 35

Subtract equation (7) from equation (6)

(6)  a+b+c+d+e+f+g+h+i = 45
(7)  a+b+c+d+e+f    +i = 35
     ----------------------
(8)              g+h   = 10

Subtract (8) from (4), swapping
h and g in (8) 

(4)              h+g+c = 11
(8)              h+g   = 10
                 ----------
                     c =  1
Substitute in (5)

(5) b+c+e+f+g = 30

    b+1+e+f+g = 30

(9)   b+e+f+g = 29   

From (9), those four digits b,e,f, and g have to be large to add to 29, so if
they were as large as they could be, 9,8,7,6, their sum would be 9+8+7+6=30,
just 1 more than 29. And the only way we can make that sum 1 smaller is to
replace the 6 by a 5, and have 9+8+7+5=29.  But we don't yet know which letter
equals which number. However we know that these two sets are equal:

(10)   {b,e,f,g} = {9,8,7,5}

We already have c=1, so the set of the remaining four letters has to equal the
set of the remaining 4 numbers.  So

(11)   {a,d,h,i} = {2,3,4,6}

Put those together so we can look at them:

(11)   {a,d,h,i} = {2,3,4,6}
(10)   {b,e,f,g} = {9,8,7,5}

Now look at (1)

(1)    a+b+c=9

We already have c=1, so substituting,

(1)    a+b+1=9
(1)    a+b=8

So a and b must have sum 8. Look back at (11) and (10). 'a' must come from (11)
and 'b' must come from (10).  So the only way that can be is for 'a' to equal 3
and 'b' to equal 5, for no other pair adds to 8.

So far we have a=3, b=5, c=1

Now the sets in (10) and (11) become

(11)   {d,h,i} = {2,4,6}
(10)   {e,f,g} = {9,8,7}

Now we look at (8)

(8) g+h   = 10

'g' must come from (10) and 'h' must come from (11). So the only way that can
be is for 'g' to equal 8 and 'h' to equal 2, for no other pair adds to 10.

Now so far we have 

a=3, b=5, c=1, g=8, h=2,

Now (11) and (10) become:

(11)   {d,i} = {4,6}
(10)   {e,f} = {9,7}

We look at (3)

(3) i+f+g+h=21.  Substituting g=8 and h=2,
    i+f+8+2=21
     i+f+10=21
        i+f=11

'i' must come from (11) and 'f' must come from (10). So the only way that can
be is for 'i' to equal 4 and 'f' to equal 7, for no other pair adds to 11.

So now we have

a=3, b=5, c=1, g=8, h=2, i=4, f=7

Now (11) and (10) become:

(11)   {d} = {6}
(10)   {e} = {9}

So d=6 and e=9.

So the solution is:

a=3, b=5, c=1, d=6, e=9, f=7, g=8, h=2, i=4

Checking:

a+b+c = 9
3+5+1 = 9

d+e+f+i = 26
6+9+7+4 = 26

i+f+g+h = 21
4+7+8+2 = 21

h+g+c = 11
2+8+1 = 11

b+c+e+f+g = 30
5+1+9+7+8 = 30

Edwin
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