SOLUTION: in the following addition problem, each letter represents a different interger from 0 - 9, inclusive. If G=0 and N=4, what are the rest of the values? TEN NINE + EIGHT _

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Question 551209: in the following addition problem, each letter represents a different interger from 0 - 9, inclusive. If G=0 and N=4, what are the rest of the values?
TEN
NINE
+ EIGHT
________
SEVEN
Answer by Edwin McCravy(20063) About Me  (Show Source):
You can put this solution on YOUR website!
  TEN =   TEN
 NINE =  NINE
EIGHT = EIGHT
-----   -----  
SEVEN = SEVEN

We put G=0 and N=4 and know there is a carry 1 to the leftmost 
column, otherwise E and S would be the same, so we put a 1 above 
the left most column, and we will put other "carrys" above when we
see what they have to be:

        1
  TEN =   TE4
 NINE =  4I4E
EIGHT = EI0HT
-----   -----  
SEVEN = SEVE4

In the rightmost column E+T has to be 10 to add to the 4 to make 14,
which means there is a carry 1 to the next to the right column:

        1  1
  TEN =   TE4
 NINE =  4I4E
EIGHT = EI0HT
-----   -----  
SEVEN = SEVE4

In the next to the rightmost column, 4 + H + the 1 to carry must
be 10 to add to E to make it 10+E. That makes H=5, and there is 1 
to carry to the middle column:


        1 11
  TEN =   TE4
 NINE =  4I4E
EIGHT = EI05T
-----   -----  
SEVEN = SEVE4

From the next from the left column, 4+I = 10+E, so I = 6+E
So E < 4 to keep I < 10.  E can't be 3 because that would make 
S be 4 in the leftmost column, and we've already used 4.  And E
can't be 0 since we've already used 0. So E is either 1 or 2.

Let's try E=1, which makes S=2 in the leftmost column, and T=9 
in the rightmost column, causing a carry 1 to the next to the
left column:

        1111
  TEN =   914
 NINE =  4I41
EIGHT = 1I059
-----   -----  
SEVEN = 21V14

That makes I have to be 6 to make the next to the leftmost column 
work:

        1111
  TEN =   914
 NINE =  4641
EIGHT = 16059
-----   -----  
SEVEN = 21V14

But that doesn't work because it makes V also have to be 6 to make
the middle column work.

So E cannot be 1. So we must go back to what we had just before we 
tried E=1, which was:

        1 11
  TEN =   TE4
 NINE =  4I4E
EIGHT = EI05T
-----   -----  
SEVEN = SEVE4

E's only other possibilty is E=2, making S=3 in the leftmost column:

        1 11
  TEN =   T24
 NINE =  4I42
EIGHT = 2I05T
-----   -----  
SEVEN = 32V24

Immediately we see that T=8 in the rightmost column,
causing there to be a carry 1 from the middle column
to the next to the leftmost column:

        1111
  TEN =   824
 NINE =  4I42
EIGHT = 2I058
-----   -----  
SEVEN = 32V24

and I must be 7 to make the next to the left column work:

        1111
  TEN =   824
 NINE =  4742
EIGHT = 27058
-----   -----  
SEVEN = 32V24

Finally V has to be 6 to make the middle column work:

        1111
  TEN =   824
 NINE =  4742
EIGHT = 27058
-----   -----  
SEVEN = 32624

Edwin