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 Subject: Math Category: Reference, Education and News > Homework Help Asked by: theginleys-ga List Price: \$5.00 Posted: 02 Mar 2005 07:03 PST Expires: 01 Apr 2005 07:03 PST Question ID: 483425
 ```I need help solving my daughter's 7th grade math problem: Substitute a digit for each letter (each letter is a different number, same letters are the same number & no two different letters are the same number.) C O R N + P O T A T O + T O M A T O ______________ C A R R O T```
 Subject: Re: Math Answered By: omnivorous-ga on 02 Mar 2005 08:11 PST Rated:
 ```Theginleys ? In order to solve this we want to start with a column with only 2 variables ? that?s column 2: 2O = A From that we know that column 5 resolves to only 2 variables ? column 5: O + 2O + 2O = R or: 5O = R --- At this point we don?t know what numbers A or O or R could be. But we can choose a number for ONE of them. And the number can be 1 or 10 or 100. It doesn?t matter, other than it can't be zero because that would make O, A and R all the same. For simplicity, let?s choose O = 1, because we know that 5O = R and also that 2O = A and it?s easy to use. Again, you could choose 10 for O, then R would be 50 and A would be 20. But 1 is just as easy, as long as you?re comfortable with negative numbers. --- So here?s what we know so far: A = 2 O = 1 R = 5 And columns 2 and 4 are solved. What else can we solve? --- Column 5: 2T = O ?R; T = -2 Column 6 : 2O = T ? N = 2 = -2 ? N ; N = -4 So we now have : A = 2 N = -4 O = 1 R = 5 T = -2 --- But that still leaves two problem columns ? C4 and C1, where we have 2 variables left in each. Column 1: T = C ? P -- or -2 = C - P Column 3: R-T = C + M ? or 7 = C + M The challenge is to select a value for C that fits ? and remember that one of our rules is that each letter has a unique number. I say, let?s assign C = 3; making P = 5 and M = 4. That doesn?t work. Let?s try C = 8; now P = 10 and M = -1 ? that works (but so too would C = 10). Now we have: A = 2 C = 8 M = -1 N = -4 O = 1 P = 10 R = 5 T = -2 --- And let?s check our columns: C1: P + T = C = 8 C2: O + O = A = 2 C3: C + T + M = R = 8 - 2 - 1 = 5 C4 : O + A + A = R = 5 C5 : R + T + T = O = 1 C6 : N + O + O = T = -4 + 1 + 1 = -2 It looks like it works ! Best regards, Omnivorous-GA```
 theginleys-ga rated this answer: `THANKS!!!! Great help!!!`

 ```Aren't the words supposed to be resolving to numbers? If so then all letters should be positive numbers. Also, the additions could be bigger than 10, for example consider the last column: N + 2O = T: If N is 5 and O is 3 then T is 1, not 11 since it should be a single digit. So we have 3 possible equations: N + 2O = T OR N + 2O = 10 - T OR N + 2O = 20 - T and so on for the other columns. Does this make sense?```
 ```Oops these should have been N + 2O = T OR N + 2O = 10 + T OR N + 2O = 20 + T```
 ```I found 3 solutions: 3 7 5 6 + 2 7 0 4 0 7 + 0 7 1 4 0 7 ______________ 3 4 5 5 7 0 6 4 3 7 + 1 4 5 9 5 4 + 5 4 0 9 5 4 ______________ 6 9 3 3 4 5 8 1 9 4 + 2 1 6 3 6 1 + 6 1 5 3 6 1 ______________ 8 3 9 9 1 6```