[Guest]

Substitute each of the capital letters in bold by a different digit from 1 to 9 to satisfy this alphametic equation, where I is a perfect square and SO > GO.

(SO)*(I)*(GO) = (O)* (DO)*(IT)

[bushindo]

Substitute each of the capital letters in bold by a different digit from 1 to 9 to satisfy this alphametic equation, where I is a perfect square and SO > GO.

(SO)*(I)*(GO) = (O)* (DO)*(IT)

Maybe it's just me, but don't the O and I cancel out of both sides of the equation? If cancelling is possible, then the resulting system isn't that formidable.

[Guest]

Maybe it's just me, but don't the O and I cancel out of both sides of the equation? If cancelling is possible, then the resulting system isn't that formidable.

This is an alphametic puzzle in conformity with the definition given here.

In accordance with this definition, each of "SO", "DO","GO" and "IT" denotes a two digit number and does not represent the products: S*O, D*O, G*O, I*T.

Therefore, the premise of the said cancellation does not arise in the current instance.

Edited August 5, 2009 by K Sengupta

[Guest]

Found 1 solution so far; checking to see if it is the only one

65*9*35 = 5*45*91

I just saw that "I" must be a perfect square... that makes life easier!

As I said (easier), the above posted soln is the only soln.

Edited August 5, 2009 by DeeGee

[Guest]

Found the same answer, mostly by brute force.

If we expand both sides of the equation, we get:

100*S*G*I + 10*(S+G)*I*O + I*O² = 100*D*I*O + 10*D*T*O + 10*I*O² + T*O²

The first 2 terms on the left side and first 3 terms on the right will result in x0 (x is 1 or more digits) so none will affect the least significant digit.

So I*O² = xN and T*O² = xN. There are 16 combinations of O, I, and T that satisfy this.

I tried to extend this to the next significant digit but the expression gets too complex and includes all of the unknowns so I don't see any potential in fllowing this line. So, a simple set of nested loops to search all combinations of S, G, and D with the 16 valid combinations of O, I, & T finds the answer given above. In retrospect, it would have been faster to use brute force alone and test every combination of the 6 unknowns.

Does anyone see a better analytical solution?

[Guest]

Does anyone see a better analytical solution?

The way I solved it was to first identify the possible values that O may take given that you have O²xI on one side and O²xT on the other for the least significant digits.

Those combinations gave you some combinations for O,I and IT (probably 16 as you said).

Since "I" could be only 1,4 or 9, its easy to easy that if "I" is one, the combination will not hold true for for "I" = 1, because IT was 16, 32 or 64 and O was 2,4 or 8 respectively. This meant that for the equation to balance, you needed to get same powers of 2 on both sides and it wasn't possible.

Now I don't remember what other eliminations I had made (based on getting equal powers of primes on both sides), but you could narrow down the possibilities to 1 or 2 combinations of O, I and IT.