[Rain]

(By logical I mean not trial and error, and don't write scripts to help either. =P)

Using the numbers 1-9, fill in the following such that they sum to 100.

A/BxC + DxExF/G + HxI = 100

[Echognome]

Is the first term A/(BC) or is it (A/B)C (=AC/? It would seem that it is the former, else we could have just expressed it as AxB/C without any confusion (and who cares whether we swap the values for A and C). However, it would be helpful to know for sure.

[fred]

Assume the first term means A/BC (and not AC/?

There are some things you can figure out by logic for sure.

For example, G cannot be either 5 or 7.

If I was going to spend a lot of time on this (most likely I am not), it would seem natural (to me at least) to break this problem down into 2 cases:

Case 1) The first 2 terms both evaluate to whole numbers
Case 2) The first 2 terms both evaluate to fractions whose sum is a whole number

If Case 1 is true you could conclude, for example, that 3 terms must either evaluate to even numbers or exactly 2 of the 3 terms must evaluate to odd numbers. In the former case then either H or I (or both) must be even...

[Rain]

I think it doesn't matter which of the multiplication/division ones you do first, but if needed:

(A/B)C + (DE)(F/G) + (HI) = 100

[Echognome]

Rain, on Oct 30 2006, 04:06 PM, said:

I think it doesn't matter which of the multiplication/division ones you do first, but if needed:

(A/B)C + (DE)(F/G) + (HI) = 100

I don't understand what you mean by "it doesn't matter".

Are you saying that:

(A/B)C = A/(BC)?

That would mean that

(4/2)3 = (2)3 = 6

= 4/(2*3) = 4/(6) = 2/3

[Rain]

Ahhh, though lets not get sidetracked.

I said it didn't matter because if I meant to put

A/(BxC) instead of A/BxC, I would have put the brackets.

[Echognome]

Ok... then how about AxB/C + DxExF/G + HxI = 100 instead. No confusion.

[hrothgar]

fred, on Oct 30 2006, 06:34 PM, said:

Assume the first term means A/BC (and not AC/?

There are some things you can figure out by logic for sure.

For example, G cannot be either 5 or 7.

If I was going to spend a lot of time on this (most likely I am not), it would seem natural (to me at least) to break this problem down into 2 cases:

Case 1) The first 2 terms both evaluate to whole numbers
Case 2) The first 2 terms both evaluate to fractions whose sum is a whole number

If Case 1 is true you could conclude, for example, that 3 terms must either evaluate to even numbers or exactly 2 of the 3 terms must evaluate to odd numbers. In the former case then either H or I (or both) must be even...

Extending on Fred's logic:

Lets assume that all three elements are whole numbers.

If we look at the numbers between one and nine, we're working with four even numbers and five odd number. I'd conjecture that

the first part of the equation is comprised of three odd numbers
the second part is 4 even numbers
the third is two odd numbers

Working from this, its pretty apparant the solution is (45 + 48 + 7)

[fred]

hrothgar, on Oct 30 2006, 04:50 PM, said:

fred, on Oct 30 2006, 06:34 PM, said:

Assume the first term means A/BC (and not AC/?

There are some things you can figure out by logic for sure.

For example, G cannot be either 5 or 7.

If I was going to spend a lot of time on this (most likely I am not), it would seem natural (to me at least) to break this problem down into 2 cases:

Case 1) The first 2 terms both evaluate to whole numbers
Case 2) The first 2 terms both evaluate to fractions whose sum is a whole number

If Case 1 is true you could conclude, for example, that 3 terms must either evaluate to even numbers or exactly 2 of the 3 terms must evaluate to odd numbers. In the former case then either H or I (or both) must be even...

Extending on Fred's logic:

Lets assume that all three sections are whole numbers.

If we look at the numebrs between one and nine, we're working with four even numbers and five odd number. I'd conjecture that

the first part of the equation is comprised of three odd numbers
the second part is 4 even numbers
the third is two odd numbers

Working from this, its pretty apparant the solution is (45 + 48 + 7)

Assuming you are right about 45+48+7, this suggests that there is more than one solution to the problem (H=7 and I=1 and vice versa).

Rain never suggested there was a unique solution, but it would be a nicer problem (at least according to my sense of aesthetics) if this was true.

[hrothgar]

hrothgar, on Oct 30 2006, 07:50 PM, said:

Working from this, its pretty apparant the solution is (45 + 48 + 7)

Bloody hell. I used 1 twice
Back to the drawing board

[FrancesHinden]

Echognome, on Oct 30 2006, 05:11 PM, said:

Rain, on Oct 30 2006, 04:06 PM, said:

I think it doesn't matter which of the multiplication/division ones you do first, but if needed:

(A/B)C + (DE)(F/G) + (HI) = 100

I don't understand what you mean by "it doesn't matter".

I take him to mean that there's only one interpretation where there is a solution.

[FrancesHinden]

fred, on Oct 30 2006, 05:54 PM, said:

Rain never suggested there was a unique solution, but it would be a nicer problem (at least according to my sense of aesthetics) if this was true.

Unless there's something unconventional about the layout, H & I are clearly interchangeable, as are A&C, and D,E&F.

I don't think that hurts the problem.

[FrancesHinden]

Has anyone specified that this problem is in base 10?

[david_c]

With this sort of problem you usually end up splitting into a number of cases, maybe managing to eliminate a few cases early on, but then eventually having to check the remaining cases by trial and error. If you've done it well, you will have proved enough things about each of your cases that there won't be many combinations you have to try by hand.

Here you can prove things like

- At least one of the denominators must be even.

- If there is a single term which has an even number in the denominator, its numerator must contain the number 4.

That sort of argument reduces the number of things you have to try to a manageable level.

[Edited:] I initially stopped after finding a solution, but now I've found another one.

4x9/6 + 7x5x2/1 + 8x3
7x8/4 + 9x6x3/2 + 5x1

This post has been edited by david_c: 2006-October-30, 15:06

[P_Marlowe]

<snip>

[ajm218]

My prog says there are 192 actual solutions, 92 of which aren't identically trivial e.g. a=b etc. :-)