Reverse Engineering Arithmetic Challenge

Two variables and two equations and not a differential in sight. I wish they all looked like this. I would add the two equations together and divide by 500 giving the simple x+y=3. Then I'd subtract the two equations and divide by 30 getting x-y=1. sub x=1+y into the x+y=3 giving y+1+y=3 or y=1 thus x=2.

I'm not an engineer so someone else will have to handle the second part of your challenge ;)

Given that the problem is a "puzzle", I'd get a feel for the ballpark value of x and y, Then I'd guess a few values. This worked, and took less time than it took to write this.

Failing that, I'd subtract the two, divide by 30 and find that x and y differ by 1. At which point the anwer is obvious, given the relationship of the other numbers.

If I had to go so far as to add the equations, and substitute, I'd give it to my son, who, still in high school, remembers how to do this stuff.

Reverse engineer it??!! And visit this torment on someone else? Surely not.

I simply noticed that the second equation had a sum 30 less than the first equation. This reduction in value was due to a decrease of 30 x and an increase of 30 y. So I thought of a new equation:

30y-30x=-30

There are infinite solutions to this, but the most obvious is x=2, y=1. Plug them in, and you find that this is correct. I've never been so great at math generalizations though... so I dont know about the last part of the challenge.

Well, the best brains have given it a shot, and fell short.

Roger Pink failed to see there are multiple answers, not just y=1 , x=2.

At least Goose recognized this, but still gave only one solution set, the same as Pink.

Since the general solution is x+y=3, multiple solution sets for (x,y) include (0,3),(3,0), (1,2),(2,1) and if you allow negative numbers (4,-1),(-1,4),(5,-2),(-2,5),(6,-3),(-3,6), et cetera, et infinitum.

To "reverse engineer" any number of simultaneous equations you can just apply simple algebra to the basic equation. For example, lets use any constant, say, 111, as a multiplier:

111x + 111y = 333

Then we can generate two more equations for our "puzzle" by splitting the equation (the sum) into two summands (or addenda) by subtraction:

111x + 111y = 333
100x + 11y = 300
------------------
11x + 100y = 33

So now we have two simultaneous equations:

100x + 11y = 300 and 11x + 100 y = 33

We could complicate this further by multiplying each by some number, or if we were really mean, choose a less obvious number other than 111 as our original multiplier, say 274, or more unlikely addenda to split the single equation. I used the simpler numbers so my method could be easily followed.

Now, why would we ever want to do such a thing, much less figure out a way to do it to others?

But wait... don't I get extra credit for NOT supplying the answer, in compliance with the statement "So we don't really want the answer... " ?? I certainly alluded to the answer without being so rude as to actually supply numbers. Granted, when I found 2 and 1, I stopped... I am, after all lazy. And to go even further, and generalize? Yikes!

I guess we do it because its fun, in a perverse sort of way.

Wait, I see I didn't generalize ENOUGH! The original puzzle said, use different constants and variables.

Let's try this:

1. Take any two variables and make an equation (#1) where the sum of the two variables equals a constant.

2. Take a different constant use it as a multiplier on each term of your equation (#1).

3. Create a new equation (#2) by changing the constants on each variable to new and different (from each other) constants and change the constant on the other side of the equals sign to a new and different (from any of the previous ones) constant as well.

4. Subtract Equation #2 from #1 to make #3.

5. By definition (and "reverse engineering") Equations #2 and #3 are simultaneous equations with (at least one) valid solution.

6. (Optional) You may further increase the complexity of the simultaneous equations, and therefore the difficulty of the solution, by applying identical algebraic operations to both sides of the equations, preferably different operations for each equation.

I apologize if I missed your point, but it seems incorrect in suggesting that multiple solution sets exist i.e "Since the general solution is x+y=3, multiple solution sets for (x,y) include (0,3),(3,0), (1,2),(2,1) and if you allow negative numbers (4,-1),(-1,4),(5,-2),(-2,5),(6,-3),(-3,6), et cetera, et infinitum."

The equations are simultaneous and any answer must apply for both equations. So to Roger's benefit, since X+Y=3 AND X-Y=1 AND the equations are simultaneous, there is only one solution. Add these two equations and you get 2X=4 so X=2. Plus, if you take one of your proposed solutions (i.e. (0,3)) and use the HP in the drawer (when no one is looking) then 265(0) + 235(3) = 705 instead of 765. Also, the equations represent straight lines and two straight lines only cross at one point therefore they only have one discreet solution. Right?

(It's been too long since the old algebra days.)

It seems the trick is to pick a base number such as 100. And then pick an amount to vary on either side such as 8. Now you have 108 and 92. To figure out the number on the right hand side of the equation, multiply your base number by 3 (i.e 100x3=300) and then either add or subtract the variance. The variance in the first X is cancelled out by Y, but the variance in the second X changes the final value. If the big number is times X then add the vairance, if the big number is times Y then subtract the variance. SOOOOO... 108X+92Y=308 and 92X+108Y=292. The only tricky part is to not pick a number that you can't multiply by 3 in your head (i.e. 1,345,785.25).

Now I have to go rest my fingers from typing. L8R.

Um, no offense, but you're wrong:

235*3=705 is not equal to 765, so (0,3) is not a solution. Neither is (3,0) (3*265=795 which is not equal to 765)

I can't believe you called me out without even bothering to check if your answers actually worked.

My answer is correct.

I goofed. My apologies to one and all. I guess that's what I get for trying to do it in my head, which is way too stuffed with other junk besides algebra!

I will admit, I was looking for it to be a "trick" question. A case of not seeing the forest because of all the trees in the way!

And yes, I should have checked my answer by putting it through one of the original equations.
Next time I will be sure to use my own H.P. (Heuristic Programming) instead of jumping on the first solution that pops up.

It's all good. Expecting a trick in the question can have that effect/affect.

The discussions are very interesting, but I think we must just get one thing straight: two independent simultaneous equations in two unknowns have only one answer set. Roger and one previous anonymous poster have that part right! As for the reverse engineering part - thats still not quite on track!

Is the solution not what was suggested by the previous coward.

Call your base value A. Now you change from that value by a delta of B.

(A+B)*X+(A-B)*Y=(3*A)+B
(A-B)*X+(A+B)*Y=(3*A)-B

As long as you pick an A and B you can do in your head, your are good to go. The original problem had an A of 250 and a B of 15.

Eureka. Well put. I'd pondered why it was easy to quickly guess correct values without doing any real math, and had this vague image of figures hinging around 250. It is nice to see it put so succinctly.

'Coward' of post #7631, you've got it very near spot-on! In my algorithm, I used a minor variation of yours that is more "programmable", in that I let the algorithm choose x and y upfront for predictable results:

Pick as a base value C, a nice round number, say 1000. Divide C into two unequal parts A and B, so that A + B = C, say A=650 and B=350. Now pick x and y values as small integers, say x=3 and y=2. The set of equations will be:

Ax + By = D
Bx + Ay = E,

where D and E are calculated using the values of the constants and variables chosen.

Note, my C is twice your A, but further the methods are equivalent, apart from my pre-knowledge of what the x, y solution would be, where I think you choose a 'delta' and then check for 'easiness' afterwards. Maybe there are even better or more general algorithms!

Here are a couple specific cases constructed with this type of method which have some interesting looking numbers.

4321 X + 1234 Y = 9876

1234 X + 4321 Y = 6789

And

321 X + 123 Y = 765

123 X + 321 Y = 567

Interestingly enough, you can get the same X,Y answers in these equations if you drop all but the first digit from each number as well. (i.e. 4321 -> 4, 765 -> 7 etc.)

Hmm, interesting numbers, you're right. In my algorithm posted above, I choose X and Y up front, so there is an almost limitless number of equations possible for that X,Y pair.

Take the two equations say the first eqn. The sum is not much greater than the constants (only around 3 times) So, I added the constants: 265+235= 500 The difference from the sum is: 765-500=265, which is one of the constant. So, value of x in this equation must be 2; by common sense you can say y = 1. Solved.

Far out, you folks complicate things. Adding the equations gives:

500x+500y=1500, therefore x+y=3, or x=3-y.

Subtract equations gives

30x-30y=30, therefore x-y=1

Combine these to get: (3-y)-y=1

So y=1