Cute Math Problem

Wow, that's a quick way to solve it. GA to you!

It one of those touchstone problems that helps you determine how a persons mind works. are they inductive, reductive, deductive, etc. in problem solving. THis is vital in case planning as it does no good to assign a task tat requires a skill the client does not possess. Setting people up for failure is never kind.

Are you rigidly tide to known rigor and rote learning or can you move outside accepted practice and still be successful.

This is the first time I have seen one in math form, usually its a full on word problem so this is interesting.

And Serious GA!

Oh no! I think this shows that I'm too structured in my thinking and possibly a bit OCD.

I do think out of the box, or at least I think I do.

I don't like the results, but I like what it tells me about myself.

If the answers are positive integers, line 3 tells us that f is between 1 and 5. Chucking that into line 2 tells us that a² is between 16 and 20. The only integer solution is a = 4. Minor detail, but it it gives an upper number to seek (in the example only 1 possibility) rather than starting at 4 in the hope we don't have to check too many possibilities. I'd never have noticed that without your approach and confabulated myself with substitutions.

TO me, this is why respect for our differences and willing collaboration are so important. It really sucks to miss the obvious solution because you were so wrapped up in the expected rigor.

We each come to the table with something, maybe not always useful, but a welcome contributor and an openness to innovation can be the difference between doing what we always done and setting the world on fire.

Interesting! I used substitution, while you made equations from a common variable.

I'm curious, where did you go to school? I wonder if the difference in approach is due to location.

I actually like your method better. It's quicker to solve vs substitution.

GA to you!

Thanks. I was really doing a substitution in disguise after a little rearranging of equations to make it easier. Fortunately, e and f occurred in 2 equations each and not together in any equation. Sometimes you just get lucky.

The hardest part is ending up with a cubic equation in a. There are methods to solve analytically, but as others have mentioned, a 9-year-old is likely to choose a positive integer value, so it would have to be a factor of the constant value in the equation (24). It turned out 4 works.

Per Ravasharek, we're both too pedantic!

I know that I like structure and I don't feel comfortable without it. As a businessman, I see entrepreneurs as amazing. The way they "see" things and have the vision to go after something that makes the logical side of me cringe.

However, I've worked with some incredible entrepreneurs. Create something out of just and idea and have the guts to go after it, even though everyone tells them their nuts! Great at starting a business, but not the best at running it. I guess that's where I come in.

I don't think a 9 year old knows much about algebra, so I think he/she would use the "take a guess" method ... I think it would be a real guess, most likely starting with a=1.

Good talking with you Rixter. Again, you're one of the guys I consider a smart one here.

I'm with you. I think it was "devised by a 9-year-old student". If it were solved by a 9-year-old, I would think that he or she would be pretty bored in the 4th grade.

Assigning values to a, c, e, and f, making up some operations, and writing down the answers is easy enough, and with any luck, the equations will be independent and can be solved. Devising is a lot easier than solving.

Guessing the answers would take a while. Assuming each variable is an integer in the range 1-10, that's 10000 permutations. The teacher could assign that and it would keep the kids quiet for some time.

Not Gauss, though!

Yeah, I was thinking about that...

https://www.nctm.org/Publications/Teaching-Children-Mathematics/Blog/The-Story-of-Gauss/

I'll read that presently, but before I do I will mention two Gauss anecdotes: 1) About 3 years old, he said "Papa, I think the reckoning is wrong" on some payroll or like matter; and 2) around 10 years old, given the problem of adding all integers from 1 through 100, he found the answer 5050 within a minute or so. (The problem was meant to take up quite a bit of time.)

[added] I see the 1 + 2 +...+100 story, but not the earlier one.

In another somewhat similar vein, consider the F. A. Kekule dream of snakes chasing tails that led to the discovery of the benzene ring.

I think I can give you a source for how he corrected his father. The story I've read is from "God Created the Integers" by Stephen Hawking. On page 564 " … at the age of three, he corrected a mistake his father made in paying out wages..."

I bought this book back in 2005 and thought I'd "just read it through"; well, that was an optimistic thought that was never completed. I started reading it again, this time more selectively based on area of interest but once again became bogged down trying to follow the math, because "just reading through" doesn't provide understanding. I then decided to read the biographies first and then, maybe, try to wade through one of the topics covered for each mathematician. Call this a future, future, probably not going to happen goal. Does that still count as a goal? ...

I haven't read that book by Hawking, so I don't know if it mentions further that "God created the integers..." was from Leopold Kronecker, who went on to say that all else [in mathematics] was the work of man.

I recall an anecdote by some chap who was an expert in the same field as Hawking. Noticing an elderly lady seated next to him on a flight he asked if she was enjoying her book in hand - A Brief History of Time. She replied 'yes, it's superb', and asked if he had read it. His reply was 'yes, though I didn't understand a word of it'.

Just a late thought, but why did the original question by the prodigy not include 'b' or 'd' ? It's vexing me more than the question itself ! I console myself by thinking that if I don't know the answer, it's because I don't understand the question.

While I noticed the lack of b and d I never gave it a thought until you mentioned it. It would have been logical to have included at least b and d and to not use e and f. So now I feel somewhat compelled to think of a reason so here it is: b and d where include as the problem developed but along the line its' creator found he had a problem and decided to drop b and d, thinking, perhaps, 4 out of 6 isn't bad...

Good talking with you...

Thanks, I really enjoy discussing technical topics with you guys and learning about new things. It's definitely a two-way street.

The cubic a^3 - 22a + 24 can be solved readily as we know it has a root a = 4, giving

a^3 - 22a + 24 = (a - 4)*(a^2 + 4a -6). The other 2 roots are 1.162 and -5.162.

I came up with a less neat solution giving a cubic in e

e^3 - 363e^2 +341e - 1113, which has root e = 7 so factors to

e^3 - 363e^2 +341e - 1113 = (e - 7)*(e^2 - 26e + 159)

So there are 3 sets of solutions, but only one giving all positive integers.

Good job! Impressive - I got lazy and assumed the two other answers were imaginary! GA to you!

Is it certain that it's not [ e^3 - 33e^2 + 341e - 1113 ] ?

You're right, typo on my part. I'd got 3.11^2 when it should be 3.11.

Thanks

My fingers routinely (operate on their own) too, so I am sympathetic to those also with (unruly?) digits.

(When we type-write a ''typo'', but hand-write an error, do we make a ''write-o'' instead? )

Similar to MACA,

Observe that [ a^2 - f = 15 ] could be solved if [ a = 4 ] and [ f = 1 ].

Then, [ c ] would equal [ 6 - ( f = 1 ) = 5 ],

and, [ e ] would equal [ 11 - ( a = 4 ) = 7 ].

Checking that [ (a = 4 ) x ( c = 5 ) - (e = 7 ) ] equals [ 20 - 7 ], which does equal [ 13 ].

QED

A 9 year old probably never heard of negative numbers, so I assume all positive integers.

a + e = 11, so neither (a) nor (e) can be over 10 (range 1 to 10 each).

f + c = 6, so neither (f) nor (c) can be over 5 (range 1 to 5 each).

But a * a - f = 15, so (a) can't be over 4 (range 1 to 4).

Trial and error solving:

Starting with a = 4, e must be 7,

then 4 * 4 - f gives (f) a value of 1.

1 + c = 6, so c = 5.

Trying in the last formula: 4 * 5 - 7 = 13 (True, no more tries necessary).

a =4, e =7, f = 1, c = 5

I think your fourth line should be:

(a) must be 4 or over.

Why? If a is less than 4, then a^2 will be 9 or less and f will need to be a negative number.

I did it by trial and error, similar to MACA, Mr G, and SG. That's why I wondered what was supposed to be so hard about it. Of course, full rigor like Rixter is harder.

I was too rusty on factorization to notice the factors a-4 and a²-4a-6.

I'm wondering if it's a generational thing. I don't know how old Rixter is, but I can tell you that I graduated college back in 1987 (undergrad) and I was taught to solve the problem by the traditional way and that hard work is good and short cuts cause problems. Well, that's not true in real life, as MACA has shown. In business and in much of life, a quicker way to accurately solve a problem is preferred.

My guess is that the younger folks are okay finding quicker solutions. It brings me back to my first encounter with my daughter's boyfriend. He would copy bits and pieces of websites to create his own. When I asked if he cared about knowing how the software was written, he looked at me like I was crazy. He told me that nobody does that anymore and everyone just takes what they want from the internet. I find that an amazing paradigm shift in programming, but I will add that he didn't have a degree (I think his education was limited to some Jr College courses) and he isn't a professional web designer.

I am not sure about generational, but that rational that you described was thrown at me also. I'm 60.

I think our generation wants to understand why things work. We also want to know how to do it.

I really liked my machine language course in college, because it showed me how a computer works - where all the 0's and 1's go and how they're both stored and used.

I wonder if they teach machine language anymore? Or Boolean Algebra? When was the last time you heard that term?

Interesting. I'm just about to turn 40 in a few days time so I'm not older generation or younger generation and that's maybe reflective of my thought process.

I was taught to work problems out methodically and in a structured way showing my working. However, I'm also a subsea engineer working in oil and gas and despite what anyone says we are here to make as much money for the company as possible. So if there is a quicker, cheaper way to solve something, that is the way I will go. I won't design a Rolls Royce if all you need is a Mini.

When you were in college, did you spend hours trying to solve a problem? The reason I'm asking is that I certainly did - on many occasions. I wanted to understand how to solve it, so I'd inefficiently spend too much time trying to solve the problem. The positive was that it gave me insight into solving other problems and it also showed me where some of my logic was flawed.

Apologies for the typo in post 5. I meant to write a²+4a-6 as the second factor. I had it correct on paper, but typed it wrong.

I used some old, old algebra to solve it. Here are my steps:

1. Four equations with four variables = solvable

2. For simplicity, we'll call the four equations eq1, eq2, eq3 and eq4. Solving for e in eq1, we have e=11-a.

3. Plug e into eq4 we get ac-(11-a)=13. ac+a-11=13 ac+a=24 a(c+1)=24 c+1=24/a finally c=24/a-1

4. Plug c into eq3. f+24/a-1=6 f+24/a=7 f=7-24/a

5. Plug f into eq2. a^2-(7-24/a)=15 Multiply both sides by a we get a^3-7a+24=15a a^3-22a+24=0

6. Factor the equation in 5. (a-4)(a^2+4a-6)=0 a=4 and two imaginary numbers. We don't need to solve imaginary numbers, so we'll use a=4.

7. f=7-24/a or f=7-24/4 f=1

8. c=24/a-1 or c=24/4-1 c=5

9. e=11-a or e=11-4 e=7

To the therapist in me it is fascinating to see the different ways people can reach the same conclusion. ITs even more fun when you can sit down and ask them to explain how they arrived at the conclusion.

I'm impressed with the approach's taken. I'm an old curmudgeon and don't get out of the box much so I just bounce around the corners. So, I looked for the tried and true method of solving for the variables by narrowing things down. It turned out "a" was my choice. It turned out that -a^3+22*a-24=0 was the result: As such a=4 was one root but it as two other real roots -5.162285 and 1.16228. So the complete set of solutions turns out to be:

a= 4 ….a= 1.16228 ….a= -5.162285

c= 5 ….c= 19.6491 ….c= -5.649186

e= 7 ….e= 9.83772 ….e= 16.162285

f= 1 ….f= -13.6491 ….f= 11.649186

You can see I have too much time on my hands but then I'm retired so time is all I have...

Those would have been expressed better as square roots than as finite but inaccurate decimal fractions. I.e., change = to ≈.

Hmmm.
If it were a real world mechanical problem, you'd struggle to find a rule graduated in square roots... dunno if you can program a CNC machine which will accept a square root as a coordinate either?
Inaccurate decimal?
Reminds me of the old one about the mathemetican and the engineer arguing in the pub.
The mathemetician bets that you can't get to the bar if you walk half the distance to the bar, and repeat half again etc.
The engineer accepts the bet, takes 4 steps, 2 steps, 1 step and then reaches out his arm to buy the beer and win the bet.
Del
BTW... I'm a pragmatist. I knew I could solve the problem, so I just looked down at the solutions.
That way I can save time and go and solve my real world 3 dimension problem involving complex curves in a piece of Yew :)

I defer to your position that I should be more accurate than I was. Initially I wrote down what I saw on my calculator but the extra effort would have made it better.

So: a= -2+SqRt(10) and a= -2-SqRt(10)

e= 13-SqRt(10) and e= 13+SqRt(10)

f= -1-4*SqRt(10) and f= -1+4*SqRt(10)

c= 7+4*SqRt(10) and c= 7-4*SqRt(10)

and lastly e= 13-SqRt(10) and e= 13+SqRt(10).

Thanks for the critique, one can always improve. I left off the solution for a=4 because, well, everyone else found it first and faster...

Here you have nailed it right. For irrational numbers such as √2, ³√7, π, etc., no terminating or repeating decimal is exact, even though more decimal places make them more nearly so. In lots of engineering, there is rounding off, approximation, etc. In electricity √3 is important, often truncated to 1.73, 1.732, or 1.7321. None of these is exact, but they are typically close enough for almost all practical applications.

And most times, instrument don't measure to as precise a number as 1.7321.

When I'm measuring distances from a ruler, I can usually get to 1.7320508, but sometimes I have trouble seeing the difference between that and 1.7320509. It's hard to see, but with a lot of staring and patience, it's possible.

For example, the other day I was working with a 30/60/90 right triangle. I knew the hyp = 4, so I wanted to make sure my angles were correct without using a protractor, so I measured the leg and I got to 3.4641015 mm. I knew it should be 3.4641016 mm, but it was really hard to see the difference with my ruler. I had to stare at it for at least 20 seconds or so to see that it was the correct distance. Then I knew I had a 30/60/90 right triangle.

That must have been fun, measuring a triangle just a bit over 1/8 inch big.

I'm glad you got a good laugh!

That's impressive! GA to you!