Little mathematical problem

  • Mathematica
  • Thread starter Werg22
  • Start date
  • Tags
    Mathematical
In summary: I was suckered in by Werg's trap. But Zurtex was the one going into the trap.In summary, Zurtex was suckered into thinking that he had made a trap for Data, when in fact he only opened the door a bit. However, the equation he gave is not divisible by prime numbers, so there is no trap.
  • #1
1,431
1
This is not for help or anything, I just to see your level of mathematical cleverness.

Knowing that a, b and c are positive whole numbers;

[tex]a^{-2} + b^{-2} = c^{-2}[/tex]

What is the sum of all possible values for a or b from 0 to 100?
 
Last edited:
Physics news on Phys.org
  • #2
Well that's the same as:

[tex]\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}[/tex]

[tex]\frac{a^2 + b^2}{(ab)^2} = \frac{1}{c^2}[/tex]

Therefore a2 + b2 must be a divisor of (ab)2 and ab=c.

The rest is a simple matter of writing a program to work the possibilities out under this and adding them together.
 
Last edited:
  • #3
Zurtex was the one going into the trap..
 
  • #4
a, b and c are positive prime numbers

Be careful!
 
  • #5
Data said:
Be careful!
You are a gentleman, Data!
I was hoping for a hurried response along the lines:
"Trap?! What trap?" :biggrin:
 
  • #6
such evil people here! :rofl:


Now, if the poster were going to be really devious, he'd say that the answer is

[tex]\sum_{ 0 \leq p_i \leq 100} p_i, \ p_i \ \mbox{prime},[/tex]

with the claim that [itex]a, \ b, \ c[/itex] don't appear in the equation he gave at all - he meant

[tex]\hat{a} + \hat{b} - 4= \hat{c} -2,[/tex]

and he also specified that you choose either [itex]a[/itex] or [itex]b[/itex] to add the values for, so it's just the sum of all primes between 0 and 100.
 
Last edited:
  • #7
arildno said:
Zurtex was the one going into the trap..
Haha, I only opened the door, I didn't go down it :tongue:
 
  • #8
Zurtex you are right on track, but not quite. You can solve it, if you have the proper thinking. One hint; square root the equation.
 
  • #9
[tex]\frac{\sqrt{(a^2+b^2)}} {ab} = \frac{1} {c} [/tex]
 
Last edited:
  • #10
Data said:
such evil people here! :rofl:


Now, if the poster were going to be really devious, he'd say that the answer is

[tex]\sum_{ 0 \leq p_i \leq 100} p_i, \ p_i \ \mbox{prime},[/tex]

with the claim that [itex]a, \ b, \ c[/itex] don't appear in the equation he gave at all - he meant

[tex]\hat{a} + \hat{b} - 4= \hat{c} -2,[/tex]

and he also specified that you choose either [itex]a[/itex] or [itex]b[/itex] to add the values for, so it's just the sum of all primes between 0 and 100.

Not at all.
 
  • #11
Is it, by any chance, zero?
ab=c, which goes against the definition of a prime.
 
  • #12
I'm terribly sorry I meant whole numbers.
 
  • #13
haha, that changes the whole question :tongue:
 
  • #14
Werg22 said:
I'm terribly sorry I meant whole numbers.
How sad; I thought you had made a really good trap Zurtex fell into, even though he strenuously asserts he only opened the (trap-) door a bit.
What he forgot, is the influence of gravity:
If you step upon a trap-door, opening it, however little, gravity pulls you down to the bottom.
 
  • #15
It would have been impossible with prime numbers, so there is no trap :p
 
  • #16
:biggrin: pfft, whatever, clearly didn't go as far as Werg, anyway I didn't make a single mistake in my answer :rolleyes:
 
  • #17
Now with the equation try awnsering it!

Now awnser? I'll give another hint. Inverse the equation

[tex]\frac{ab} {\sqrt{a^{2} + {b^2}}} = c[/tex]

If no one can awnser by the end of the day I will awnser it myself :frown:
 
Last edited:
  • #18
Ok here how to solve it.

First, if a, b and c are whole numbers then [tex]{\sqrt{a^{2}[/tex] has to be a certain number that will devid ab in a whole number. Now the result of [tex]{\sqrt{a^{2}[/tex] is either whole or decimal... not quite. Phytagorus with whole numbers NEVER give decimal numbers that are perfect
 
  • #19
Ok here how to solve it.

First, if a, b and c are whole numbers then [tex]{\sqrt{a^{2} + b^{2}}[/tex] has to be a certain number that will divide ab in a whole number. Now the result of [tex]{\sqrt{a^{2} + b^{2}}[/tex] is either whole or decimal... not quite. No product of whole numbers is devidable by their Pythagoaras result into another whole number. So, now you know that [tex]{\sqrt{a^{2} + b^{2}}[/tex] has to be a perfect Pythagoras. But witch numbers when multiplied give a number that is dividable by those two numbers Pythagoras? Only 5 factors. Witch numbers give 5 factors perfect Pythagoras? Factors of 3 and 4. So a and b

have to be:

Factors of 5
And factors of 3 or 4

[tex]3(5)=15[/tex]

[tex]4(5)=20[/tex]

Every perfect Pythagoras with factors of 3 and 4 have common factor witch we will name x.

[tex]a=15x[/tex]
[tex]b=20x[/tex]

[tex]\frac{15x(20x)} {\sqrt{(15x)^{2} + {(20x)^2}}} = c[/tex]

[tex]\frac{300 x^2} {\sqrt{225 x^2 + 400 x^2}}} = c[/tex]

[tex]\frac{300 x^2} {\sqrt{x^2 625}}} = c[/tex]

[tex]\frac{300 x^2} {25x}= c[/tex]

[tex]12x= c[/tex]

All the possible values for a and be would be

[tex]15(x+1+1+...)+20(x+1+1...)[/tex]

The maximum value for x in the case of 15 is 6 since 15*6=90 and 15*7=105
The maximum value for x in the case of 20 is 5 since 20*5=100 and 20*6=120

The sum for all numbers to 0 to 6 is

[tex]\frac{6^2 + 6} {2}=21[/tex]

and 5

[tex]\frac{5^2 + 5} {2}=15[/tex]

SO

[tex]15(21) + 20(15)=615[/tex]

But there is ONE exception. That is 65.

[tex]\frac{65(156)} {\sqrt{65^{2} + {156^2}}} = 169[/tex]

[tex]615+65=680[/tex]

680 is the sum of all possible values for a and b. I must admit this problem disturbed a while and it took me almost a week to solve it! I had to rediscover all of this. I guess a quite clever mathematician would have been able to solve it right away...
 
Last edited:

1. What is a "little mathematical problem"?

A "little mathematical problem" is a term used to describe a small or simple mathematical question or puzzle that can be solved using basic mathematical concepts and operations.

2. How difficult are "little mathematical problems"?

The difficulty of a "little mathematical problem" can vary depending on the individual's mathematical abilities and the complexity of the problem itself. However, these problems are typically designed to be approachable and solvable by a wide range of people.

3. What types of math are involved in "little mathematical problems"?

Most "little mathematical problems" involve basic arithmetic, algebra, geometry, and sometimes even calculus. Some may also require logical thinking and problem-solving skills.

4. Can "little mathematical problems" be solved without a calculator?

Yes, most "little mathematical problems" can be solved without a calculator. These problems are designed to be solved using basic mathematical concepts and operations, so a calculator is not necessary.

5. Where can I find "little mathematical problems" to solve?

There are many resources available online such as math websites, blogs, and social media pages that share "little mathematical problems" for people to solve. You can also create your own problems or find them in puzzle books and magazines.

Suggested for: Little mathematical problem

Replies
7
Views
2K
Replies
3
Views
176
Replies
6
Views
2K
Replies
5
Views
4K
Replies
1
Views
882
Replies
9
Views
1K
Replies
2
Views
3K
Replies
1
Views
789
Replies
2
Views
1K
Back
Top