a hard equation, please help!

[Harryvo]

A3. The equation a^3 + b^3 + c^3 = 2008 has a solution in which a; b; and c aredistinct even positive integers. Find a + b + c.
Multiple choice: A 20, B 22, C 24, D 26, E 28

I would say 22 but I just guessed because it is beyond my ability. Thank you for your times.

 

[Dr.Peterson]

A3. The equation a^3 + b^3 + c^3 = 2008 has a solution in which a; b; and c are distinct even positive integers. Find a + b + c.
Multiple choice: A 20, B 22, C 24, D 26, E 28

I would say 22 but I just guessed because it is beyond my ability. Thank you for your times.

Guessing will be part of it; did you also check?? What did you guess a, b, and c are?

An important key is that you are told the numbers are all even. So you could restate the equation, letting a = 2x, b = 2y, c = 2z, as

(2x)^3 + (2y)^3 + (2z)^3 = 2008

Simplify that, then list the cubes of the first few positive integers and look for a solution. Guessing among just few choices is reasonable.

 

[pka]

A3. The equation a^3 + b^3 + c^3 = 2008 has a solution in which a; b; and c aredistinct even positive integers. Find a + b + c.
Multiple choice: A 20, B 22, C 24, D 26, E 28

Guessing will be part of it; did you also check?? What did you guess a, b, and c are?
An important key is that you are told the numbers are all even. So you could restate the equation, letting a = 2x, b = 2y, c = 2z, as
\(\displaystyle (2x)^3 + (2y)^3 + (2z)^3 = 2008\). Guessing among just few choices is reasonable.

If we follow the hint. \(\displaystyle 2008=2^3\cdot 251\) so \(\displaystyle 8x^3 + 8y^3 + 8z^3 = 8\cdot 251\) or \(\displaystyle x^3 + y^3 + z^3 = 251\)

 

[JeffM]

1

A3. The equation a^3 + b^3 + c^3 = 2008 has a solution in which a; b; and c aredistinct even positive integers. Find a + b + c.
Multiple choice: A 20, B 22, C 24, D 26, E 28

I would say 22 but I just guessed because it is beyond my ability. Thank you for your times.

This is a Diophantine equation, and they are indeed hard. See the relevant wiki article.

https://en.m.wikipedia.org/wiki/Diophantine_equation.

Moreover, it is a vexing problem because it requires a lot of unnecessary work to determine whether there are multiple solutions to the cubic. There are, but it turns out to make no difference to the sum asked for

One way to solve these beasts is to use every bit of information that you have to reduce the amount of trial and error required. If you follow Dr. P's advice to take into account that the solutions wanted are all even numbers, you get:

\(\displaystyle a = 2x,\ b = 2y, \ c = 2z \text { and } x,\ y,\ z \in \mathbb Z.\).

If you further notice that a, b, and c are all positive, you get

\(\displaystyle 2 \le a \implies 1 \le x.\)

Now here is a general technique that applies to many such questions. These numbers may or may not be equal, but they certainly can be ordered. And because labelling the unknowns is arbitrary, we can say

\(\displaystyle 1 \le x \le y \le z.\)

Finally, we know

\(\displaystyle a^3 + b^3 + c^3 = 2008 \implies 8x^3 + 8y^3 + 8z^3 = 2008 \implies x^3 + y^3 + z^3 = 251.\)

Now, unless you are knowledgeable about solving Diophantine equations, you will need trial and error, but you can be clever about it.

The cubes start small, but grow fast.

4^3 = 64 and 3 * 64 = 192 < 251 so z > 4. Do you see why?

7^3 = 343 > 226 so z < 7. Do you see why?

Thus z = 5 or z = 6.

Moreover, we might try z = 6 first because it will put the greater constraints on y.

Give it a go to complete it. Remember that x and y must be integers and that

\(\displaystyle 1 \le x \le y \le z.\)

 

[topsquark]

Nice! :cool:

-Dan

 

[Steven G]

This is a Diophantine equation, and they are indeed hard. See the relevant wiki article.

https://en.m.wikipedia.org/wiki/Diophantine_equation.

Moreover, it is a vexing problem because it requires a lot of unnecessary work to determine whether there are multiple solutions to the cubic. There are, but it turns out to make no difference to the sum asked for

One way to solve these beasts is to use every bit of information that you have to reduce the amount of trial and error required. If you follow Dr. P's advice to take into account that the solutions wanted are all even numbers, you get:

\(\displaystyle a = 2x,\ b = 2y, \ c = 2z \text { and } x,\ y,\ z \in \mathbb Z.\).

If you further notice that a, b, and c are all positive, you get

\(\displaystyle 2 \le a \implies 1 \le x.\)

Now here is a general technique that applies to many such questions. These numbers may or may not be equal, but they certainly can be ordered. And because labelling the unknowns is arbitrary, we can say

\(\displaystyle 1 \le x \le y \le z.\)

Finally, we know

\(\displaystyle a^3 + b^3 + c^3 = 2008 \implies 8x^3 + 8y^3 + 8z^3 = 2008 \implies x^3 + y^3 + z^3 = 251.\)

Now, unless you are knowledgeable about solving Diophantine equations, you will need trial and error, but you can be clever about it.

The cubes start small, but grow fast.

4^3 = 64 and 3 * 64 = 192 < 251 so z > 4. Do you see why?

7^3 = 343 > 226 so z < 7. Do you see why?

Thus z = 5 or z = 6.

Moreover, we might try z = 6 first because it will put the greater constraints on y.

Give it a go to complete it. Remember that x and y must be integers and that

\(\displaystyle 1 \le x \le y \le z.\)

Hi Jeff. You say that 7^3 = 343 > 226 so z < 7. I am sorry but I do not see where the 226 came from. Can you explain?


EDIT: OK, I see that the solution is (?.??.z) and that ?^3 + ??^3 =35. I also see that 251-35 = 226. So were you ahead of yourself when you wrote 226 when you really meant 251. Or am I still missing something. BTW, I put question marks as I do not want to give the solution to the OP

 

[pka]

This is a Diophantine equation, and they are indeed hard. See the relevant wiki article.

https://en.m.wikipedia.org/wiki/Diophantine_equation.

Moreover, it is a vexing problem because it requires a lot of unnecessary work to determine whether there are multiple solutions to the cubic. There are, but it turns out to make no difference to the sum asked for

One way to solve these beasts is to use every bit of information that you have to reduce the amount of trial and error required. If you follow Dr. P's advice to take into account that the solutions wanted are all even numbers, you get:

\(\displaystyle a = 2x,\ b = 2y, \ c = 2z \text { and } x,\ y,\ z \in \mathbb Z.\).

If you further notice that a, b, and c are all positive, you get

\(\displaystyle 2 \le a \implies 1 \le x.\)

Now here is a general technique that applies to many such questions. These numbers may or may not be equal, but they certainly can be ordered. And because labelling the unknowns is arbitrary, we can say

\(\displaystyle 1 \le x \le y \le z.\)

Finally, we know

\(\displaystyle a^3 + b^3 + c^3 = 2008 \implies 8x^3 + 8y^3 + 8z^3 = 2008 \implies x^3 + y^3 + z^3 = 251.\)

Now, unless you are knowledgeable about solving Diophantine equations, you will need trial and error, but you can be clever about it.

The cubes start small, but grow fast.

4^3 = 64 and 3 * 64 = 192 < 251 so z > 4. Do you see why?

7^3 = 343 > 226 so z < 7. Do you see why?

Thus z = 5 or z = 6.

Moreover, we might try z = 6 first because it will put the greater constraints on y.

Give it a go to complete it. Remember that x and y must be integers and that

\(\displaystyle 1 \le x \le y \le z.\)

It is just not that complicated.
I posted this

If we follow the hint. \(\displaystyle 2008=2^3\cdot 251\) so \(\displaystyle 8x^3 + 8y^3 + 8z^3 = 8\cdot 251\) or \(\displaystyle x^3 + y^3 + z^3 = 251\)

.
Now \(\displaystyle 251-1=250=2\cdot 5^3\) let \(\displaystyle x=1,~y=5,~\&~z=5\).
Thus \(\displaystyle (2\cdot 1)^3+(2\cdot 5)^3+(2\cdot 5)^3=2008\)

 

[JeffM]

It is just not that complicated.
I posted this .
Now \(\displaystyle 251-1=250=2\cdot 5^3\) let \(\displaystyle x=1,~y=5,~\&~z=5\).
Thus \(\displaystyle (2\cdot 1)^3+(2\cdot 5)^3+(2\cdot 5)^3=2008\)

I do not think that suggesting that solving Diophantine equations by inspection generally works or that assuming most students are familiar with the cubes of the integers is helpful. Students need to be shown that systematic approaches to problems that initially seem impenetrable can frequently succeed.

I agree of course that if you have at your mental finger tips the fact that 5^3 = 125, then it is "obvious" that 2 * 5^3 + 1 = 251 = 2 * 5^3 + 1^3. I do not think that kind of explanation does anything but induce students to say "I just don't get math." If a student finds something hard and is told that it is actually easy as pie, the student makes the reasonable conclusion that all math is only for people with a great talent for it. And of course your answer is incomplete because you have ignored the restriction that the solution is restricted to positive integers and that there may be, and in fact are, more than one set of solutions in the positive integers.

$6^3 + 3^3 + 2^3 = 216 + 27 + 8 = 251.$

There are two solutions to the cubic in positive integers. It happens that both

$2(5 + 5 + 1) = 22 = 2(6 + 3 + 2).$

Now that equality is interesting, but I certainly do not know that the equality of the sum follows generally from the equality of the sum of the cubes.

 

[Denis]

Now \(\displaystyle 251-1=250=2\cdot 5^3\) let \(\displaystyle x=1,~y=5,~\&~z=5\).
Thus \(\displaystyle (2\cdot 1)^3+(2\cdot 5)^3+(2\cdot 5)^3=2008\)

But problem states a,b,c are distinct even integers.

So 2*2, 2*3 and 2*6, right?

Only 1 solution.

 

[Dr.Peterson]

A3. The equation a^3 + b^3 + c^3 = 2008 has a solution in which a; b; and c are distinct even positive integers. Find a + b + c.
Multiple choice: A 20, B 22, C 24, D 26, E 28

I would say 22 but I just guessed because it is beyond my ability. Thank you for your times.

I do not think that suggesting that solving Diophantine equations by inspection generally works or that assuming most students are familiar with the cubes of the integers is helpful. Students need to be shown that systematic approaches to problems that initially seem impenetrable can frequently succeed.

I agree of course that if you have at your mental finger tips the fact that 5^3 = 125, then it is "obvious" that 2 * 5^3 + 1 = 251 = 2 * 5^3 + 1^3. I do not think that kind of explanation does anything but induce students to say "I just don't get math." If a student finds something hard and is told that it is actually easy as pie, the student makes the reasonable conclusion that all math is only for people with a great talent for it. And of course your answer is incomplete because you have ignored the restriction that the solution is restricted to positive integers and that there may be, and in fact are, more than one set of solutions in the positive integers.

$6^3 + 3^3 + 2^3 = 216 + 27 + 8 = 251.$

There are two solutions to the cubic in positive integers. It happens that both

$2(5 + 5 + 1) = 22 = 2(6 + 3 + 2).$

Now that equality is interesting, but I certainly do not know that the equality of the sum follows generally from the equality of the sum of the cubes.

Did anyone but me notice that word "distinct" in the problem? I didn't even consider trying the 5, 5, 1 case, so I (rightly) missed this interesting fact.

(EDIT: Now I know that Denis saw it.)

As often happens, we don't know the context of the problem; my impression is that no special knowledge is expected, and that using the word Diophantine might be as scary as anything else we could do. That's part of my reason for just giving a hint, and waiting to see if there was any fruit. It also seemed important to point out that trial and error can be a valid part of a mathematical solution, without being specific as to how to do it efficiently. Often, a big part of helping someone is to assure them that it can be done with what they know, and tell them just to try something.

 

[JeffM]

But problem states a,b,c are distinct even integers.

So 2*2, 2*3 and 2*6, right?

Only 1 solution.

Oh, you are right. I missed that entirely. I should have said

\(\displaystyle 1 \le x < y < z.\)

The OP gave no clue as to the course being taken or level of math achieved, but, based on the way the question was phrased and where it was placed (in calculus), I wonder why the problem was posed to this student.