I got this nice little math puzzle elsewhere. I got it with a hint but try without the hint first. Here it goes:

P.S. Without a calculator, of course.

Allow me to give the hint: 216 = 6³

The hint for me was that this problem is going to have a simple solution and is therefore constrained so as to have a simple solution.

Edited January 17Jan 17 by KJW

38 minutes ago, KJW said:

Allow me to give the hint: 216 = 6³

The hint for me was that this problem is going to have a simple solution and is therefore constrained so as to have a simple solution.

Yes, 216 is specially related to the other numbers in the puzzle.

23 minutes ago, Genady said:

Yes, 216 is specially related to the other numbers in the puzzle.

For me, 216 = 6³ was the hard part of the problem without a calculator. Everything else is just applying logarithm identities.

16 minutes ago, KJW said:

For me, 216 = 6³ was the hard part of the problem without a calculator. Everything else is just applying logarithm identities.

This is correct.

1 hour ago, KJW said:

For me, 216 = 6³ was the hard part of the problem without a calculator. Everything else is just applying logarithm identities.

Here is another puzzle from the "Without a Calculator" set:

I know of two different ways to solve it and I'm curious about your way.

And another one, the last one of this kind that I seem to remember:

Find two 4-digit numbers that multiply to give 4^8 + 6^8 + 9^8.

27 minutes ago, Genady said:

Find two 4-digit numbers that multiply to give 4^8 + 6^8 + 9^8

4⁸ + 6⁸ + 9⁸ = (2⁸ + 3⁸)² – (6⁴)² = (2⁸ + 3⁸ + 6⁴) (2⁸ + 3⁸ – 6⁴) = 8113 x 5521

16 minutes ago, KJW said:

4⁸ + 6⁸ + 9⁸ = (2⁸ + 3⁸)² – (6⁴)² = (2⁸ + 3⁸ + 6⁴) (2⁸ + 3⁸ – 6⁴) = 8113 x 5521

Yep, that's the only way I know.

How about the one above it?

1 minute ago, Genady said:

  18 minutes ago, KJW said:

4⁸ + 6⁸ + 9⁸ = (2⁸ + 3⁸)² – (6⁴)² = (2⁸ + 3⁸ + 6⁴) (2⁸ + 3⁸ – 6⁴) = 8113 x 5521

Yep, that's the only way I know.

How about the one above it?

I haven't figured that one out yet. It seems to me that 50 = 7² + 1² = (7 + i) (7 – i), but how that is used is not clear to me.

Just now, KJW said:

I haven't figured that one out yet. It seems to me that 50 = 7² + 1² = (7 + i) (7 – i), but how that is used is not clear to me.

If it is, this will be a third way 😉.

[math]\text{Let:}\ \ \ \ u=\sqrt[3]{7+\sqrt{50}}+\sqrt[3]{7-\sqrt{50}}[/math]

[math]\text{Let:}\ \ \ \ a=\sqrt[3]{7+\sqrt{50}}[/math]

[math]\text{Let:}\ \ \ \ b=\sqrt[3]{7-\sqrt{50}}[/math]

[math](a+b)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab(a+b)[/math]

[math](a+b)^3-3ab(a+b)-(a^3+b^3)=0[/math]

[math]a^3+b^3=14[/math]

[math]ab=-1[/math]

[math]u=a+b[/math]

[math]\text{Therefore:}\ \ \ \ u^3+3u-14=0[/math]

[math]u=\sqrt[3]{7+\sqrt{50}}+\sqrt[3]{7-\sqrt{50}}=2[/math]

[If the above LaTeX doesn't render, please refresh browser]

[math]\text{Also:}\ \ \ \ \dfrac{u^3+3u-14}{u-2}=u^2+2u+7[/math]

[math]u=-1\pm\sqrt{6}i[/math]

6 hours ago, KJW said:

Let:    u=7+50−−√−−−−−−−√3+7−50−−√−−−−−−−√3

Let:    a=7+50−−√−−−−−−−√3

Let:    b=7−50−−√−−−−−−−√3

(a+b)3=a3+3a2b+3ab2+b3=a3+b3+3ab(a+b)

(a+b)3−3ab(a+b)−(a3+b3)=0

a3+b3=14

ab=−1

u=a+b

Therefore:    u3+3u−14=0

u=7+50−−√−−−−−−−√3+7−50−−√−−−−−−−√3=2

[If the above LaTeX doesn't render, please refresh browser]

This is the same as how I've solved it:

So, x=2.

Then, somebody has showed me different way. I'll give you a hint:

(I make these LaTeX expressions outside because I don't remember what the tags in SFN are.)

4 hours ago, KJW said:

Also:    u3+3u−14u−2=u2+2u+7

u=−1±6–√i

However, these are not solutions.

Edited January 18Jan 18 by Genady

3 hours ago, Genady said:

Then, somebody has showed me different way. I'll give you a hint:

Perhaps I'm narrowly focusing on the wrong area, but I don't see how this helps. Does it enable a solution without solving a non-trivial cubic equation?

3 hours ago, Genady said:

  7 hours ago, KJW said:

[math]u=-1\pm\sqrt{6}i[/math]

However, these are not solutions.

Actually, they are solutions of the cubic equation. Bear in mind that numbers have three cube roots. Real numbers have one real and two conjugate complex cube roots. So there are three distinct values for [math]\sqrt[3]{7+\sqrt{50}}[/math] and [math]\sqrt[3]{7-\sqrt{50}}[/math]. There are also three distinct values for [math]ab=\sqrt[3]{-1}[/math], each leading to a distinct cubic equation. But it is not clear how many distinct values there are for [math]\sqrt[3]{7+\sqrt{50}}+\sqrt[3]{7-\sqrt{50}}[/math]. It should be noted that [math]a^3+b^3=14[/math] and [math]ab=-1[/math] determine a unique pair of values for [math]a^3[/math] and [math]b^3[/math], so the cubic equation in [math]u=a+b[/math] is not producing spurious values for [math]a[/math] and [math]b[/math]. Also, the above complex solutions demonstrate that there is indeed only one real solution for [math]u=a+b[/math].

Edited January 18Jan 18 by KJW

6 minutes ago, KJW said:

Does it enable a solution without solving a non-trivial cubic equation?

Yes, it does. Here is the next step (attach this to the previous chain of equalities):

10 minutes ago, KJW said:

Actually, they are solutions of the cubic equation. Bear in mind that numbers have three cube roots. Real numbers have one real and two conjugate complex cube roots. So there are three distinct values for 7+50−−√−−−−−−−√3 and 7−50−−√−−−−−−−√3, There are also three distinct values for ab=−1−−−√3, each leading to a distinct cubic equation. But it is not clear how many distinct values there are for 7+50−−√−−−−−−−√3+7−50−−√−−−−−−−√3. It should be noted that a3+b3=14 and ab=−1 determine a unique pair of values for a3 and b3, so the cubic equation in u=a+b is not producing spurious values for a and b.

Yes, I see what you mean.

BTW, would you please remind me of the tags/delimiters for inserting LaTeX in SFn?

2 minutes ago, Genady said:

BTW, would you please remind me of the tags/delimiters for inserting LaTeX in SFn?

I use [ math] [ /math] (remove the space after "[")

Edited January 18Jan 18 by KJW

4 minutes ago, KJW said:

I use [ math] [ /math] (remove the space after "["

Thank you!

1 minute ago, Genady said:

  5 minutes ago, KJW said:

I use [ math] [ /math] (remove the space after "[")

Thank you!

There is another way, but I don't use it and am not exactly sure of what it is.

11 minutes ago, Genady said:

Yes, it does. Here is the next step (attach this to the previous chain of equalities):

Ok.