What's the largest number?

Question

I have problems with the problem: What's the largest number?

\begin{align} &60^{\frac{1}{3}} \mbox{ or } 2 + 7^{\frac{1}{3}}& \end{align}

I tried using factorization, but I didn't get a good result and I dislike using an estimate.

Thanks.

Mariano Suárez-Álvarez · Answer 1 · 2013-03-15T20:41:01.083

5

Compare the cubes of your numbers, which are $60$ and $$(2+7^{1/3})^3=8+3\cdot2^2\cdot 7^{1/3}+3\cdot 2\cdot 7^{2/3}+7=15+12\cdot 7^{1/3}+6\cdot 7^{2/3}.$$ Now $7^{1/3}<44/23$...

edited Mar 15 '13 at 20:41

answered Mar 15 '13 at 20:21

Mariano Suárez-Álvarez

135,076

2

That is still quite painful though, unfortunately. – dinoboy Mar 15 '13 at 20:22
1

The cubes are within 0.1 of each other, so something more probably needs to be said about how to deal with the two remaining cube roots. It would be the perfect tip if the problem said $63^{1/3}$, though! – rschwieb Mar 15 '13 at 20:25
1

@dinoboy, you should probably adapt your definition of painful to something more realistic! – Mariano Suárez-Álvarez Mar 15 '13 at 20:33

score 2 · Answer 2 · answered Mar 15 '13 at 21:14

2

$$ (2+\root3\of7)^3<(2+\root3\of7)^3+(2-\root3\of7)^3=16+12\root3\of{49}. $$ Here $$ 16+12\root3\of{49}<60\Leftrightarrow\root3\of{49}<\frac{11}3. $$ The last inequality holds because both sides are positive and cubing gives a true inequality $$ 49\cdot27=1323<1331=11^3. $$

answered Mar 15 '13 at 21:14

Jyrki Lahtonen

133,153

Adding a small quantity to one side simplified it somewhat, but didn't affect the conclusion, so it worked. I was very fortunate in that sense (or the problem was designed with such a trick in mind). – Jyrki Lahtonen Mar 15 '13 at 21:17
The trick is justified / motivated because you know that $(2+\sqrt[3]{7})^3$ is not an integer, while 60 is. – Calvin Lin Mar 16 '13 at 00:24
@Calvin: Yeah, but I would have felt more certain about success (I'm sure you would have too), if I were dealing with an algebraic number of degree two. Adding a conjugate and all that. Here there were no small conjgates around, but simplification was still to be expected. – Jyrki Lahtonen Mar 16 '13 at 05:43

score 0 · Answer 3 · answered Mar 15 '13 at 20:23

0

Cube both sides of the inequality.

answered Mar 15 '13 at 20:23

Jonathan Rich

662

I am not following this hint, can you provide more detail as to what you meant? – Amzoti Mar 15 '13 at 20:50

What's the largest number?

3 Answers3