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There is a related answer on stackoverflow but I trust this community more for math questions.

The problem is: Prove that $13^3=2197$. Generalize your answer.

How can I approach such a problem?

Problem by Donald Knuth, The Art Of Computer Programming,

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closed as off-topic by Bill Dubuque, DonAntonio, Macavity, M Turgeon, Henry Swanson Apr 24 '14 at 18:33

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Bill Dubuque, DonAntonio, Macavity, M Turgeon, Henry Swanson
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Are there any hints on how he wants you to approach the problem? For most people, 'put it in a calculator' would be a valid proof. –  Alex Zorn Apr 24 '14 at 17:39
Not really. He actually states that "This is a horrible kind of problem that the author has tried to avoid" so the calculator might actually be a good choice :) –  ierax Apr 24 '14 at 17:41
I'm not sure I understand what you want us to achieve then. Calculate $13^3$ and you're done. –  David Apr 24 '14 at 17:42
This is a joke exercise from near the beginning of the book, just above an "exercise" where you are supposed to prove Fermat's last theorem. It's there to give an example of the type of problems you won't see later on. –  Cocopuffs Apr 24 '14 at 17:45
Context is essential here. The exercise appears between the following two: $\tag*{}$ Exercise $1.0.2\!:\ $ Of what value can the exercises in a textbook be to the reader? $\tag*{}$ Exercise $1.0.4\!:\ $ Prove that when $n$ is an integer, $n > 2,\,$ the equation $\,x^n+y^n = z^n$ has no solution in positive integers $\,x, y, z.$ –  Bill Dubuque Apr 24 '14 at 17:47

5 Answers 5

up vote 13 down vote accepted

The author did not really want you to solve the problem, but to see that this kind of problem is undesireable and not satisfying. To see this you should consider the full bock of exercises of that chapter "Notes on the Exrcises".

For one, solving the base problem consists of merely computing a simple expression, for example by multiplying $13\cdot 13\cdot 13$. Next, there is nothing remarkable about this specific result. Then the "Genaralize your answer" strikes us in the face: What should one generalize here? That $13^4$, $13^5$ and so on can als be computed? So while the difficulty score 14 suggests that you should be able to solve the problem in a few minutes, Knuth's parenthesized comment should tell you not to take this specific problem too seriously.

In fact, the fourth exercise (to prove FLT) is rated M50, i.e. as mathematically oriented (open at the time of his writing) research project, thus rather being an example of how the rating system works than a seriously intended exercise. So, once you have mastered the (recommended by the author!) exercise 1, and maybe also exercise 2, the remaining two exercises of this chapter (after all it is titled "Notes on the Exercises" - so we could expect something meta going on here) are not to be considered as exercises, but rather as examples of how exercises might look (or in these cases maybe: should not look). A final obseration in support of this: These exercises belong to the front matter, not to the text per se! (Which doesn't keep Knuth from adding answer hints for exercises 1 and 4 in the answers section)

Remark: I base the above on what I have in the shelf, which is the second edition. But I'm pretty sure it applies to all editions.

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The downvoter really should explain, since this answer is the only current answer that is on target. –  Bill Dubuque Apr 24 '14 at 17:52

Solve ${13}^3$ by using the definition of exponention ${13}^3$ comes out to be $2197$.
Since $L.H.S=R.H.S$ the given equality is true.

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$13^3=13\cdot 13\cdot 13=169\cdot 13=2197.$ This shows that $13^3=2197.$

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It's not a serious exercise - see my comment to the question. –  Bill Dubuque Apr 24 '14 at 17:48

The notation $13^3$ is shorthand for $13 \times 13 \times 13$.

You can do this in steps, e.g. $13 \times 13 \times 13 = (13 \times 13) \times 13$

To find $13 \times 13$ we can do "block multiplication":

$$13 \times 13 = 10\times 13 + 3\times 13 = 130 + 39 = 169$$

It follows that $13^3 = 169 \times 13$. We can, again, use "block multiplication":

$$\begin{eqnarray*} 169 \times 13 &=& 100 \times 13 + 60 \times 13 + 9 \times 13 \\ \\ &=&1300 + (60 \times 10 +60\times 3)+(9\times 10 + 9\times 3) \\ \\ &=&1300 + 600 + 180 + 90 +27 \\ \\ &=&2197 \end{eqnarray*}$$

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After a bit of thinking I came to the conclusion that this is a prime factorization (both 13 and 3 are primes). Look up fermat's little theorem.

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