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$$\frac{57^{46}}{17}$$

What is best and quickest way to evaluate an approximate answer to this by hand?

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Do you want the result $57^{46} / 17 \approx 3.5 \times 10^{79}$, the quotient, or the remainder? –  KennyTM Oct 25 '10 at 13:00
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Some of the answers below are interpreting the question as being $57^{46}$ modulo $17$, or in other words the remainder upon division by $17$. The OP should of course clarify, but given that s/he asks for an "approximate answer", it sounds like s/he really wants the quotient. –  Pete L. Clark Oct 25 '10 at 14:15
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@Pete, if the OP has the remainder in mind, an easy approximate answer would be $0\pm 9$ :) –  Mariano Suárez-Alvarez Oct 25 '10 at 14:29
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@Mariano: I can get a sharper bound with $8\pm 8$. =) –  Willie Wong Oct 25 '10 at 15:48
    
@Mariano: agreed. –  Pete L. Clark Oct 26 '10 at 5:35

9 Answers 9

By Euler's theorem $57^{16}\equiv1 \pmod{17}$.

It could be done using this.

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I think the question is for the quotient, not the remainder, and this answer has been given before. –  Ross Millikan May 9 '13 at 19:34

With mental arithmetic alone I would try the following. Undoubtedly you have needed the value of $$ \tan\frac\pi6=\frac1{\sqrt3} \approx0.57\ldots $$ enough many times to have it memorized. Therefore $57\approx 100/\sqrt3$. This means that $57^{46}\approx 10^{92}/3^{23}$. Because $\sqrt{10}\approx3$ we also have $\log_{10}3\approx 0.5$ Thereforer $3^{23}\approx10^{11.5}$. So $57^{46}$ is probably somewhere between $10^{80}$ and $10^{81}$. Dividing this by 17 gives a result most likely between $10^{79}$ and $10^{80}$. If such a ballpark figure is enough, then we're done. If you want a significant digit, I need to work a bit harder, and use more accurate estimates.

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The below concerns methods for obtaining the most significant digits, the roughest approximation to the magnitude or mantissa of the quantity you want. Other posts can help you use the some results of number theory to find the least significant digits (in any base). Arbitrary precision methods for hand calculation are also available, or perhaps you can find someone like Gauss to make the exact calculation in their head.

With a slide rule (if you literally need it by hand and can accept only a few decimal places of accuracy) or hand calculator (if 10 or so places will suffice), you can explicitly approximate such a number using the principle of logarithms. If $x=\frac{57^{46}}{17}$, then $$ \eqalign{ \log x &= 46\log 57-\log 17 \\ &\approx 46(4.04305126783455)-(2.83321334405622) \approx 183.147144976333 \qquad\text{base }e \\ &\approx 46(1.75587485567249)-(1.23044892137827) \approx 79.5397944395563 \qquad\text{base }10 } $$ so that $$ \eqalign{ x&=10^{\log_{10}x}=e^{\log_{e}x} \\ &\approx e^{183.147144976333} \\ &\approx 10^{79.5397944395563} \\ &\approx 3.46572771657154 \times 10^{79} } $$ The values of the logarithms vary depending on which base you are using. Calculators usually have the natural logarithm written as $\ln=\log_e$ (logarithm naturale: in French, the adjective follows the noun), whereas computers often use the "natural" base $e\approx 2:71828182845905$. The unspecified base is unfortunately ambiguous, and is a frequent source of confusion much like radians and degrees for $\sin$ and $\cos$. Whether $\log$ means $\log_{10}$ or $\ln$ must be made clear by a statement as there is no universally accepted convention.

Computers might return the final answer as $3.46572771657154$ E $79$, where the E is an abbreviation for scientific notation "$\times10$^". In this notation, the quantity $x$ is represented as a mantissa $M$ and an exponent $E$: $$ x \approx M \times 10^E $$ where $M$ is typically either in the interval $[0,1)$ or $[1,10)$ and $E$ is any integer (up to the storage size used by the data type on the computer or built-in size on the calculator). More info if you search for IEEE float/double.

The accuracy of the above calculation is practically as good as the size (number of digits of accuracy) of the mantissa.

Depending on the exponent storage size of your calculator or computer system, you could also just directly calculate the number, and the mantissa will be automatically truncated to the available precision.

There are also of course methods of calculating this to more than ten places by pen(cil) and paper; is this what you want?

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The result is exactly: $ 34657277165715299429841134098673896005946880584360167911317357675111364398875720+\frac{9}{17} $

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The question asks for the quickest way to calculate an approximate answer by hand. Did you do this by hand, and quickly? –  Graphth Apr 6 '12 at 16:01
    
(-1) for the answer and (+1) for @Grapth's comment... –  Gottfried Helms Apr 6 '12 at 19:38
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This answer is hilarious! –  Jase Nov 10 '12 at 4:45
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@Jase: I am a mental arithmetic amateur. –  dot dot Nov 10 '12 at 9:16

Why not simply do the computation? The most time-consuming part is to compute the $46$th power of $57$, but that can be done quite fast using repeated squaring in 6 multiplications. That results in a number with 81 digits. Dividing by 17 using long division is faster, and it takes about 80 easy steps.

When you are done, you can check your result with Google.

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Does Google compute that for you? For me it doesn't. –  Rasmus Oct 25 '10 at 14:46
    
@Rasmus: I've fixed the link (I don't know what was wrong...) –  Mariano Suárez-Alvarez Oct 25 '10 at 15:01
    
Ah, nice! $\mbox{}$ –  Rasmus Oct 25 '10 at 15:46

HINT $\rm \mod\ 17:\ \ 57\equiv 6,\ \ 6^{46}\equiv 1/6^2,\ \ 1/6\equiv 3\ \ $ which immediately yields the result

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It really depends what you mean by "approximate answer". Since you tagged arithmetic...

Notice 57 = 3 . 19, you can say that $57^{46} / 17 = 3 \cdot 57^{45} \cdot \frac{19}{17}$. So if you know what $57^{45}$ is, you can approximate to within 2% if you multiply that by 3.3.

Now, compute by hand $57^5 = 601692057$, which is less than 0.3% off from $6\cdot 10^8$. Observe that $57^{45} = (57^5)^9$. Using the binomial theorem you see that you can approximate that by $(6\cdot 10^8)^9 = 10 077 696 \cdot 10^{72}$ to within 3%. Rounding off the lower digits won't matter much in the error, so you have $57^{45} \sim 10^{79}$ to 3%.

So putting it all together you have that to within 5%,

$$ 57^{46} \sim 3.3 \cdot 10^{79} $$

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Actually, I was the one who tagged it arithmetic, since the original tags didn't exactly fit... in any event, this looks something like what Fermi would have done if asked this question. +1! –  J. M. Oct 25 '10 at 12:23
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Hum, I must have made a small error propagating the error. Wolfram tells me that the actual answer is roughtly 3.466 . 10^79, so the error is 5.03%, so not within 5%. But I hope that's close enough to be forgiven. –  Willie Wong Oct 25 '10 at 12:41

You can do it by Fermats little theorem as well.

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Euler's theorem gives you the remainder.

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