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I'd like to estimate $7^{100}$ without using these new-fangled computers everyone is talking about, but I think I need your help.

Please estimate $7^{100}$ without the use of a computer. My friend Steven told me the value is approximately

$3.23448 \times 10^{84}$,

but he refused to show his work, so I'm inclined not to trust him.

Please give your answer in scientific notation, e.g., $4.03 \times 10^7$. I'd like the estimate to be within $\pm 50$% of the correct answer, but getting the correct number of digits would be great.


This puzzle was inspired by this one.

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6 Answers 6

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A simple upper bound:

$7^2 = 49$ is just under $50$, so $7^{100} = (7^2)^{50}$ is somewhat less than $50^{50} = (100/2)^{50}$.

$2^{10} = 1024$ is just over $1000 = 10^3$, so $2^{50}$ is somewhat over $10^{15}$.

So $$7^{50} < \frac{100^{50}}{10^{15}} = \frac{\left(10^2\right)^{50}}{10^{15}} = \frac{10^{100}}{10^{15}} = 10^{85} = 1\ \text{e}\ 85$$

And a semi-simple lower bound:

$7^6 = 117649$ is somewhat more than $10^5$, so $7^{96} = \left(7^6\right)^{16}$ is somewhat more than $\left(10^5\right)^{16} = 10^{80}$.

$7^{100} = 7^{96} \times 7^4$ is somewhat more than $10^{80} \times 2401 = 2.40 1\ \mathrm{e}\ 83$.

Or, borrowing an answer from the previous problem: If you have

a log table giving $\log_{10}(7) \approx 0.845$

then you can determine that

$\log_{10}(7^{100}) = 100 \times \log_{10}(7) \approx 84.5$

which produces an estimate of

$\sqrt{10} \times 10^{84} \approx 3.162\ \mathrm{e}\ 84$

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  • $\begingroup$ Hi, just some small advice on formatting. For log in math mode, you can use \log. Otherwise, great answer! +1 $\endgroup$ Commented Oct 6, 2022 at 5:20
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Once again, I solved this in my head, without writing or looking anything up.

$7^{100} = 49 ^ {50} = \left(1-\frac{1}{50}\right)^{50} 50^{50} $

We use the approximation

$ \left(1-\frac{1}{50}\right)^{50} \approx 1/e $

and also rewrite and estimate

$ 50^{50} = \left(\frac{100}{2} \right) ^ {50} = 10^{100} / 2^{50} = 10^{100} / (2^{10}) ^ 5 = 10^{100} / 1024 ^ 5 \approx 10^{100} / 10^{15} = 10^{85}$

to get

$7^{100} \approx \frac{10^{85}}{e} \approx \frac{10^{85}}{3} \approx 3 \cdot 10^{84}$

which is quite close and gives the right digit count of

85 digits

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Using my strategy from before:

Using the logarithm tables, we get $$\log_{10}(7.000)=0.8451\\\log_{10}(7.000^{100})=84.51\\7^{100}=10^{84}\cdot10^{0.51}\\7^{100}=3.236\cdot10^{84}$$.

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I'll start with the approximation:

$7^4 = 2401 \approx 2400 = 2^3*3*10^2 \\ 7^{100} \approx 2^{75} 3^{25} 10^{50}$

Next, let's estimate that

$3^4 = 81 \approx 80 =2^3 10 \\ 3^{25} \approx 3*2^{18} 10^6 \\ 7^{100} \approx 3*2^{93} 10^{56}$

Our last step is

$2^{10} = 1024 \approx 1000 = 10^3 \\ 7^{100} \approx 2.4 * 10^{84}$

To refine this estimate, let's do some first-order analysis.

$2401 \to 2400$: Off by 1/2400, used 25 times. Error: about 1% low. $81 \to 80$: Off by 1/80, used 6 times. Error: 7.5% low. $1024 \to 1000$: Off by 2.4%, used 9 times. Error: 21.6% low. In total, our first estimate is about 30% low.

Therefore, the final estimate is

$2.4 * 1.3 = 3.12 \\ 3.12 * 10^{84}$

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  • $\begingroup$ This is nice. Simply adding the errors is technically incorrect but it is a very good approximation for the actual compound error, so your estimate gets much more accurate with that. $\endgroup$
    – quarague
    Commented Oct 6, 2022 at 13:35
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Simplify the following: 7^10

Compute 7^10 by repeated squaring. For example
$a^7 = a a^6 = a (a^3)^2 = a (a a^2)^2.$
$\Rightarrow 7^{10} = (7^5)^2 = (7×7^4)^2 = (7(7^2)^2)^2$

Evaluate $7^2$.
$7^2 = 49 \Rightarrow (7×49^2)^2$

Evaluate $49^2$.

    4 9
  × 4 9
-------
  4 4 1
1 9 6 0
-------
2 4 0 1


$(7×2401)^2$

Multiply 7 and 2401 together.
$7×2401 = 16807 \Rightarrow 16807^2$

Evaluate $16807^2$.

         1 6 8 0 7
       × 1 6 8 0 7
------------------
       1 1 7 6 4 9
     0 0 0 0 0 0 0
   1 3 4 4 5 6 0 0
 1 0 0 8 4 2 0 0 0
 1 6 8 0 7 0 0 0 0
 -----------------
 2 8 2 4 7 5 2 4 9


Answer: $282475249$

Hence,

$7^{100} = ((7^{10})^2)^5= (79792266297612001)^5 = 3234476509624757991344647769100216810857203198904625400933895331391691459636928060001$, which has 85 digits.

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  • $\begingroup$ I did some editing to fix the formatting, but since markdown tables don't work in spoilers, I couldn't quite figure out how to format your manual multiplications. Perhaps you could turn them into pictures that you can embed. Just leave the results as text, please! $\endgroup$
    – xyldke
    Commented Oct 5, 2022 at 15:29
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    $\begingroup$ How did you do the final calculation? $\endgroup$
    – xyldke
    Commented Oct 5, 2022 at 15:29
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    $\begingroup$ @xyldke You can use preformatted text inside spoiler blocks to preserve column alignment. $\endgroup$
    – fljx
    Commented Oct 5, 2022 at 16:12
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A close enough lower bound:

$7^2 = 49$
$7^4 > 50*48 = 2400$
$7^8 > 2500*2300 = 5.75*10^6$
$7^{16} > 6 * 10^6 * 5.5 * 10^6 = 33 * 10^{12}$
$7^{32} > 1000 * 10^{24}$

$7^{100} = 7^4 *7^{32} * 7^{32} * 7^{32} > 2400 * 10^{81}$

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