# How many digits does $7^{100}$ have?

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.

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

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

• Hi, just some small advice on formatting. For log in math mode, you can use \log. Otherwise, great answer! +1 Commented Oct 6, 2022 at 5:20

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

Using my strategy from before:

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

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

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

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

• 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. Commented Oct 6, 2022 at 13:35

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.

• 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! Commented Oct 5, 2022 at 15:29
• How did you do the final calculation? Commented Oct 5, 2022 at 15:29
• @xyldke You can use preformatted text inside spoiler blocks to preserve column alignment.
– fljx
Commented Oct 5, 2022 at 16:12

A close enough lower bound:

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

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