The Universe of Discourse

Tue, 28 Feb 2023

More about the seventh root of a 14-digit number

I recently explained how to quickly figure out the seventh root of the number !!19203908986159!! without a calculator, or even without paper if you happen to know a few things.

The key insight is that the answer has only two digits. To get the tens digit, I just estimated the size. But Roger Crew pointed out that there is another way. Suppose the number we want to find, !!n!!, is written as !!10p+q!!. We already know that !!q=9!!, so write this as !!n = 10(p+1)-1!! as in the previous article. Then expanding with the binomial theorem as before:

$$ \begin{align} n^7 & = \sum_{k=0}^7 \binom 7k\; (10(p+1))^{7-k}\; (-1)^k \\ & = (10(p+1))^ 7 + \ldots + \binom71 (10(p+1)) - 1 \\ \end{align} $$

All the terms except the last two are multiples of 100, because they are divisible by !!(10(p+1))^i!! for !!i\ge 2!!. So if we consider this equation mod-!!100!!, those terms all vanish, leaving:

$$ \begin{align} 19203908986159 & \equiv \binom71 (10(p+1)) - 1 & \pmod{100} \\ 59 & \equiv 70(p+1) - 1 & \pmod{100} \\ 90 & \equiv 70p & \pmod{100} \\ 9 & \equiv 7p & \pmod{10} \\ \end{align} $$

and the (only, because !!\gcd(7, 10) = 1!!) solution to this has !!p=7!! since !!7\cdot 7=49!!.

This does seem cleaner somehow, and my original way seems to depend on a lucky coincidence between the original number being close to !!2\cdot 10^{14}!! and my being able to estimate !!8^7 = 2^{21} \approx 2000000!! because !!2^{10}!! is luckily close to !!1000!!. On the other hand, I did it the way I did it, so in some sense it was good enough. As longtime Gentle Readers know already, I am a mathematical pig-slaughterer.

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