Roots of Arbitrary Numbers [closed]

Question

One day, when I was bored in maths class, I learned of a neat trick for solving the real cube root of a number!

Let's use the number \$79,507\$ as an example.

First, take digit in the one's place and compare it to this table:

\begin{array} {|r|r|} \hline \text{Extracted Digit} &\text{Resulting Digit} \\ \hline \text{1} &\text{1} \\ \text{2} &\text{8} \\ \text{3} &\text{7} \\ \text{4} &\text{4} \\ \text{5} &\text{5} \\ \text{6} &\text{6} \\ \text{7} &\text{3} \\ \text{8} &\text{2} \\ \text{9} &\text{9} \\ \text{0} &\text{0} \\ \hline \end{array}

In this example, the Resulting Digit will be \$3\$ since the extracted digit is \$7\$.

Next, remove all digits that are less than \$10^3\$:

$$ 79507 → 79 $$

Then, find the largest perfect cube that does not exceed the input:

$$ 64 < 79 $$

\$64=4^3\$, thus the next digit needed is \$4\$.

Finally, multiply the digit found in the previous step by \$10\$ and add the Resulting Digit found in the first step:

$$ 10*4+3=43 $$

Thus, the cube root of \$79,507\$ equals \$43\$.

However, there a neat quirk about this trick: it doesn't apply to only cubed numbers. In fact, it works with all \$n>1\$ where \$n\bmod2\ne0\$.

The steps mentioned above can be summed up in this generalization for an \$n\$ power:

• Step 1) Take the digit in the one's place in the input. Compare it to the one's place digit of the \$n\$th powers of \$1\$ to \$10\$, then use the corresponding digit.

• Step 2) Remove all digits of the input less than \$10^n\$. Compare the resulting number to the perfect powers definied in Step 1. Use the \$n\$th root of the largest perfect power less than said number. (Largest perfect power can exceed \$10^n\$)

• Step 3) Multiply the number from Step 2 by 10 then add the number from Step 1. This will be the final result.

Task

Given two positive integers \$n\$ and \$m\$, return the \$n\$th root of \$m\$.

Input:

• Two positive integers \$n\$ and \$m\$.

• \$m\$ is guaranteed to be a perfect \$n\$th power of an integer.

• \$n\$ is guaranteed to be odd and greater than \$2\$. (This method doesn't work if \$n\$ is even.)

Output:

• The values calculated in steps 1 and 2.

• The \$n\$th root of \$m\$.

• Output can be on multiples lines or a list, whichever is more convenient.

Rules:

• This is code-golf, so the fewer bytes, the better!

• Standard I/O rules apply.

• The output must be calculated using the aforementioned method.

• No builtins allowed that already calculate this. A prime example being TI-BASIC's x√ command.

Examples:

Input      | Output
-------------------
3, 79507   | 3
           | 4
           | 43
3, 79507   | [3, 4, 43]
5, 4084101 | 1
           | 2
           | 21
5, 4084101 | [1, 2, 21]

Answer 1

05AB1E, 18 bytes

Saved 1 byte thanks to Kevin Cruijssen

mθ=²¹°÷Dݹm@O<=ìï=

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Answer 2

JavaScript (ES7),  58  56 bytes

Saved 11 many bytes thanks to @Emigna

Takes input as (m)(n). Returns the 2 intermediate values and the final result as an array.

m=>g=(n,k)=>k**n>m/10**n?[u=m**n%10,--k,k*10+u]:g(n,-~k)

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---

JavaScript (Node.js), 62 bytes

This version uses BigInts as I/O to support larger inputs.

m=>g=(n,k=0n)=>k**n>m/10n**n?[u=m**n%10n,--k,k*10n+u]:g(n,++k)

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Answer 3

R, 74 bytes

function(m,n)c(a<-match(m%%10,(1:9)^n%%10,0),b<-(m/10^n)^(1/n)%/%1,a+10*b)

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match(m%%10,(1:9)^n%%10,0) compares the last digit of m to the reference table (and outputs 0 if no match is found), thus performing step 1.

(m/10^n)^(1/n)%/%1 gives the output of step 2.