# Reach A and B by multiplying them with K and K^2 at every step

We are given two numbers A and B, we need to write a program to determine if A and B can be reached starting from (1, 1) following the given steps. Start from (1, 1) and at every step choose a random number K and multiply K to any one of the two numbers obtained in the previous step and K^{2} to the other number.

**Examples:**

Input : A = 3, B = 9 Output : yesExplanation: Starting from A = 1 and B = 1. We choose k=3 and multiply 3 with the first number to get A=3 and multiply k^{2}=9 to the second- number to get B=9. Input : A = 60, B = 450 Output : yesExplanation: Starting from A = 1 and B = 1, Step 1: multiply k=3 and k^{2}to get 3 and 9 Step 2: Multiply k=5 and k^{2}= 25 to get to 15 and 225 Step 3: Multiply k^{2}=4 and k=2 to get to A=60 and B=450

The idea to solve this problem is to observe closely that at each step we are multiplying k and k^{2} to the numbers. So if A and B can be reached, it will have **k^3 at every step as factors in A*B**. In simple words, if the number **A*B is a perfect cube and it divides A and B both**, only then the number can be reached starting from 1 and 1 by performing given steps.

Below is the implementation of above idea:

## C++

`// CPP program to determine if ` `// A and B can be reached starting ` `// from 1, 1 following the given steps. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// function to check is it is possible to reach ` `// A and B starting from 1 and 1 ` `bool` `possibleToReach(` `int` `a, ` `int` `b) ` `{ ` ` ` `// find the cuberoot of the number ` ` ` `int` `c = cbrt(a * b); ` ` ` ` ` `// divide the number by cuberoot ` ` ` `int` `re1 = a / c; ` ` ` `int` `re2 = b / c; ` ` ` ` ` `// if it is a perfect cuberoot and divides a and b ` ` ` `if` `((re1 * re1 * re2 == a) && (re2 * re2 * re1 == b)) ` ` ` `return` `true` `; ` ` ` `else` ` ` `return` `false` `; ` `} ` ` ` `int` `main() ` `{ ` ` ` `int` `A = 60, B = 450; ` ` ` ` ` `if` `(possibleToReach(A, B)) ` ` ` `cout << ` `"yes"` `; ` ` ` `else` ` ` `cout << ` `"no"` `; ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to determine if ` `// A and B can be reached starting ` `// from 1, 1 following the given ` `// steps. ` `class` `GFG { ` ` ` ` ` `// function to check is it is ` ` ` `// possible to reach A and B ` ` ` `// starting from 1 and 1 ` ` ` `static` `boolean` `possibleToReach(` `int` `a, ` `int` `b) ` ` ` `{ ` ` ` ` ` `// find the cuberoot of the number ` ` ` `int` `c = (` `int` `)Math.cbrt(a * b); ` ` ` ` ` `// divide the number by cuberoot ` ` ` `int` `re1 = a / c; ` ` ` `int` `re2 = b / c; ` ` ` ` ` `// if it is a perfect cuberoot and ` ` ` `// divides a and b ` ` ` `if` `((re1 * re1 * re2 == a) && ` ` ` `(re2 * re2 * re1 == b)) ` ` ` `return` `true` `; ` ` ` `else` ` ` `return` `false` `; ` ` ` `} ` ` ` ` ` `// Driver code ` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{ ` ` ` `int` `A = ` `60` `, B = ` `450` `; ` ` ` ` ` `if` `(possibleToReach(A, B)) ` ` ` `System.out.println(` `"yes"` `); ` ` ` `else` ` ` `System.out.println(` `"no"` `); ` ` ` `} ` `} ` ` ` `// This code is contributed by ` `// Smitha Dinesh Semwal ` |

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## Python 3

`# Python 3 program to determine if ` `# A and B can be reached starting ` `# from 1, 1 following the given steps. ` `import` `numpy as np ` ` ` `# function to check is it is possible to ` `# reach A and B starting from 1 and 1 ` `def` `possibleToReach(a, b): ` ` ` ` ` `# find the cuberoot of the number ` ` ` `c ` `=` `np.cbrt(a ` `*` `b) ` ` ` ` ` `# divide the number by cuberoot ` ` ` `re1 ` `=` `a ` `/` `/` `c ` ` ` `re2 ` `=` `b ` `/` `/` `c ` ` ` ` ` `# if it is a perfect cuberoot and ` ` ` `# divides a and b ` ` ` `if` `((re1 ` `*` `re1 ` `*` `re2 ` `=` `=` `a) ` `and` ` ` `(re2 ` `*` `re2 ` `*` `re1 ` `=` `=` `b)): ` ` ` `return` `True` ` ` `else` `: ` ` ` `return` `False` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `A ` `=` `60` ` ` `B ` `=` `450` ` ` ` ` `if` `(possibleToReach(A, B)): ` ` ` `print` `(` `"yes"` `) ` ` ` `else` `: ` ` ` `print` `(` `"no"` `) ` ` ` `# This code is contributed by ita_c ` |

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## C#

`// C# program to determine if ` `// A and B can be reached starting ` `// from 1, 1 following the given ` `// steps. ` `using` `System; ` ` ` `public` `class` `GFG{ ` ` ` ` ` `// function to check is it is ` ` ` `// possible to reach A and B ` ` ` `// starting from 1 and 1 ` ` ` `public` `static` `bool` `possibleToReach(` `int` `a, ` `int` `b) ` ` ` `{ ` ` ` ` ` `// find the cuberoot of the number ` ` ` `int` `c = (` `int` `)Math.Pow(a * b, (` `double` `) 1 / 3); ` ` ` `// divide the number by cuberoot ` ` ` `int` `re1 = a / c; ` ` ` `int` `re2 = b / c; ` ` ` ` ` `// if it is a perfect cuberoot ` ` ` `// and divides a and b ` ` ` `if` `((re1 * re1 * re2 == a) && ` ` ` `(re2 * re2 * re1 == b)) ` ` ` `return` `true` `; ` ` ` `else` ` ` `return` `false` `; ` ` ` `} ` ` ` `// Driver Code ` ` ` `static` `public` `void` `Main (String []args) ` ` ` `{ ` ` ` ` ` `int` `A = 60, B = 450; ` ` ` ` ` `if` `(possibleToReach(A, B)) ` ` ` `Console.WriteLine(` `"Yes"` `); ` ` ` `else` ` ` `Console.WriteLine(` `"No"` `); ` ` ` `} ` `} ` ` ` `// This code is contributed by Ajit. ` |

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## PHP

`<?php ` `// PHP program to determine if ` `// A and B can be reached starting ` `// from 1, 1 following the given steps. ` ` ` `// function to check is it is ` `// possible to reach A and B ` `// starting from 1 and 1 ` `function` `possibleToReach(` `$a` `, ` `$b` `) ` `{ ` ` ` ` ` `// find the cuberoot ` ` ` `// of the number ` ` ` `$c` `=(` `$a` `* ` `$b` `); ` ` ` ` ` `// divide the number ` ` ` `// by cuberoot ` ` ` `$re1` `= ` `$a` `/ ` `$c` `; ` ` ` `$re2` `= ` `$b` `/ ` `$c` `; ` ` ` ` ` `// if it is a perfect cuberoot ` ` ` `// and divides a and b ` ` ` `if` `((` `$re1` `* ` `$re1` `* ` `$re2` `== ` `$a` `) && ` ` ` `(` `$re2` `* ` `$re2` `* ` `$re1` `== ` `$b` `)) ` ` ` `return` `1; ` ` ` `else` ` ` `return` `-1; ` `} ` ` ` ` ` `// Driver Code ` ` ` `$A` `= 60; ` ` ` `$B` `= 450; ` ` ` `if` `(possibleToReach(` `$A` `, ` `$B` `)) ` ` ` `echo` `"yes"` `; ` ` ` `else` ` ` `echo` `"no"` `; ` ` ` `// This code is contributed by aj_36 ` `?> ` |

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Output:

Yes

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