# Check if N can be obtained from 1 by repetitively multiplying by 10 or 20

Given an integer N, the task is to determine whether it is possible to obtain the value N from 1 by repetitively multiplying by 10 or 20. Print Yes if possible or No otherwise.
Examples:

Input: N = 200
Output: YES
Explanation:
1 * 10 -> 10 * 20 -> 200
Input: N = 90
Output: NO

Approach:
Follow the steps below to solve the problem:

1. Count the number of trailing zeroes.
2. After removing the trailing zeroes, if the remaining N cannot be expressed as a power of 2, print NO.
3. Otherwise, if log2N <= Count of trailing zeroes, print Yes.

Below is the implementation of the above approach:

## C++

 `// C++ program to check if N  ` `// can be obtained from 1 by  ` `// repetitive multiplication  ` `// by 10 or 20  ` `#include ` `using` `namespace` `std; ` ` `  `// Function to check if N can  ` `// be obtained or not  ` `void` `Is_possible(``long` `long` `int` `N) ` `{ ` `    ``int` `C = 0; ` `    ``int` `D = 0; ` `     `  `    ``// Count and remove trailing  ` `    ``// zeroes  ` `    ``while` `(N % 10 == 0) ` `    ``{ ` `        ``N = N / 10; ` `        ``C += 1; ` `    ``} ` `     `  `    ``// Check if remaining  ` `    ``// N is a power of 2  ` `    ``if``(``pow``(2, (``int``)log2(N)) == N)  ` `    ``{ ` `        ``D = (``int``)log2(N); ` `         `  `        ``// To check the condition  ` `        ``// to print YES or NO  ` `        ``if` `(C >= D) ` `            ``cout << ``"YES"``; ` `        ``else` `            ``cout << ``"NO"``;  ` `    ``}  ` `    ``else` `        ``cout << ``"NO"``;  ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``long` `long` `int` `N = 2000000000000; ` `     `  `    ``Is_possible(N);  ` `} ` ` `  `// This code is contributed by Stream_Cipher `

## Java

 `// Java program to check if N  ` `// can be obtained from 1 by  ` `// repetitive multiplication  ` `// by 10 or 20  ` `import` `java.util.*;  ` ` `  `class` `GFG{ ` `     `  `static` `void` `Is_possible(``long` `N) ` `{ ` `    ``long` `C = ``0``; ` `    ``long` `D = ``0``; ` `     `  `    ``// Count and remove trailing  ` `    ``// zeroes  ` `    ``while` `(N % ``10` `== ``0``) ` `    ``{ ` `        ``N = N / ``10``; ` `        ``C += ``1``; ` `    ``} ` `     `  `    ``// Check if remaining  ` `    ``// N is a power of 2 ` `    ``if``(Math.pow(``2``, (``long``)(Math.log(N) /  ` `                         ``(Math.log(``2``)))) == N)  ` `    ``{ ` `        ``D = (``long``)(Math.log(N) / (Math.log(``2``))); ` `         `  `        ``// To check the condition  ` `        ``// to prlong YES or NO  ` `        ``if` `(C >= D) ` `            ``System.out.print(``"YES"``); ` `        ``else` `            ``System.out.print(``"NO"``);  ` `    ``}  ` `    ``else` `        ``System.out.print(``"NO"``);  ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[])  ` `{  ` `    ``long` `N = 2000000000000L; ` `    ``Is_possible(N); ` `} ` `} ` ` `  `// This code is contributed by Stream_Cipher `

## Python

 `# Python Program to check if N  ` `# can be obtained from 1 by ` `# repetitive multiplication  ` `# by 10 or 20 ` ` `  `import` `math ` ` `  `# Function to check if N can ` `# be obtained or not ` `def` `Is_possible(N): ` ` `  `    ``C ``=` `0` `    ``D ``=` `0` ` `  `    ``# Count and remove trailing ` `    ``# zeroes ` `    ``while` `( N ``%` `10` `=``=` `0``): ` `        ``N ``=` `N ``/` `10` `        ``C ``+``=` `1` ` `  `    ``# Check if remaining ` `    ``# N is a power of 2 ` `    ``if` `( math.log(N, ``2``)  ` `    ``-` `int``(math.log(N, ``2``)) ``=``=` `0``): ` ` `  `        ``D ``=` `int``(math.log(N, ``2``)) ` ` `  `        ``# To check the condition ` `        ``# to print YES or NO ` `        ``if` `(C >``=` `D): ` `            ``print``(``"YES"``) ` `             `  `        ``else``: ` `            ``print``(``"NO"``) ` `     `  `    ``else``: ` `        ``print``(``"NO"``) ` `             `  `# Driver Program ` `N ``=` `2000000000000` `Is_possible(N) `

## C#

 `// C# program to check if N  ` `// can be obtained from 1 by  ` `// repetitive multiplication  ` `// by 10 or 20  ` `using` `System;  ` ` `  `class` `GFG{ ` `     `  `static` `void` `Is_possible(``long` `N) ` `{ ` `    ``long` `C = 0; ` `    ``long` `D = 0; ` `     `  `    ``// Count and remove trailing  ` `    ``// zeroes  ` `    ``while` `(N % 10 == 0) ` `    ``{ ` `        ``N = N / 10; ` `        ``C += 1; ` `    ``} ` `     `  `    ``// Check if remaining  ` `    ``// N is a power of 2 ` `    ``if``(Math.Pow(2, (``long``)(Math.Log(N) /  ` `                         ``(Math.Log(2)))) == N)  ` `    ``{ ` `        ``D = (``long``)(Math.Log(N) / (Math.Log(2))); ` `         `  `        ``// To check the condition  ` `        ``// to prlong YES or NO  ` `        ``if` `(C >= D) ` `            ``Console.WriteLine(``"YES"``); ` `        ``else` `            ``Console.WriteLine(``"NO"``);  ` `    ``}  ` `    ``else` `        ``Console.WriteLine(``"NO"``);  ` `} ` ` `  `// Driver Code  ` `public` `static` `void` `Main()  ` `{  ` `    ``long` `N = 2000000000000L; ` `     `  `    ``Is_possible(N); ` `} ` `} ` ` `  `// This code is contributed by Stream_Cipher `

Output:

```YES
```

Time Complexity: O(log10(N))
Auxiliary Space: O(1)

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