# Finding n-th number made of prime digits (2, 3, 5 and 7) only

Given a number ‘n’, we need to find the nth number whose each digit is a prime number i.e 2, 3, 5, 7….In other words you have to find nth number of this sequence. 2, 3, 5, 5, 22, 23……
Given that the nth number such found will be less then equal to 10^18
Examples :

```Input  : 10
Output : 33
2, 3, 5, 7, 22, 23, 25,
27, 32, 33

Input  : 21
Output : 222
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

There are four prime digits 2, 3, 5 and 7. First observation is that the number of numbers of x length and made of prime digits are because for each position you have 4 choices so total number is 4^x.
So total count of such numbers whose length is = 1 to len (i.e. 2 or 3 or more) will be 4*((4len – 1)/3). (This is sum of G.P with first term 4 and common ratio 4)

The algorithm is mainly divided in two steps.

1. We find the number of digits in n-th number using above observation. We start from len = 0 and keep incrementing it while value of it is smaller than 4*((4len – 1)/3).
2. Now we know number of digits in n-th number. We also know the count of numbers with with (len-1) digits. Let this count be ‘prev_count’. Now we one by one find digits in our result. First fix 2 at i-th place (assuming all the places upto i-1 are already filled), we have 4^(len – i) numbers possible and to check if 2 is the right candidate or not check if count of numbers after putting 2 is greater than or equal to n or not. If it is true then 2 is the right candidate if this is not true this means if we fix 2 at ith place only prev_count + 4^(len-i) numbers can be covered. So increase prev_count by 4^(len-i) and repeat this step for 3 check if 3 fits at ith place or not .If not go for 5. If 5 also does not fit, go for 7. It is guaranteed that 7 will fit it if 2, 3 and 5 do not fit, because we are sure that the length of nth such number is len only.

Below is implementation of above steps.

## C/C++

 `// C++ implementation for finding nth number ` `// made of prime digits only ` `#include ` `using` `namespace` `std; ` ` `  `// Prints n-th number where each digit is a ` `// prime number ` `void` `nthprimedigitsnumber(``long` `long` `n) ` `{ ` `    ``// Finding the length of n-th number ` `    ``long` `long` `len = 1; ` ` `  `    ``// Count of numbers with len-1 digits ` `    ``long` `long` `prev_count = 0; ` `    ``while` `(``true``) { ` `        ``// Count of numbers with i digits ` `        ``long` `long` `curr_count = prev_count + ``pow``(4, len); ` ` `  `        ``// if i is the length of such number ` `        ``// then n<4*(4^(i-1)-1)/3 and n>= 4*(4 ^ i-1)/3 ` `        ``// if a valid i is found break the loop ` `        ``if` `(prev_count < n && curr_count >= n) ` `            ``break``; ` ` `  `        ``// check for i + 1 ` `        ``len++; ` ` `  `        ``prev_count = curr_count; ` `    ``} ` ` `  `    ``// Till now we have covered 'prev_count' numbers ` ` `  `    ``// Finding ith digit at ith place ` `    ``for` `(``int` `i = 1; i <= len; i++) { ` `        ``// j = 1 means 2 j = 2 means ...j = 4 means 7 ` `        ``for` `(``long` `long` `j = 1; j <= 4; j++) { ` `            ``// if prev_count + 4 ^ (len-i) is less ` `            ``// than n, increase prev_count by 4^(x-i) ` `            ``if` `(prev_count + ``pow``(4, len - i) < n) ` `                ``prev_count += ``pow``(4, len - i); ` ` `  `            ``// else print the ith digit and break ` `            ``else` `{ ` `                ``if` `(j == 1) ` `                    ``cout << ``"2"``; ` `                ``else` `if` `(j == 2) ` `                    ``cout << ``"3"``; ` `                ``else` `if` `(j == 3) ` `                    ``cout << ``"5"``; ` `                ``else` `if` `(j == 4) ` `                    ``cout << ``"7"``; ` `                ``break``; ` `            ``} ` `        ``} ` `    ``} ` `    ``cout << endl; ` `} ` ` `  `// Driver function ` `int` `main() ` `{ ` `    ``nthprimedigitsnumber(10); ` `    ``nthprimedigitsnumber(21); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation for finding nth number ` `// made of prime digits only ` ` `  `import` `static` `java.lang.Math.pow; ` ` `  `class` `Test { ` ` `  `    ``// Prints n-th number where each digit is a ` `    ``// prime number ` `    ``static` `void` `nthprimedigitsnumber(``long` `n) ` `    ``{ ` `        ``// Finding the length of n-th number ` `        ``long` `len = ``1``; ` ` `  `        ``// Count of numbers with len-1 digits ` `        ``long` `prev_count = ``0``; ` `        ``while` `(``true``) { ` `            ``// Count of numbers with i digits ` `            ``long` `curr_count = (``long``)(prev_count + pow(``4``, len)); ` ` `  `            ``// if i is the length of such number ` `            ``// then n<4*(4^(i-1)-1)/3 and n>= 4*(4 ^ i-1)/3 ` `            ``// if a valid i is found break the loop ` `            ``if` `(prev_count < n && curr_count >= n) ` `                ``break``; ` ` `  `            ``// check for i + 1 ` `            ``len++; ` ` `  `            ``prev_count = curr_count; ` `        ``} ` ` `  `        ``// Till now we have covered 'prev_count' numbers ` ` `  `        ``// Finding ith digit at ith place ` `        ``for` `(``int` `i = ``1``; i <= len; i++) { ` `            ``// j = 1 means 2 j = 2 means ...j = 4 means 7 ` `            ``for` `(``long` `j = ``1``; j <= ``4``; j++) { ` `                ``// if prev_count + 4 ^ (len-i) is less ` `                ``// than n, increase prev_count by 4^(x-i) ` `                ``if` `(prev_count + pow(``4``, len - i) < n) ` `                    ``prev_count += pow(``4``, len - i); ` ` `  `                ``// else print the ith digit and break ` `                ``else` `{ ` `                    ``if` `(j == ``1``) ` `                        ``System.out.print(``"2"``); ` `                    ``else` `if` `(j == ``2``) ` `                        ``System.out.print(``"3"``); ` `                    ``else` `if` `(j == ``3``) ` `                        ``System.out.print(``"5"``); ` `                    ``else` `if` `(j == ``4``) ` `                        ``System.out.print(``"7"``); ` `                    ``break``; ` `                ``} ` `            ``} ` `        ``} ` `        ``System.out.println(); ` `    ``} ` ` `  `    ``// Driver method ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``nthprimedigitsnumber(``10``); ` `        ``nthprimedigitsnumber(``21``); ` `    ``} ` `} `

## Python3

 `# Python3 implementation for  ` `# finding nth number made of ` `# prime digits only ` `import` `math ` ` `  `# Prints n-th number where  ` `# each digit is a prime number ` `def` `nthprimedigitsnumber(n): ` `     `  `    ``# Finding the length ` `    ``# of n-th number ` `    ``len` `=` `1``; ` ` `  `    ``# Count of numbers  ` `    ``# with len-1 digits ` `    ``prev_count ``=` `0``; ` `    ``while``(``1``):  ` `         `  `        ``# Count of numbers  ` `        ``# with i digits ` `        ``curr_count ``=` `(prev_count ``+`  `                      ``math.``pow``(``4``, ``len``)); ` ` `  `        ``# if i is the length of such ` `        ``# number then n<4*(4^(i-1)-1)/3 ` `        ``# and n>= 4*(4 ^ i-1)/3 if a valid ` `        ``# i is found break the loop ` `        ``if` `(prev_count < n ``and`  `            ``curr_count >``=` `n): ` `            ``break``; ` ` `  `        ``# check for i + 1 ` `        ``len` `+``=` `1``; ` ` `  `        ``prev_count ``=` `curr_count; ` ` `  `    ``# Till now we have covered ` `    ``# 'prev_count' numbers ` ` `  `    ``# Finding ith digit at ith place ` `    ``for` `i ``in` `range` `(``1``, ``len` `+` `1``): ` `         `  `        ``# j = 1 means 2 j = 2 ` `        ``# means ...j = 4 means 7 ` `        ``for` `j ``in` `range``(``1``, ``5``):  ` `             `  `            ``# if prev_count + 4 ^ (len-i)  ` `            ``# is less than n, increase  ` `            ``# prev_count by 4^(x-i) ` `            ``if` `(prev_count ``+` `pow``(``4``, ``len` `-` `i) < n): ` `                ``prev_count ``+``=` `pow``(``4``, ``len` `-` `i); ` ` `  `            ``# else print the ith ` `            ``# digit and break ` `            ``else``: ` `                ``if` `(j ``=``=` `1``): ` `                    ``print``(``"2"``, end ``=` `""); ` `                ``elif` `(j ``=``=` `2``): ` `                    ``print``(``"3"``, end ``=` `""); ` `                ``elif` `(j ``=``=` `3``): ` `                    ``print``(``"5"``, end ``=` `""); ` `                ``elif` `(j ``=``=` `4``): ` `                    ``print``(``"7"``, end ``=` `""); ` `                ``break``; ` `    ``print``(); ` ` `  `# Driver Code ` `nthprimedigitsnumber(``10``); ` `nthprimedigitsnumber(``21``); ` ` `  `# This code is contributed by mits `

## C#

 `// C# implementation for finding nth  ` `// number made of prime digits only ` `using` `System; ` ` `  `public` `class` `GFG { ` `     `  `    ``// Prints n-th number where each  ` `    ``// digit is a prime number ` `    ``static` `void` `nthprimedigitsnumber(``long` `n) ` `    ``{ ` `        ``// Finding the length of n-th number ` `        ``long` `len = 1; ` ` `  `        ``// Count of numbers with len-1 digits ` `        ``long` `prev_count = 0; ` `        ``while` `(``true``) { ` `             `  `            ``// Count of numbers with i digits ` `            ``long` `curr_count = (``long``)(prev_count +  ` `                               ``Math.Pow(4, len)); ` ` `  `            ``// if i is the length of such number ` `            ``// then n<4*(4^(i-1)-1)/3 and n>= 4*(4 ^ i-1)/3 ` `            ``// if a valid i is found break the loop ` `            ``if` `(prev_count < n && curr_count >= n) ` `                ``break``; ` ` `  `            ``// check for i + 1 ` `            ``len++; ` ` `  `            ``prev_count = curr_count; ` `        ``} ` ` `  `        ``// Till now we have covered 'prev_count' numbers ` ` `  `        ``// Finding ith digit at ith place ` `        ``for` `(``int` `i = 1; i <= len; i++) { ` `             `  `            ``// j = 1 means 2 j = 2 means ...j = 4 means 7 ` `            ``for` `(``long` `j = 1; j <= 4; j++) { ` `                 `  `                ``// if prev_count + 4 ^ (len-i) is less ` `                ``// than n, increase prev_count by 4^(x-i) ` `                ``if` `(prev_count + Math.Pow(4, len - i) < n) ` `                    ``prev_count += (``long``)Math.Pow(4, len - i); ` ` `  `                ``// else print the ith digit and break ` `                ``else` `{ ` `                    ``if` `(j == 1) ` `                        ``Console.Write(``"2"``); ` `                    ``else` `if` `(j == 2) ` `                        ``Console.Write(``"3"``); ` `                    ``else` `if` `(j == 3) ` `                        ``Console.Write(``"5"``); ` `                    ``else` `if` `(j == 4) ` `                        ``Console.Write(``"7"``); ` `                    ``break``; ` `                ``} ` `            ``} ` `        ``} ` `        ``Console.WriteLine(); ` `    ``} ` ` `  `    ``// Driver method ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``nthprimedigitsnumber(10); ` `        ``nthprimedigitsnumber(21); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

## PHP

 `= 4*(4 ^ i-1)/3 if a valid ` `        ``// i is found break the loop ` `        ``if` `(``\$prev_count` `< ``\$n` `&&  ` `            ``\$curr_count` `>= ``\$n``) ` `            ``break``; ` ` `  `        ``// check for i + 1 ` `        ``\$len``++; ` ` `  `        ``\$prev_count` `= ``\$curr_count``; ` `    ``} ` ` `  `    ``// Till now we have covered ` `    ``// 'prev_count' numbers ` ` `  `    ``// Finding ith digit at ith place ` `    ``for` `(``\$i` `= 1; ``\$i` `<= ``\$len``; ``\$i``++)  ` `    ``{ ` `        ``// j = 1 means 2 j = 2 ` `        ``// means ...j = 4 means 7 ` `        ``for` `(``\$j` `= 1; ``\$j` `<= 4; ``\$j``++)  ` `        ``{ ` `            ``// if prev_count + 4 ^ (len-i)  ` `            ``// is less than n, increase  ` `            ``// prev_count by 4^(x-i) ` `            ``if` `(``\$prev_count` `+  ` `                 ``pow(4, ``\$len` `- ``\$i``) < ``\$n``) ` `                ``\$prev_count` `+= pow(4, ``\$len` `- ``\$i``); ` ` `  `            ``// else print the ith ` `            ``// digit and break ` `            ``else`  `            ``{ ` `                ``if` `(``\$j` `== 1) ` `                    ``echo` `"2"``; ` `                ``else` `if` `(``\$j` `== 2) ` `                    ``echo` `"3"``; ` `                ``else` `if` `(``\$j` `== 3) ` `                    ``echo` `"5"``; ` `                ``else` `if` `(``\$j` `== 4) ` `                    ``echo` `"7"``; ` `                ``break``; ` `            ``} ` `        ``} ` `    ``} ` `     `  `echo` `"\n"``; ` `} ` ` `  `// Driver Code ` `nthprimedigitsnumber(10); ` `nthprimedigitsnumber(21); ` ` `  `// This code is contributed by ajit ` `?> `

Output :

```33
222
```

Alternate Solution (Works in O(Log n)

```In this post, a O(log n) solution is discussed
which is based on below pattern in numbers. The
numbers can be seen
""
/                |                    |                 \
2                 3                    5                  7
/ |  | \           / | |  \             /  | | \          /  | |  \
22 23 25 27        32 33 35 37         52 53 55 57        72 73 75 77
/||\/||\/||\/||\   /||\/||\/||\/||\   /||\/||\/||\/||\   /||\/||\/||\/||\

We can notice following :
1st. 5th, 9th. 13th, ..... numbers have 2 as last digit.
2nd. 6th, 10th. 14th, ..... numbers have 3 as last digit.
3nd. 7th, 11th. 15th, ..... numbers have 5 as last digit.
4th. 8th, 12th. 16th, ..... numbers have 7 as last digit.
```

## C++

 `// CPP program to find n-th number with ` `// prime digits 2, 3 and 7 ` `#include ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `string nthprimedigitsnumber(``int` `number) ` `{ ` `    ``int` `rem; ` `    ``string num; ` `    ``while` `(number) { ` `        ``// remainder for check element position ` `        ``rem = number % 4; ` `        ``switch` `(rem) { ` ` `  `        ``// if number is 1st position in tree ` `        ``case` `1: ` `            ``num.push_back(``'2'``); ` `            ``break``; ` ` `  `        ``// if number is 2nd position in tree ` `        ``case` `2: ` `            ``num.push_back(``'3'``); ` `            ``break``; ` ` `  `        ``// if number is 3rd position in tree ` `        ``case` `3: ` `            ``num.push_back(``'5'``); ` `            ``break``; ` ` `  `        ``// if number is 4th position in tree ` `        ``case` `0: ` `            ``num.push_back(``'7'``); ` `            ``break``; ` `        ``} ` ` `  `        ``if` `(number % 4 == 0) ` `           ``number--; ` ` `  `        ``number = number / 4; ` `    ``} ` `    ``reverse(num.begin(), num.end()); ` `    ``return` `num; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `number = 21; ` `    ``cout << nthprimedigitsnumber(10) << ``"\n"``; ` `    ``cout << nthprimedigitsnumber(21) << ``"\n"``; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find n-th number with  ` `// prime digits 2, 3 and 7  ` `import` `java.util.*; ` `class` `GFG{ ` `static` `String nthprimedigitsnumber(``int` `number)  ` `{  ` `    ``int` `rem;  ` `    ``String num=``""``;  ` `    ``while` `(number>``0``) {  ` `        ``// remainder for check element position  ` `        ``rem = number % ``4``;  ` `        ``switch` `(rem) {  ` ` `  `        ``// if number is 1st position in tree  ` `        ``case` `1``:  ` `            ``num+=``'2'``;  ` `            ``break``;  ` ` `  `        ``// if number is 2nd position in tree  ` `        ``case` `2``:  ` `            ``num+=``'3'``;  ` `            ``break``;  ` ` `  `        ``// if number is 3rd position in tree  ` `        ``case` `3``:  ` `            ``num+=``'5'``;  ` `            ``break``;  ` ` `  `        ``// if number is 4th position in tree  ` `        ``case` `0``:  ` `            ``num+=``'7'``;  ` `            ``break``;  ` `        ``}  ` ` `  `       ``if` `(number % ``4` `== ``0``) ` `           ``number--; ` ` `  `        ``number = number / ``4``;  ` `    ``}  ` `     `  `    ``return` `new` `StringBuilder(num).reverse().toString();  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``int` `number = ``21``;  ` `    ``System.out.println(nthprimedigitsnumber(``10``));  ` `    ``System.out.println(nthprimedigitsnumber(``21``));  ` `}  ` `} ` `// This code is contributed by mits `

## Python3

 `# Python3 program to find n-th number  ` `# with prime digits 2, 3 and 7  ` `def` `nthprimedigitsnumber(number): ` ` `  `    ``num ``=` `"";  ` `    ``while` `(number > ``0``):  ` `         `  `        ``# remainder for check element position  ` `        ``rem ``=` `number ``%` `4``;  ` `         `  `        ``# if number is 1st position in tree ` `        ``if` `(rem ``=``=` `1``): ` `            ``num ``+``=` `'2'``;  ` ` `  `        ``# if number is 2nd position in tree  ` `        ``if` `(rem ``=``=` `2``):  ` `            ``num ``+``=` `'3'``;  ` ` `  `        ``# if number is 3rd position in tree  ` `        ``if` `(rem ``=``=` `3``):  ` `            ``num ``+``=` `'5'``;  ` ` `  `        ``# if number is 4th position in tree  ` `        ``if` `(rem ``=``=` `0``):  ` `            ``num ``+``=` `'7'``;  ` ` `  `        ``if` `(number ``%` `4` `=``=` `0``): ` `            ``number ``=` `number ``-` `1` ` `  `        ``number ``=` `number ``/``/` `4``;  ` ` `  `    ``return` `num[::``-``1``];  ` ` `  `# Driver code  ` `number ``=` `21``;  ` `print``(nthprimedigitsnumber(``10``));  ` `print``(nthprimedigitsnumber(number));  ` ` `  `# This code is contributed by mits `

## C#

 `// C# program to find n-th number with  ` `// prime digits 2, 3 and 7  ` `using` `System; ` `class` `GFG{ ` `static` `string` `nthprimedigitsnumber(``int` `number)  ` `{  ` `    ``int` `rem;  ` `    ``string` `num=``""``;  ` `    ``while` `(number>0) {  ` `        ``// remainder for check element position  ` `        ``rem = number % 4;  ` `        ``switch` `(rem) {  ` ` `  `        ``// if number is 1st position in tree  ` `        ``case` `1:  ` `            ``num+=``'2'``;  ` `            ``break``;  ` ` `  `        ``// if number is 2nd position in tree  ` `        ``case` `2:  ` `            ``num+=``'3'``;  ` `            ``break``;  ` ` `  `        ``// if number is 3rd position in tree  ` `        ``case` `3:  ` `            ``num+=``'5'``;  ` `            ``break``;  ` ` `  `        ``// if number is 4th position in tree  ` `        ``case` `0:  ` `            ``num+=``'7'``;  ` `            ``break``;  ` `        ``}  ` ` `  `       ``if` `(number % 4 == 0) ` `           ``number--; ` ` `  `        ``number = number / 4;  ` `    ``}  ` `    ``char``[] st = num.ToCharArray(); ` `    ``Array.Reverse(st); ` `    ``return` `new` `string``(st);  ` `}  ` ` `  `// Driver code  ` `static` `void` `Main()  ` `{  ` `    ``int` `number = 21;  ` `    ``Console.WriteLine(nthprimedigitsnumber(10));  ` `    ``Console.WriteLine(nthprimedigitsnumber(number));  ` `}  ` `} ` `// This code is contributed by mits `

## PHP

 ` 0)  ` `    ``{  ` `        ``// remainder for check element position  ` `        ``\$rem` `= ``\$number` `% 4;  ` `        ``switch` `(``\$rem``)  ` `        ``{  ` ` `  `            ``// if number is 1st position in tree  ` `            ``case` `1:  ` `                ``\$num` `.= ``'2'``;  ` `                ``break``;  ` `     `  `            ``// if number is 2nd position in tree  ` `            ``case` `2:  ` `                ``\$num` `.= ``'3'``;  ` `                ``break``;  ` `     `  `            ``// if number is 3rd position in tree  ` `            ``case` `3:  ` `                ``\$num` `.= ``'5'``;  ` `                ``break``;  ` `     `  `            ``// if number is 4th position in tree  ` `            ``case` `0:  ` `                ``\$num` `.= ``'7'``;  ` `                ``break``;  ` `        ``}  ` ` `  `       ``if` `(``\$number` `% 4 == 0) ` `           ``\$number``--; ` ` `  `        ``\$number` `= (int)(``\$number` `/ 4);  ` `    ``}  ` ` `  `    ``return` `strrev``(``\$num``);  ` `}  ` ` `  `// Driver code  ` `\$number` `= 21;  ` `print``(nthprimedigitsnumber(10) . ``"\n"``);  ` `print``(nthprimedigitsnumber(``\$number``));  ` ` `  `// This code is contributed by mits `

Output :

```33
222
```

This article is contributed by Ayush Jha and Devanshu agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.