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Compute sum of digits in all numbers from 1 to n
• Difficulty Level : Hard
• Last Updated : 22 Apr, 2021

Given a number n, find the sum of digits in all numbers from 1 to n.
Examples:

```Input: n = 5
Output: Sum of digits in numbers from 1 to 5 = 15

Input: n = 12
Output: Sum of digits in numbers from 1 to 12 = 51

Input: n = 328
Output: Sum of digits in numbers from 1 to 328 = 3241```

Naive Solution:
A naive solution is to go through every number x from 1 to n and compute the sum in x by traversing all digits of x. Below is the implementation of this idea.

## C++

 `// A Simple C++ program to compute sum of digits in numbers from 1 to n``#include``using` `namespace` `std;` `int` `sumOfDigits(``int` `);` `// Returns sum of all digits in numbers from 1 to n``int` `sumOfDigitsFrom1ToN(``int` `n)``{``    ``int` `result = 0; ``// initialize result` `    ``// One by one compute sum of digits in every number from``    ``// 1 to n``    ``for` `(``int` `x = 1; x <= n; x++)``        ``result += sumOfDigits(x);` `    ``return` `result;``}` `// A utility function to compute sum of digits in a``// given number x``int` `sumOfDigits(``int` `x)``{``    ``int` `sum = 0;``    ``while` `(x != 0)``    ``{``        ``sum += x %10;``        ``x   = x /10;``    ``}``    ``return` `sum;``}` `// Driver Program``int` `main()``{``    ``int` `n = 328;``    ``cout << ``"Sum of digits in numbers from 1 to "` `<< n << ``" is "``         ``<< sumOfDigitsFrom1ToN(n);``    ``return` `0;``}`

## Java

 `// A Simple JAVA program to compute sum of``// digits in numbers from 1 to n``import` `java.io.*;` `class` `GFG {``    ` `    ``// Returns sum of all digits in numbers``    ``// from 1 to n``    ``static` `int` `sumOfDigitsFrom1ToN(``int` `n)``    ``{``        ``int` `result = ``0``; ``// initialize result``     ` `        ``// One by one compute sum of digits``        ``// in every number from 1 to n``        ``for` `(``int` `x = ``1``; x <= n; x++)``            ``result += sumOfDigits(x);``     ` `        ``return` `result;``    ``}``     ` `    ``// A utility function to compute sum``    ``// of digits in a given number x``    ``static` `int` `sumOfDigits(``int` `x)``    ``{``        ``int` `sum = ``0``;``        ``while` `(x != ``0``)``        ``{``            ``sum += x % ``10``;``            ``x   = x / ``10``;``        ``}``        ``return` `sum;``    ``}``     ` `    ``// Driver Program``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `n = ``328``;``        ``System.out.println(``"Sum of digits in numbers"``                          ``+``" from 1 to "` `+ n + ``" is "``                          ``+ sumOfDigitsFrom1ToN(n));``    ``}``}` `/*This code is contributed by Nikita Tiwari.*/`

## Python3

 `# A Simple Python program to compute sum``# of digits in numbers from 1 to n` `# Returns sum of all digits in numbers``# from 1 to n``def` `sumOfDigitsFrom1ToN(n) :` `    ``result ``=` `0`   `# initialize result`` ` `    ``# One by one compute sum of digits``    ``# in every number from 1 to n``    ``for` `x ``in` `range``(``1``, n``+``1``) :``        ``result ``=` `result ``+` `sumOfDigits(x)`` ` `    ``return` `result` `# A utility function to compute sum of``# digits in a given number x``def` `sumOfDigits(x) :``    ``sum` `=` `0``    ``while` `(x !``=` `0``) :``        ``sum` `=` `sum` `+` `x ``%` `10``        ``x   ``=` `x ``/``/` `10``    ` `    ``return` `sum`  `# Driver Program``n ``=` `328``print``(``"Sum of digits in numbers from 1 to"``, n, ``"is"``, sumOfDigitsFrom1ToN(n))`  `# This code is contributed by Nikita Tiwari.`

## C#

 `// A Simple C# program to compute sum of``// digits in numbers from 1 to n` `using` `System;` `public` `class` `GFG {``    ` `    ``// Returns sum of all digits in numbers``    ``// from 1 to n``    ``static` `int` `sumOfDigitsFrom1ToN(``int` `n)``    ``{``        ` `        ``// initialize result``        ``int` `result = 0;``    ` `        ``// One by one compute sum of digits``        ``// in every number from 1 to n``        ``for` `(``int` `x = 1; x <= n; x++)``            ``result += sumOfDigits(x);``    ` `        ``return` `result;``    ``}``    ` `    ``// A utility function to compute sum``    ``// of digits in a given number x``    ``static` `int` `sumOfDigits(``int` `x)``    ``{``        ``int` `sum = 0;``        ` `        ``while` `(x != 0)``        ``{``            ``sum += x % 10;``            ``x = x / 10;``        ``}``        ` `        ``return` `sum;``    ``}``    ` `    ``// Driver Program``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 328;``        ` `        ``Console.WriteLine(``"Sum of digits"``               ``+ ``" in numbers from 1 to "``                             ``+ n + ``" is "``                ``+ sumOfDigitsFrom1ToN(n));``    ``}``}` `// This code is contributed by shiv_bhakt.`

## PHP

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## Javascript

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Output

`Sum of digits in numbers from 1 to 328 is 3241`

Efficient Solution:
Above is a naive solution. We can do it more efficiently by finding a pattern.
Let us take few examples.

```sum(9) = 1 + 2 + 3 + 4 ........... + 9
= 9*10/2
= 45

sum(99)  = 45 + (10 + 45) + (20 + 45) + ..... (90 + 45)
= 45*10 + (10 + 20 + 30 ... 90)
= 45*10 + 10(1 + 2 + ... 9)
= 45*10 + 45*10
= sum(9)*10 + 45*10

sum(999) = sum(99)*10 + 45*100```

In general, we can compute sum(10d – 1) using the below formula

`   sum(10d - 1) = sum(10d-1 - 1) * 10 + 45*(10d-1) `

In the below implementation, the above formula is implemented using dynamic programming as there are overlapping subproblems.
The above formula is one core step of the idea. Below is the complete algorithm

Algorithm: sum(n)

```1) Find number of digits minus one in n. Let this value be 'd'.
For 328, d is 2.

2) Compute some of digits in numbers from 1 to 10d - 1.
Let this sum be w. For 328, we compute sum of digits from 1 to
99 using above formula.

3) Find Most significant digit (msd) in n. For 328, msd is 3.

4) Overall sum is sum of following terms

a) Sum of digits in 1 to "msd * 10d - 1".  For 328, sum of
digits in numbers from 1 to 299.
For 328, we compute 3*sum(99) + (1 + 2)*100.  Note that sum of
sum(299) is sum(99) + sum of digits from 100 to 199 + sum of digits
from 200 to 299.
Sum of 100 to 199 is sum(99) + 1*100 and sum of 299 is sum(99) + 2*100.
In general, this sum can be computed as w*msd + (msd*(msd-1)/2)*10d

b) Sum of digits in msd * 10d to n.  For 328, sum of digits in
300 to 328.
For 328, this sum is computed as 3*29 + recursive call "sum(28)"
In general, this sum can be computed as  msd * (n % (msd*10d) + 1)
+ sum(n % (10d))```

Below is the implementation of the above algorithm.

## C++

 `// C++ program to compute sum of digits in numbers from 1 to n``#include``using` `namespace` `std;` `// Function to computer sum of digits in numbers from 1 to n``// Comments use example of 328 to explain the code``int` `sumOfDigitsFrom1ToN(``int` `n)``{``    ``// base case: if n<10 return sum of``    ``// first n natural numbers``    ``if` `(n<10)``      ``return` `n*(n+1)/2;` `    ``// d = number of digits minus one in n. For 328, d is 2``    ``int` `d = ``log10``(n);` `    ``// computing sum of digits from 1 to 10^d-1,``    ``// d=1 a=0;``    ``// d=2 a=sum of digit from 1 to 9 = 45``    ``// d=3 a=sum of digit from 1 to 99 = a*10 + 45*10^1 = 900``    ``// d=4 a=sum of digit from 1 to 999 = a*10 + 45*10^2 = 13500``    ``int` `*a = ``new` `int``[d+1];``    ``a = 0, a = 45;``    ``for` `(``int` `i=2; i<=d; i++)``        ``a[i] = a[i-1]*10 + 45*``ceil``(``pow``(10,i-1));` `    ``// computing 10^d``    ``int` `p = ``ceil``(``pow``(10, d));` `    ``// Most significant digit (msd) of n,``    ``// For 328, msd is 3 which can be obtained using 328/100``    ``int` `msd = n/p;` `    ``// EXPLANATION FOR FIRST and SECOND TERMS IN BELOW LINE OF CODE``    ``// First two terms compute sum of digits from 1 to 299``    ``// (sum of digits in range 1-99 stored in a[d]) +``    ``// (sum of digits in range 100-199, can be calculated as 1*100 + a[d]``    ``// (sum of digits in range 200-299, can be calculated as 2*100 + a[d]``    ``//  The above sum can be written as 3*a[d] + (1+2)*100` `    ``// EXPLANATION FOR THIRD AND FOURTH TERMS IN BELOW LINE OF CODE``    ``// The last two terms compute sum of digits in number from 300 to 328``    ``// The third term adds 3*29 to sum as digit 3 occurs in all numbers``    ``//                from 300 to 328``    ``// The fourth term recursively calls for 28``    ``return` `msd*a[d] + (msd*(msd-1)/2)*p + ``           ``msd*(1+n%p) + sumOfDigitsFrom1ToN(n%p);``}` `// Driver Program``int` `main()``{``    ``int` `n = 328;``    ``cout << ``"Sum of digits in numbers from 1 to "` `<< n << ``" is "``         ``<< sumOfDigitsFrom1ToN(n);``    ``return` `0;``}`

## Java

 `// JAVA program to compute sum of digits``// in numbers from 1 to n``import` `java.io.*;``import` `java.math.*;` `class` `GFG{``    ` `    ``// Function to computer sum of digits in``    ``// numbers from 1 to n. Comments use``    ``// example of 328 to explain the code``    ``static` `int` `sumOfDigitsFrom1ToN(``int` `n)``    ``{``        ``// base case: if n<10 return sum of``        ``// first n natural numbers``        ``if` `(n < ``10``)``          ``return` `(n * (n + ``1``) / ``2``);``     ` `        ``// d = number of digits minus one in``        ``// n. For 328, d is 2``        ``int` `d = (``int``)(Math.log10(n));``     ` `        ``// computing sum of digits from 1 to 10^d-1,``        ``// d=1 a=0;``        ``// d=2 a=sum of digit from 1 to 9 = 45``        ``// d=3 a=sum of digit from 1 to 99 =``        ``// a*10 + 45*10^1 = 900``        ``// d=4 a=sum of digit from 1 to 999 =``        ``// a*10 + 45*10^2 = 13500``        ``int` `a[] = ``new` `int``[d+``1``];``        ``a[``0``] = ``0``; a[``1``] = ``45``;``        ``for` `(``int` `i = ``2``; i <= d; i++)``            ``a[i] = a[i-``1``] * ``10` `+ ``45` `*``                 ``(``int``)(Math.ceil(Math.pow(``10``, i-``1``)));``     ` `        ``// computing 10^d``        ``int` `p = (``int``)(Math.ceil(Math.pow(``10``, d)));``     ` `        ``// Most significant digit (msd) of n,``        ``// For 328, msd is 3 which can be obtained``        ``// using 328/100``        ``int` `msd = n / p;``     ` `        ``// EXPLANATION FOR FIRST and SECOND TERMS IN``        ``// BELOW LINE OF CODE``        ``// First two terms compute sum of digits from``        ``// 1 to 299``        ``// (sum of digits in range 1-99 stored in a[d]) +``        ``// (sum of digits in range 100-199, can be``        ``// calculated as 1*100 + a[d]``        ``// (sum of digits in range 200-299, can be``        ``// calculated as 2*100 + a[d]``        ``//  The above sum can be written as 3*a[d] +``        ``// (1+2)*100``     ` `        ``// EXPLANATION FOR THIRD AND FOURTH TERMS IN``        ``// BELOW LINE OF CODE``        ``// The last two terms compute sum of digits in``        ``// number from 300 to 328. The third term adds``        ``// 3*29 to sum as digit 3 occurs in all numbers``        ``// from 300 to 328. The fourth term recursively``        ``// calls for 28``        ``return` `(msd * a[d] + (msd * (msd - ``1``) / ``2``) * p + ``              ``msd * (``1` `+ n % p) + sumOfDigitsFrom1ToN(n % p));``    ``}``     ` `    ``// Driver Program``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `n = ``328``;``        ``System.out.println(``"Sum of digits in numbers "` `+``                          ``"from 1 to "` `+n + ``" is "` `+``                          ``sumOfDigitsFrom1ToN(n));``    ``}``}` `/*This code is contributed by Nikita Tiwari.*/`

## Python3

 `# PYTHON 3 program to compute sum of digits``# in numbers from 1 to n``import` `math` `# Function to computer sum of digits in``# numbers from 1 to n. Comments use example``# of 328 to explain the code``def` `sumOfDigitsFrom1ToN( n) :``  ` `    ``# base case: if n<10 return sum of``    ``# first n natural numbers``    ``if` `(n<``10``) :``        ``return` `(n``*``(n``+``1``)``/``2``)`` ` `    ``# d = number of digits minus one in n.``    ``# For 328, d is 2``    ``d ``=` `(``int``)(math.log10(n))`` ` `    ``"""computing sum of digits from 1 to 10^d-1,``    ``d=1 a=0;``    ``d=2 a=sum of digit from 1 to 9 = 45``    ``d=3 a=sum of digit from 1 to 99 = a*10``    ``+ 45*10^1 = 900``    ``d=4 a=sum of digit from 1 to 999 = a*10``    ``+ 45*10^2 = 13500"""``    ``a ``=` `[``0``] ``*` `(d ``+` `1``)``    ``a[``0``] ``=` `0``    ``a[``1``] ``=` `45``    ``for` `i ``in` `range``(``2``, d``+``1``) :``        ``a[i] ``=` `a[i``-``1``] ``*` `10` `+` `45` `*` `(``int``)(math.ceil(math.``pow``(``10``,i``-``1``)))`` ` `    ``# computing 10^d``    ``p ``=` `(``int``)(math.ceil(math.``pow``(``10``, d)))`` ` `    ``# Most significant digit (msd) of n,``    ``# For 328, msd is 3 which can be obtained``    ``# using 328/100``    ``msd ``=` `n``/``/``p`` ` `    ``"""EXPLANATION FOR FIRST and SECOND TERMS IN``    ``BELOW LINE OF CODE``    ``First two terms compute sum of digits from 1 to 299``    ``(sum of digits in range 1-99 stored in a[d]) +``    ``(sum of digits in range 100-199, can be calculated``    ``as 1*100 + a[d]. (sum of digits in range 200-299,``    ``can be calculated as 2*100 + a[d]``    ``The above sum can be written as 3*a[d] + (1+2)*100`` ` `    ``EXPLANATION FOR THIRD AND FOURTH TERMS IN BELOW``    ``LINE OF CODE``    ``The last two terms compute sum of digits in number``    ``from 300 to 328. The third term adds 3*29 to sum``    ``as digit 3 occurs in all numbers from 300 to 328.``    ``The fourth term recursively calls for 28"""``    ``return` `(``int``)(msd ``*` `a[d] ``+` `(msd``*``(msd``-``1``) ``/``/` `2``) ``*` `p ``+` `           ``msd ``*` `(``1` `+` `n ``%` `p) ``+` `sumOfDigitsFrom1ToN(n ``%` `p))` `# Driver Program``n ``=` `328``print``(``"Sum of digits in numbers from 1 to"``,``      ``n ,``"is"``,sumOfDigitsFrom1ToN(n))`  `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# program to compute sum of digits``// in numbers from 1 to n` `using` `System;` `public` `class` `GFG {``    ` `    ``// Function to computer sum of digits in``    ``// numbers from 1 to n. Comments use``    ``// example of 328 to explain the code``    ``static` `int` `sumOfDigitsFrom1ToN(``int` `n)``    ``{``        ` `        ``// base case: if n<10 return sum of``        ``// first n natural numbers``        ``if` `(n < 10)``            ``return` `(n * (n + 1) / 2);``    ` `        ``// d = number of digits minus one in``        ``// n. For 328, d is 2``        ``int` `d = (``int``)(Math.Log(n) / Math.Log(10));``    ` `        ``// computing sum of digits from 1 to 10^d-1,``        ``// d=1 a=0;``        ``// d=2 a=sum of digit from 1 to 9 = 45``        ``// d=3 a=sum of digit from 1 to 99 =``        ``// a*10 + 45*10^1 = 900``        ``// d=4 a=sum of digit from 1 to 999 =``        ``// a*10 + 45*10^2 = 13500``        ``int``[] a = ``new` `int``[d+1];``        ``a = 0; a = 45;``        ` `        ``for` `(``int` `i = 2; i <= d; i++)``            ``a[i] = a[i-1] * 10 + 45 *``                ``(``int``)(Math.Ceiling(Math.Pow(10, i-1)));``    ` `        ``// computing 10^d``        ``int` `p = (``int``)(Math.Ceiling(Math.Pow(10, d)));``    ` `        ``// Most significant digit (msd) of n,``        ``// For 328, msd is 3 which can be obtained``        ``// using 328/100``        ``int` `msd = n / p;``    ` `        ``// EXPLANATION FOR FIRST and SECOND TERMS IN``        ``// BELOW LINE OF CODE``        ``// First two terms compute sum of digits from``        ``// 1 to 299``        ``// (sum of digits in range 1-99 stored in a[d]) +``        ``// (sum of digits in range 100-199, can be``        ``// calculated as 1*100 + a[d]``        ``// (sum of digits in range 200-299, can be``        ``// calculated as 2*100 + a[d]``        ``// The above sum can be written as 3*a[d] +``        ``// (1+2)*100``    ` `        ``// EXPLANATION FOR THIRD AND FOURTH TERMS IN``        ``// BELOW LINE OF CODE``        ``// The last two terms compute sum of digits in``        ``// number from 300 to 328. The third term adds``        ``// 3*29 to sum as digit 3 occurs in all numbers``        ``// from 300 to 328. The fourth term recursively``        ``// calls for 28``        ``return` `(msd * a[d] + (msd * (msd - 1) / 2) * p +``            ``msd * (1 + n % p) + sumOfDigitsFrom1ToN(n % p));``    ``}``    ` `    ``// Driver Program``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 328;``        ``Console.WriteLine(``"Sum of digits in numbers "` `+``                             ``"from 1 to "` `+n + ``" is "` `+``                               ``sumOfDigitsFrom1ToN(n));``    ``}``}` `// This code is contributed by shiv_bhakt.`

## PHP

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## Javascript

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Output:

`Sum of digits in numbers from 1 to 328 is 3241`

The efficient algorithm has one more advantage that we need to compute the array ‘a[]’ only once even when we are given multiple inputs.

Improvement:
The above implementation takes O(d2) time as each recursive call calculates dp[] array once again. First call takes O(d), second call takes O(d-1), third call O(d-2) and so on. We don’t need to recalculate dp[] array in each recursive call. Below is the modified implementation which works in O(d) time. Where d is a number of digits in input number.

## C++

 `// C++ program to compute sum of digits``// in numbers from 1 to n``#include``using` `namespace` `std;` `int` `sumOfDigitsFrom1ToNUtil(``int` `n, ``int` `a[])``{``    ``if` `(n < 10)``        ``return` `(n * (n + 1) / 2);``    ` `    ``int` `d = (``int``)(``log10``(n));``    ``int` `p = (``int``)(``ceil``(``pow``(10, d)));``    ``int` `msd = n / p;``    ` `    ``return` `(msd * a[d] + (msd * (msd - 1) / 2) * p +``            ``msd * (1 + n % p) +``            ``sumOfDigitsFrom1ToNUtil(n % p, a));``}` `// Function to computer sum of digits in``// numbers from 1 to n``int` `sumOfDigitsFrom1ToN(``int` `n)``{``    ``int` `d = (``int``)(``log10``(n));``    ``int` `a[d + 1];``    ``a = 0; a = 45;``    ` `    ``for``(``int` `i = 2; i <= d; i++)``        ``a[i] = a[i - 1] * 10 + 45 *``               ``(``int``)(``ceil``(``pow``(10, i - 1)));` `    ``return` `sumOfDigitsFrom1ToNUtil(n, a);``}`` ` `// Driver code``int` `main()``{``    ``int` `n = 328;``    ` `    ``cout << ``"Sum of digits in numbers from 1 to "``         ``<< n << ``" is "``<< sumOfDigitsFrom1ToN(n);``}` `// This code is contributed by ajaykr00kj`

## Java

 `// JAVA program to compute sum of digits``// in numbers from 1 to n``import` `java.io.*;``import` `java.math.*;` `class` `GFG{``    ` `    ``// Function to computer sum of digits in``    ``// numbers from 1 to n``    ``static` `int` `sumOfDigitsFrom1ToN(``int` `n)``    ``{``        ``int` `d = (``int``)(Math.log10(n));``        ``int` `a[] = ``new` `int``[d+``1``];``        ``a[``0``] = ``0``; a[``1``] = ``45``;``        ``for` `(``int` `i = ``2``; i <= d; i++)``            ``a[i] = a[i-``1``] * ``10` `+ ``45` `*``                ``(``int``)(Math.ceil(Math.pow(``10``, i-``1``)));``    ` `        ``return` `sumOfDigitsFrom1ToNUtil(n, a);``    ``}``   ` `    ``static` `int` `sumOfDigitsFrom1ToNUtil(``int` `n, ``int` `a[])``    ``{``        ``if` `(n < ``10``)``            ``return` `(n * (n + ``1``) / ``2``);``      ` `        ``int` `d = (``int``)(Math.log10(n));``        ``int` `p = (``int``)(Math.ceil(Math.pow(``10``, d)));``        ``int` `msd = n / p;``        ``return` `(msd * a[d] + (msd * (msd - ``1``) / ``2``) * p +``            ``msd * (``1` `+ n % p) + sumOfDigitsFrom1ToNUtil(n % p, a));``    ``}``    ` `    ``// Driver Program``    ``public` `static` `void` `main(String args[])``    ``{``        ``int` `n = ``328``;``        ``System.out.println(``"Sum of digits in numbers "` `+``                        ``"from 1 to "` `+n + ``" is "` `+``                        ``sumOfDigitsFrom1ToN(n));``    ``}``}` `/*This code is contributed by Narendra Jha.*/`

## Python3

 `# Python program to compute sum of digits``# in numbers from 1 to n``import` `math` `# Function to computer sum of digits in``# numbers from 1 to n``def` `sumOfDigitsFrom1ToN(n):``    ` `    ``d ``=` `int``(math.log(n, ``10``))``    ``a ``=` `[``0``]``*``(d ``+` `1``)``    ``a[``0``] ``=` `0``    ``a[``1``] ``=` `45``    ``for` `i ``in` `range``(``2``, d ``+` `1``):``        ``a[i] ``=` `a[i ``-` `1``] ``*` `10` `+` `45` `*` `\``                ``int``(math.ceil(``pow``(``10``, i ``-` `1``)))``        ` `    ``return` `sumOfDigitsFrom1ToNUtil(n, a)` `def` `sumOfDigitsFrom1ToNUtil(n, a):``    ``if` `(n < ``10``):``        ``return` `(n ``*` `(n ``+` `1``)) ``/``/` `2``    ` `    ``d ``=` `int``(math.log(n,``10``))``    ``p ``=` `int``(math.ceil(``pow``(``10``, d)))``    ``msd ``=` `n ``/``/` `p``    ``return` `(msd ``*` `a[d] ``+` `(msd ``*` `(msd ``-` `1``) ``/``/` `2``) ``*` `p ``+``    ``msd ``*` `(``1` `+` `n ``%` `p) ``+` `sumOfDigitsFrom1ToNUtil(n ``%` `p, a))` `# Driver code``n ``=` `328``print``(``"Sum of digits in numbers from 1 to"``,n,``"is"``,sumOfDigitsFrom1ToN(n))` `# This code is contributed by shubhamsingh10`

## C#

 `// C# program to compute sum of digits``// in numbers from 1 to n``using` `System;` `class` `GFG``{``    ` `    ``// Function to computer sum of digits in``    ``// numbers from 1 to n``    ``static` `int` `sumOfDigitsFrom1ToN(``int` `n)``    ``{``        ``int` `d = (``int``)(Math.Log10(n));``        ``int` `[]a = ``new` `int``[d+1];``        ``a = 0; a = 45;``        ``for` `(``int` `i = 2; i <= d; i++)``            ``a[i] = a[i-1] * 10 + 45 *``                ``(``int``)(Math.Ceiling(Math.Pow(10, i-1)));``    ` `        ``return` `sumOfDigitsFrom1ToNUtil(n, a);``    ``}``    ` `    ``static` `int` `sumOfDigitsFrom1ToNUtil(``int` `n, ``int` `[]a)``    ``{``        ``if` `(n < 10)``            ``return` `(n * (n + 1) / 2);``        ` `        ``int` `d = (``int``)(Math.Log10(n));``        ``int` `p = (``int``)(Math.Ceiling(Math.Pow(10, d)));``        ``int` `msd = n / p;``        ``return` `(msd * a[d] + (msd * (msd - 1) / 2) * p +``            ``msd * (1 + n % p) + sumOfDigitsFrom1ToNUtil(n % p, a));``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main(String []args)``    ``{``        ``int` `n = 328;``        ``Console.WriteLine(``"Sum of digits in numbers "` `+``                        ``"from 1 to "` `+n + ``" is "` `+``                        ``sumOfDigitsFrom1ToN(n));``    ``}``}` `// This code contributed by Rajput-Ji`

Output :

`Sum of digits in numbers from 1 to 328 is 3241`