# Smith Number

Given a number n, the task is to find out whether this number is smith or not. A Smith Number is a composite number whose sum of digits is equal to the sum of digits in its prime factorization.
Examples:

```Input  : n = 4
Output : Yes
Prime factorization = 2, 2  and 2 + 2 = 4
Therefore, 4 is a smith number

Input  : n = 6
Output : No
Prime factorization = 2, 3  and 2 + 3 is
not 6. Therefore, 6 is a smith number

Input   : n = 666
Output  : Yes
Prime factorization = 2, 3, 3, 37 and
2 + 3 + 3 + (3 + 7) = 6 + 6 + 6 = 18
Therefore, 666 is a smith number

Input   : n = 13
Output  : No
Prime factorization = 13 and 13 = 13,
But 13 is not a smith number as it is not
a composite number
```

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

The idea is first find all prime numbers below a limit using Sieve of Sundaram (This is especially useful when we want to check multiple numbers for Smith). Now for every input to be checked for Smith, we go through all prime factors of it and find sum of digits of every prime factor. We also find sum of digits in given number. Finally we compare two sums. If they are same, we return true.

## C++

```// C++ program to to check whether a number is
// Smith Number or not.
#include<bits/stdc++.h>
using namespace std;
const int MAX  = 10000;

// array to store all prime less than and equal to 10^6
vector <int> primes;

// utility function for sieve of sundaram
void sieveSundaram()
{
// In general Sieve of Sundaram, produces primes smaller
// than (2*x + 2) for a number given number x. Since
// we want primes smaller than MAX, we reduce MAX to half
// This array is used to separate numbers of the form
// i+j+2ij from others where 1 <= i <= j
bool marked[MAX/2 + 100] = {0};

// Main logic of Sundaram. Mark all numbers which
// do not generate prime number by doing 2*i+1
for (int i=1; i<=(sqrt(MAX)-1)/2; i++)
for (int j=(i*(i+1))<<1; j<=MAX/2; j=j+2*i+1)
marked[j] = true;

// Since 2 is a prime number
primes.push_back(2);

// Print other primes. Remaining primes are of the
// form 2*i + 1 such that marked[i] is false.
for (int i=1; i<=MAX/2; i++)
if (marked[i] == false)
primes.push_back(2*i + 1);
}

// Returns true if n is a Smith number, else false.
bool isSmith(int n)
{
int original_no = n;

// Find sum the digits of prime factors of n
int pDigitSum = 0;
for (int i = 0; primes[i] <= n/2; i++)
{
while (n % primes[i] == 0)
{
// If primes[i] is a prime factor,
// add its digits to pDigitSum.
int p = primes[i];
n = n/p;
while (p > 0)
{
pDigitSum += (p % 10);
p = p/10;
}
}
}

// If n!=1 then one prime factor still to be
// summed up;
if (n != 1 && n != original_no)
{
while (n > 0)
{
pDigitSum = pDigitSum + n%10;
n = n/10;
}
}

// All prime factors digits summed up
// Now sum the original number digits
int sumDigits = 0;
while (original_no > 0)
{
sumDigits = sumDigits + original_no % 10;
original_no = original_no/10;
}

// If sum of digits in prime factors and sum
// of digits in original number are same, then
// return true. Else return false.
return (pDigitSum == sumDigits);
}

// Driver code
int main()
{
// Finding all prime numbers before limit. These
// numbers are used to find prime factors.
sieveSundaram();

cout << "Printing first few Smith Numbers"
" using isSmith()n";
for (int i=1; i<500; i++)
if (isSmith(i))
cout << i << " ";
return 0;
}
```

## Java

```// Java program to to check whether a number is
// Smith Number or not.

import java.util.Vector;

class Test
{

static int MAX  = 10000;

// array to store all prime less than and equal to 10^6
static Vector <Integer>  primes = new Vector<>();

// utility function for sieve of sundaram
static void sieveSundaram()
{
// In general Sieve of Sundaram, produces primes smaller
// than (2*x + 2) for a number given number x. Since
// we want primes smaller than MAX, we reduce MAX to half
// This array is used to separate numbers of the form
// i+j+2ij from others where 1 <= i <= j
boolean marked[] = new boolean[MAX/2 + 100];

// Main logic of Sundaram. Mark all numbers which
// do not generate prime number by doing 2*i+1
for (int i=1; i<=(Math.sqrt(MAX)-1)/2; i++)
for (int j=(i*(i+1))<<1; j<=MAX/2; j=j+2*i+1)
marked[j] = true;

// Since 2 is a prime number

// Print other primes. Remaining primes are of the
// form 2*i + 1 such that marked[i] is false.
for (int i=1; i<=MAX/2; i++)
if (marked[i] == false)
}

// Returns true if n is a Smith number, else false.
static boolean isSmith(int n)
{
int original_no = n;

// Find sum the digits of prime factors of n
int pDigitSum = 0;
for (int i = 0; primes.get(i) <= n/2; i++)
{
while (n % primes.get(i) == 0)
{
// If primes[i] is a prime factor,
// add its digits to pDigitSum.
int p = primes.get(i);
n = n/p;
while (p > 0)
{
pDigitSum += (p % 10);
p = p/10;
}
}
}

// If n!=1 then one prime factor still to be
// summed up;
if (n != 1 && n != original_no)
{
while (n > 0)
{
pDigitSum = pDigitSum + n%10;
n = n/10;
}
}

// All prime factors digits summed up
// Now sum the original number digits
int sumDigits = 0;
while (original_no > 0)
{
sumDigits = sumDigits + original_no % 10;
original_no = original_no/10;
}

// If sum of digits in prime factors and sum
// of digits in original number are same, then
// return true. Else return false.
return (pDigitSum == sumDigits);
}

// Driver method
public static void main(String[] args)
{
// Finding all prime numbers before limit. These
// numbers are used to find prime factors.
sieveSundaram();

System.out.println("Printing first few Smith Numbers" +
" using isSmith()");

for (int i=1; i<500; i++)
if (isSmith(i))
System.out.print(i + " ");
}
}
```

## Python

```# Python program to to check whether a number is
# Smith Number or not.

import math

MAX  = 10000

# array to store all prime less than and equal to 10^6
primes = []

# utility function for sieve of sundaram
def sieveSundaram ():
#In general Sieve of Sundaram, produces primes smaller
# than (2*x + 2) for a number given number x. Since
# we want primes smaller than MAX, we reduce MAX to half
# This array is used to separate numbers of the form
# i+j+2ij from others where 1 <= i <= j
marked  = [0] * ((MAX/2)+100)
# Main logic of Sundaram. Mark all numbers which
# do not generate prime number by doing 2*i+1
i = 1
while i <= ((math.sqrt (MAX)-1)/2) :
j = (i* (i+1)) << 1
while j <= MAX/2 :
marked[j] = 1
j = j+ 2 * i + 1
i = i + 1
# Since 2 is a prime number
primes.append (2)

# Print other primes. Remaining primes are of the
# form 2*i + 1 such that marked[i] is false.
i=1
while i <= MAX /2 :
if marked[i] == 0 :
primes.append( 2* i + 1)
i=i+1

#Returns true if n is a Smith number, else false.
def isSmith( n) :
original_no = n

#Find sum the digits of prime factors of n
pDigitSum = 0;
i=0
while (primes[i] <= n/2 ) :

while n % primes[i] == 0 :
#If primes[i] is a prime factor ,
# add its digits to pDigitSum.
p = primes[i]
n = n/p
while p > 0 :
pDigitSum += (p % 10)
p = p/10
i=i+1
# If n!=1 then one prime factor still to be
# summed up
if not n == 1 and not n == original_no :
while n > 0 :
pDigitSum = pDigitSum + n%10
n=n/10

# All prime factors digits summed up
# Now sum the original number digits
sumDigits = 0
while original_no > 0 :
sumDigits = sumDigits + original_no % 10
original_no = original_no/10

#If sum of digits in prime factors and sum
# of digits in original number are same, then
# return true. Else return false.
return pDigitSum == sumDigits
#-----end of function isSmith------

#Driver method
# Finding all prime numbers before limit. These
# numbers are used to find prime factors.
sieveSundaram();
print "Printing first few Smith Numbers using isSmith()"
i = 1
while i<500 :
if isSmith(i) :
print i,
i=i+1

#This code is contributed by Nikita Tiwari
```

Output:

```Printing first few Smith Numbers using isSmith()
4 22 27 58 85 94 121 166 202 265 274 319 346 355 378 382 391 438 454 483
```

This article is contributed by Sahil Chhabra(KILLER). 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.

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