Happy New year wishes to all my readers

May this new year bring many opportunities your way, To explore every joy of life & turning all your dreams into reality & all your efforts into great achievements.

Code for string library functions

                   String Copy

char *strcpy( char            *s1, const char   *s2)

{

      char        *s = s1;

      while ((*s1++ = *s2++));

      return s;

}

String Length

int strlen(const char *s)

{

      int   l = 0;

      while (*s++) l++;

      return l;

}

String concat

char *strcat(char *s1, char *s2)

{

      char  *s = s1;

      while (*s1) s1++;

      while (*s2 ) {

            *s1++ = *s2++;

            n--;

      }

      *s1='\0';

      return s;

}

String compare

int strcmp(const char   *s1,const char    *s2)

{

      while ((*s1 || *s2) && n--)

            if (*s1++ != *s2++)

                  return (*--s1 - *--s2);

      return 0;

}

To lower

char tolower(char c)

{

      if (isupper(c))

        return c + ('a' - 'A');

        else

        return c;

}

To upper

char toupper(char c)

{

  if (islower(c))

      return c - ('a' - 'A');

  else

      return c;

}

Proofs of Divisibility Tests

Here is a basic fact: Suppose you have a positive integer x which, when you write its digits, looks like:

am · · · a4a3a2a1a0.

So a0  is the digit in the one’s place,a1  is the digit in the 10’s place, a2  is the digit in the 100’s place, etc. Then the number x equals

x  =  a0 + a1 · 10 + a2 · 100 + a3 · 1000 + a4 · 10000 + · · · + am · 10m

=  a0 + a1 · 10 + a2 · 102 + a3 · 103 + a4 · 104 + · · · + am · 10m.

Most of the divisibility tests and theirproofs come from examiningthis expression mod

n for some n, and replacingthe various powers of 10 with their equivalents mod n:

1.    Proof of Test for Divisibility by 2. Observe that 10 divided by 2 has a remainderof 0.

So 10 ≡ 0 (mod 2).Then 10k ≡ 0k ≡ 0 (mod 2) for k = 1, 2, 3, . . .. Hence

x  ≡  a0 + a1 · 0 + a2 · 0 + a3 · 0 + a4 · 0 + · · · + am · 0

≡  a0 (mod 2).

Therefore x is divisible by 2 if and only if its last digit a0 is divisibleby 2, which happens if and only if the last digit is one of 0, 2, 4, 6, 8.

2.    Proof of Test for Divisibility by 4. Observethat 100 divided by 4 has a remainder of 0. So 100 ≡ 0 (mod 4). Hence 10k ≡ 0 (mod 4) for k = 2, 3, 4, . . .. Then

x  ≡  a0 + a1 · 10 + a2 · 0 + a3 · 0 + a4 · 0 + · · · + am · 0

≡  a0 + a1 · 10 (mod 4).

Therefore x isdivisible by 4 if and only if the number a0 + a1 · 10is divisible 4. But a0 + a1 · 10 is the number formed by keeping only the last two digits of x. So x is divisible by 4 if and only if the number formed by droppingall but the last two digits of x is divisible by 4.

3.    Proof of Test for Divisibility by 9. Observe that 10 divided by 9 has a remainderof 1.

So 10 ≡ 1 (mod 9). Hence 10k ≡ 1k ≡ 1 (mod 9) for k = 1, 2, 3, 4, . . .. Then

x  ≡  a0 + a1 · 1 + a2 · 1 + a3 · 1 + a4 · 1 + · · · + am · 1

≡  a0 + a1 + a2 + a3 + · · · + am (mod 9).

Therefore x is divisible by 9 if and only if the sum of its digits is divisible by 9.

4.    Proof of Test for Divisibility by 10. Observe that 10 dividedby 10 has a remainder of 0. So 10 ≡ 0 (mod 10). Hence 10k ≡ 0k ≡ 0 (mod 10) for k = 1, 2, 3, . . .. Then

x  ≡  a0 + a1 · 0 + a2 · 0 + a3 · 0 + a4 · 0 + · · · + am · 0

≡  a0 (mod 10).

Therefore x is divisibleby 10 if and only if its last digit a0  is divisibleby 10, which happens if and only if the last digit is 0.

5.    Proof of Test for Divisibility by 11.Observe that 10 ≡ −1 (mod 11). Hence 10k ≡

(−1)k (mod 11) for k = 1, 2, 3, 4, . . .. Then

x  ≡  a0 + a1 · (−1) + a2 · (−1)2 + a3 · (−1)3 + a4 · (−1)4 + · · · + am · (−1)m

≡  a0 − a1 + a2 − a3 + a4 + · · · + am(−1)m (mod 11).

Therefore x is divisible by 11 if andonly if alternating sum of itsdigits a0 − a1 + a2 −

a3 + a4 + · · · + am(−1)m  is divisible by 11.

6.    Proof of Test for Divisibility by 12. Since 12 = 223 involves more than one prime,let’s start a different way. If a number is divisibleby 12, then in its prime factorization it must contain 223, possibly alongwith other primefactors (perhaps even some more 2’s and 3’s). Thereforethe number must be divisible by both 3 and 4. Conversely, if a number is divisible by both 3 and 4, then in its prime factorization it must contain 223, possibly along with other primefactors (perhaps even some more 2’s and 3’s). Therefore, a number is divisible by 12 if and only if it is divisibleby both 3 and 4, and this is our divisibility test.