**Problem
I.**

Prove that an integer is divisible by 9 if and only if the sum of its digits is divisible by 9.

Any integer n can be written in the following form:

n = d_{0} + 10 * d_{1} + 100 * d_{2}
+ … + 10^{k}d_{k}, where d_{0} is the least significant
digit and d_{k} is the most significant digit in the decimal notation
of the number n. For instance, if n is 2759, then d_{0} = 9, d_{1}
= 5, d_{2} = 7, d_{3} = 2 and k is 3.

n = d_{0} + 10 * d_{1} + 100 * d_{2}
+ … + 10^{k}d_{k} = d_{0} + (9 + 1)d_{1} + (99
+ 1)d_{2} + … + (99…9 + 1)d_{k} = (d_{0} + d_{1}
+ d_{2} + … + d_{k}) + (9d_{1} + 99d_{2} + … +
99…9d_{k})

The number in the last parenthesis is divisible by 9. Thus
number n is divisible by 9 if and only if the number (d_{0} + d_{1}
+ d_{2} + … + d_{k}) is divisible by 9, but this number is the
sum of the digits of number n.

**Problem
II.**

Show that an integer is divisible by 11 if and only if the alternate sum of its digits is divisible by 11.

As in previous problem, we can write:

n = d_{0} + 10 * d_{1} + 100 * d_{2}
+ … + 10^{k}d_{k}.

Alternate sum of the digits of number n is: d_{k} –
d_{k-1} + d_{k-2} - … + (-1)^{k}d_{0}

Let us consider the following expression:

E = (10 + 1) * (d_{0} + 10(d_{1} – d_{0})
+ 100(d_{2} – d_{1} – d_{0}) + … + 10^{k}(d_{k}
– d_{k-1} + d_{k-2} - … + (-1)^{k}d_{0})) – 10^{k+1}(d_{k}
– d_{k-1} + d_{k-2} - … + (-1)^{k}d_{0}).

After performing careful multiplication, one can see that E equals to n.

The term that has a factor of (10 + 1) or 11, is divisible
by 11. Thus n is divisible by 11 whenever 10^{k+1}(d_{k} – d_{k-1}
+ d_{k-2} - … + (-1)^{k}d_{0}) is divisible by 11. Since 10^{k+1} is not divisible by 11
and 11 is a prime, we can make a conclusion that n is divisible by 11 if and
only if d_{k} – d_{k-1} + d_{k-2} - … + (-1)^{k}d_{0}
is divisible by 11. This number is exactly an alternate sum of digits of n.

**Problem
III.**

Show that any prime number greater than 3 has a remainder of 1 or 5 when divisible by 6.

Any integer n can be represented in the following form:

n = 6 * q + r, where 0 £ r < 6

Thus, we have 6 choices for r: 0, 1, 2, 3, 4, 5.

Recalling that n is a prime and therefore is not divisible by 2 or 3, we can analyze these 6 choices:

1) r is 0, then n = 6 * q, hence n is divisible by 2, which is not possible since n is a prime

2) r is 1, this is possible

3) r is 2, then n = 6 * q + 2, hence n is divisible by 2, which is not possible since n is a prime

4) r is 3, then n = 6 * q + 3, hence n is divisible by 3, which is not possible since n is a prime

5) r is 4, then n = 6 * q + 4, hence n is divisible by 2, which is not possible since n is a prime

6) r is 5, this is possible

We can see that the only possible remainders for n divided by 6 are 1 and 5.