Number Theory Problems Discussed in Class

 

Problem I.

 

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

 

Solution

 

Any integer n can be written in the following form:

 

n = d0 + 10 * d1 + 100 * d2 + + 10kdk, where d0 is the least significant digit and dk is the most significant digit in the decimal notation of the number n. For instance, if n is 2759, then d0 = 9, d1 = 5, d2 = 7, d3 = 2 and k is 3.

 

n = d0 + 10 * d1 + 100 * d2 + + 10kdk = d0 + (9 + 1)d1 + (99 + 1)d2 + + (999 + 1)dk = (d0 + d1 + d2 + + dk) + (9d1 + 99d2 + + 999dk)

 

The number in the last parenthesis is divisible by 9. Thus number n is divisible by 9 if and only if the number (d0 + d1 + d2 + + dk) 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.

 

Solution

 

As in previous problem, we can write:

 

n = d0 + 10 * d1 + 100 * d2 + + 10kdk.

 

Alternate sum of the digits of number n is: dk dk-1 + dk-2 - + (-1)kd0

 

Let us consider the following expression:

 

E = (10 + 1) * (d0 + 10(d1 d0) + 100(d2 d1 d0) + + 10k(dk dk-1 + dk-2 - + (-1)kd0)) 10k+1(dk dk-1 + dk-2 - + (-1)kd0).

 

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 10k+1(dk dk-1 + dk-2 - + (-1)kd0) is divisible by 11. Since 10k+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 dk dk-1 + dk-2 - + (-1)kd0 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.

 

Solution

 

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.