2. How many of the integers $123,234,345,456,567$ are multiples of 3 ?
A 1
В 2
C 3
D 4
E 5
Solution: $\mathbf{E}$
We could simply divide each of the given numbers by 3 , and check that there is no remainder in each case. It is quicker to use the fact that

An integer is divisible by 3 if and only if the sum of its digits is a multiple of 3 .
Now $1+2+3=6,2+3+4=9,3+4+5=12,4+5+6=15$ and $5+6+7=18$, which are all multiples of 3. So 123, 234, 345, 456 and 567 are all multiples of 3 .

Extension Problems
The fact that an integer is divisible by 3 if and only if the sum of its digits is a multiple of 3 , follows from the fact that
The remainder when a positive integer is divided by 9 is equal to the remainder when the sum of its digits is divisible by 9 .

For example, consider the number 2011. The sum of its digits is 4 , which leaves remainder 4 when we divide by 9 . We can deduce that 2011 has remainder 4 when we divide by 9 . When the sum of the digits of a number is itself quite large we can repeat the process.

For example, consider the number $777 \ldots 777$ with 2011 s. What is the remainder when this number is divided by 9 ? The sum of its digits is $2011 \times 7=14077$. The sum of the digits of 14077 is 19 , which has remainder 1 when divided by 9 . We can deduce that the number 777... 777 has remainder 1 when divided by 9 .
2.1 What is the remainder when the following numbers are divided by 9 ?
a) 123456789 ;
b) 9876543210 ;
c) $555 \ldots 555$ with $20115 \mathrm{~s}$.
2.2 Can you prove the above fact about the remainder when a positive integer is divided by 9 ?
2.3 We didn't really need to work out all the sums of the digits of the given numbers. The digits of each of these numbers are three successive integers. So all we need do is to make use of the fact that
The sum of three successive integers is always a multiple of 3 .

Can you prove this?