1. ## House Points

A math teacher, Mr. Stan, has his class do mental arithmetic every lesson. He gets each student to choose a set of consecutive numbers, none of which end in zero. The student is to add these numbers and divide the total by the last digit of each number. For each whole number answer, the student earns one house point.

Example, Andy chooses 15, 16, 17 with a total of 48 and calculates 48/5, 48/6 and 48/7. Just one of these divisions gibes a whole number, so he wins one house point.

a) Show that if Andy chooses eight consecutive numbers he wins at least one house point.

b) Belinda chooses five consecutive numbers and wins five house points. What is the smallest total her five numbers can have. Explain.

c) Caroline chooses six consecutive numbers. Explain why she wins no more than two house points.

Again, all working would be appreciated, thanks guys.

2. Originally Posted by Bartimaeus

a) Show that if Andy chooses eight consecutive numbers he wins at least one house point.
The end digits of those 8 consecutive numbers can be from 1 to 8 or from 2 to 9.

We can write that sum of 8 consecutive numbers as:

(for 1-8)
(k10^n + 1) + (k10^n + 2) + (k10^n + 3) + (k10^n + 4) + (k10^n + 5) + (k10^n + 6) + (k10^n + 7) + (k10^n + 8) =
k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 =
k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + 36
or as (for 2-9)
k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + k10^n + 44

As I understand 8 consecutive numbers must be greater then 10.

Since in 8 consecutive numbers there is always 2 as last digit then k10^n is divisible by 2 for k>0 and n>0 and so is 36 (or 44) divisible by 2 so we get whole number.

Similar can be done for b) and c).

3. okay, thanks
are there any other ways to do it?