House Points

[kaytae]

Can anyone help with this problem? I really don't understand and the teacher's on sick leave, so I really need help.

Year 10 Maths teacher Mr Stan Deliver has his class practising mental arithmetic. He gets each student to choose a set of consectutive numbers, none of which end in zero. The student is to add these numbers and then divide the total by the last digit of each number. For each whole number answer the student earns one house point.

a) Belinda chooses five consectutive numbers and wins five house points. What is the smallest total she could have had? Explain.

By the way, I would prefer it if you could include working out and explain how you got your answers because I'd prefer to understand the problem rather than just have the answer. Thankyou

[Logophobic]

I know this question has been posted before, but I cannot seem to find it. I believe the numbers were 22-26, and the explanation was clear. Try the search function, I know it's here somewhere.

[Logophobic]

Original thread

-b
We are looking for 5 consectuvie numbers, none of those numbers shall be a multiple of 10.
The possible sequences are 1 to 5, 2 to 6, 3 to 7, 4 to 8 and 5 to 9 and all the ones with added multiples of 10 (multiplicator*10)!
There sums will be 15, 20, 25, 30 or 35 or plus 50*multiplicator.
Since no odd sum could be divided evenly by a even ending digit, the sequences with 1 to 5, 3 to 7 and 5 to 9 can be neglected.
So we are looking for the least common multiple of the ending digits.
For 2 to 6 it would be 60, but 60 is no possible sum, the next one is 120. 120 is the sum of 22,23,24,25,26.
For 4 to 8 the least common multiple is 840, that is also no possible sum, so the first one in here is 1680.
So the solution is 22,23,24,25,26!

[kaytae]

thankyou!