Desperate: maths Help please :)?

Question

Year 9 Maths Teacher Mr Stan Deliver has his class practicing mental arithmetic. He gets each student to choose a set of consecutive numbers, none of which ends 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 a student earns a house point.

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

I need help with this questions and how to do it:

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

Answer 1

Note that the sum of the numbers 1 through 6 is 21. Therefore, by choosing any set of six consecutive numbers, the sum will be 6n + 21, where the first number chosen is n + 1.

So, what single-digits numbers can 6n + 21 be evenly divisible by? 1, of course. This number is odd, so it cannot be divisible by 2, 4, 6, or 8. The number is definitely divisible by 3. It could be divisible by 5 on the face of it, because while 6n cannot end in a 9, it could end in a 4, giving 6n + 21 a final digit of 5. However, this is actually impossible. To get 6n ending with 4, n must end in 4 or 9. If n ends in 4, the first number chosen ends in 5 and the sixth ends in zero, which is not acceptable. If n ends in 9, the first number chosen ends in zero, which again is not acceptable. I will claim that it is possible for 6n + 21 to be divisible by 7 or 9, and I offer no proof. So 6n + 1 is definitely divisible by 1 and 3 and may be divisible by 7 or 9, but definitely no others.

Now, what are the possibilities for the six final digits that must be divided by? We know that 0 is not among them. So they can be 1 through 6, 2 through 7, 3 through 8, or 4 through 9. Note that these ranges include either 1 and 3, 3 and 7, or 7 and 9, but never more than two of the possible divisors. So Caroline cannot earn more than two house points.

Answer 2

Let's suppose Caroline starts with the number a.

Then her six consecutive numbers are a, a+1, a+2, a+3, a+4, and a+5 and none of them can end with a zero.

These six numbers add up to 6a+15

6a+15=3(2a+5) showing that the sum is a multiple of 3 and must also be an odd number since 2a+5 is odd.

With a sequence of 6 consecutive numbers, none of which end in zero, we can be sure of the following:

There are exactly 3 even and 3 odd numbers in the sequence
The number, a, cannot end in 0 or 5 so it is not a multiple of 5.
Thus the sum, 6a+15, cannot be a multiple of 5.
There is exactly 1 number in the sequence that ends with a 5.

Since the sum is odd, none of the 3 even numbers can divide it.

Since one of the 3 odd numbers in the sequence must end in 5 and 6a+15 is not a multiple of 5, there is at least one odd number that cannot divide the sum.

That leaves only 2 numbers out of 6 at most that can divide the sum and earn a house point.

Answer 3

Let a be her smallest number, then she chose a,a+1,a+2,a+3,a+4,a+5.
So the sum is 6a +15. That number is uneven, no matter what a is, so its not divisible by 2,4,6,8.

Furthermore a can't have 5 (and 0) as the first number (otherwise the last one a+5 would end with 0). So its not divisible by 5.
So 6a +15 is not divisible by 5 (else a would be divisible by 5, because (6a +15) - (5a+15) would be (as a sum of two numbers that are divisible by 5) divisible by 5.

So, the sum is not divisible by 2,4,5,6,8.

Lets look at the last number of a:
a could end by 1 => we get the end numbers 1,2,3,4,5,6
=> only 1 and 3 can possibly divide the sum => max. 2 house points.

a could end by 2 => we get the end numbers 2,3,4,5,6,7
=> only 3 and 7 can possibly divide the sum => max. 2 house points.

a could end by 3 => we get the end numbers 3,4,5,6,7,8
=> only 3 and 7 can possibly divide the sum => max. 2 house points.

a could end by 4 => we get the end numbers 4,5,6,7,8,9
=> only 7 and 9 can possibly divide the sum => max. 2 house points.

a couldn't end by any larger number, because else one of the numbers would end with 0.

So we are done already.

Answer 4

The handiest factor you must take into account is whilst you transfer anything to the opposite facet of the equals signal, you are making it bad (or constructive whether it is already bad) so 3x + 5x + 7 = -nine 3x + 5x = -nine -7 8x = -sixteen x = -two

Answer 5

and why is it you don't ask the teacher? or a fellow classmate?

just look at things this way,, school will be out soon,, and you'll have all summer to practice maths

Answer 6

want some points try these 121,122,123,124,125,126 all 6 of those number will give you a point.