Math homework help- only yr7 but got us foxed!

anonymousie

So far DS DH (and DH and I have A level maths!)and and I have been totally foxed by this one, which is just one part of quite a few bits of maths H/w that DS has (including filling out "magic squares" and stuff), so the answer shouldn't take too long to find out....

How many ways can the number 105 be made from the sum of two or more consecutive positive integers.

We have an answer just by trial and error of adding lists of numbers (we think it is 7) but for the life of me I don't know why it is 7 or what it has got to do with any sort of mathmatical rules/patterns etc.

Any clever brains out there can help?
THanks
Anon

tankgirl1

How many ways can the number 105 be made from the sum of two or more consecutive positive integers.

i'm ashamed to say i have no idea what that means

(gcse grade B top set maths :rotfl:- i really must have a brain like a sieve!)

stacey21_2

This page might help:

http://www.qbyte.org/puzzles/p092s.html

silvercar

looks like a trial and error question:

52+53

34+35+36

19+20+21+22+23

15+16+17+18+19+20

12+13+14+15+16+17+18

6+7+8+9+10+11+12+13+14+15

1+2+3+4+5+6+7+8+9+10+11+12+13+14

dmg24

Consecutive positive integers are whole numbers that follow on from each other.

From stacey's link:

the number of ways in which n may be expressed as a sum of one or more consecutive positive integers is equal to the number of positive odd divisors of n.

1, 3, 5, 7, 15, 21, 35

So it looks like 7 is correct! x

PParka

Can't believe this is for year 7 homework....but here goes:

The sum of x consecutive integers equal x times their average value.
If the number of integers is odd then the average is balanced by the numbers on either side.
If the number if integers is even then the average will end in 0.5
By in turn dividing 105 by 2, then 3, then 4, then 5 etc. you can see which result in a whole number or end in 0.5. These are the combinations that will add up to 105.

So the answers are:
1 to 14
6 to 15
12 to 18
15 to 20
19 to 23
34 to 36
52 to 53

Total combinations: 7

Clear as mud

randomtask_2

Hats off to PParka - I somehow have an A Level in maths (which went in one ear and quickly left the other) so it while I might have cracked it with the "mathematical proof" eventually, I can sleep easy with your solution.

I agree though, mighty tricky homework for a Year 7.

tomstickland

If you did A level maths then you could say
sequence of numbers is
x
x+1
x+2
x+3
and so on
so the total for n numbers is
nx + n(n-1)/2
where the second part is the sum of an Arithmetic progression.
This has to equal the total, which I'll call y
so
y = nx + n(n-1)/2

So
x = (y - n(n-1)/2) / n
which is
x = y/n - (n-1)/2
So you need to choose a value of n that divides into y by an integer amount and is also an odd number. Well, that works, but it's not definitive, since there might be an n that divides in by an integer plus a half times and is even.
Hmmm.
7 works anyway.

maninthestreet

tankgirl1 wrote: »

i'm ashamed to say i have no idea what that means

(gcse grade B top set maths :rotfl:- i really must have a brain like a sieve!)

Integer = whole number, i.e 1,2,3 etc

Positive= greater than 0

Consecutive = one number immediately follows the other, i.e 1 and 2, 2 and 3, 3 and 4, etc.

For 2 consective postive integers, the problem can specified as the two simultaneous equations:

A + B = 105

and:

B - A = 1

The solution is relatively simple for two integers, it's A=52 and B=53.

For 3 consective integers, the problem can be specified as the three simultaneous equations:

A + B + C = 105

and;

B - A = 1

and:

C - B = 1

These equations have the solution A=34, B=35, C=36.


For 4 consecutive integers, A,B,C,D, the equations are;

A + B + C + D = 105

and:

B = A + 1

and:

C = A + 2

and:

D = A + 3

These equations have no solutions for consecutive postive integers.

For 5 consecutive integers, A,B,C,D,E the equations are;

A + B + C + D + E = 105

and:

B = A + 1

and:

C = A + 2

and:

D = A + 3

and:

E = A + 4

These equations have solutions of 19,20,21,22,23.


Given that the sum of the integer sequence of 1,2,3....(n-1),n is given by the equation n(n+1)/2, we have for any n = k integers :

kA + (k-1)k/2 = 105

=> 2kA + k(k-1) = 210

=> 2kA + k**2 - K = 210

so for k=6, 12A +36 - 6 = 210

=> 12A + 30 = 210

=> 12A = 180

=> A = 15

So 15,16,17,18,19,20 is a solution.


For k=7, 12,13,14,15,16,17,18 is a solution.


For k=8, and k=9, there are no solutions that involve all consecutive postive integers.


For k=10, 6,7,8,9,10,11,12,13,14,15 is a solution.

For k = 11, there is no solution.

For k =12, there is no solution.

For k = 13, there is no solution.

For K =14, 1,2,3,4,5,6,7,8.9,10,11,12,13,14 is a solution.

For k = 15 or more there are no solutions (k**2 - k = 210)

So the answer to the original question is 7.

krazykidskate

Can anyone think of a practical application for this?
That's what I'd be asking the teacher.

tomstickland

I use mathematics at work all the time and find that things I learnt at school allow me to solve all sorts of problems.
The most satisfying is using it to solve gambling loophole offers where I can make profit from it.