How to sort numbers from 1 to 20 into 6 piles where sum in each pile is the same?
This question, my son got in school and I can't figure out what is the correct approach to solve this.
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this communityHow to sort numbers from 1 to 20 into 6 piles where sum in each pile is the same?
This question, my son got in school and I can't figure out what is the correct approach to solve this.
If $n$ is the sum of one pile, then we must have
$$1+2+3+\ldots+19+20=6n\;.$$
The sum on the left is $\dfrac{20\cdot21}2=210$, so we must have $6n=210$ and $n=35$. Now you just have to break the integers from $1$ through $20$ into groups that total $35$ each. You can do this in any way you like. For instance, we could start by letting one group be $\{15,20\}$. Another could then be $\{16,19\}$, and $\{17,18\}$ could be a third. That leaves the integers from $1$ through $14$ to be divided into three groups. $14+13=27$, and $35-27=8$, so you could let the fourth group be $\{8,13,14\}$, leaving $\{1,2,3,4,5,6,7,9,10,11,12\}$ to be split into two groups.
But it really doesn’t matter how you do it: just keep forming groups that sum to $35$, and when you’ve formed five of them, whatever numbers remain must sum to $35$ to give you your sixth group.
This question was given to 9 year olds at a school whom I'm instructed have never done algebra or set theory Neither have they learn t anything about bodmas
I don't quite know how they expect them to solve this problem but I was asked by a work college to assist him with this question because he dint know how to solve this ( just shows you the standards in the UK now ) terrible that we have adults who cannot do this
No to worry maths problem explained in full for you all
How to derive the sum of consecutive numbers the key is consecutive in this case only
Integers are whole numbers without fractional or decimal components.
If a math problem requires you to sum a certain number of integers from 1 to a given value N, it's not necessary to add each and every value by hand. Instead, use the equation (N(N + 1))/2, where N is the highest number in your series
Th example given is 1 to 20
From the above our highest number N is in this case == 20
We plug 20 into our equation (N(N + 1))/2
Substituting for N we have 20 * (20+1) / 2 From this we perform what is inside the brackets first This is where you apply BODMAS
This stands for brackets , of , division , multiplication , addition and subtraction
With any maths like this we have to perform a series of calculations in this order
This is how we perform these equations hence we have 20 * ((20+1)=21) = 20*21
Now we must divide this by 2
So 20 * 21 = 420 / 2 = 210
But we also need 6 piles of 210 from previous question
This gives us 210 = 6X could call the variable N Doesn't matter
So we divide 210 / 6 and 6x / 6
6X or 6N / 6 leaves with X or N
Thus X or N = 210/6 giving you 35 N or x which ever you prefer = 35
Our universal set of numbers comprises {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 }
I might as well show you these at the same time here
This can be represented in what we term an Array having 20 elements that is 1 to 20
Our first element is or element 0 is 1 ( Set theory applicable to arrays in IT)
Our last element or element 19 is thus 20
Starting with the highest numbers
From our sum of numbers / 6 we subtract
35 -20 leaves you with 15
Our first set is hence {20,15} = one pile
We subtract this set from our universal set {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20 }
This leaves us with a resultant set of {1,2,3,4,5,6,7,8,9,10,11,12,13,14,16,17,18,19}
Note we have extracted {20,15}
Next we perform the same again taking the next highest number available which is 19
35 - 19 answer = 16 so next set is {19,16} Our next complete pile {19,16 }
So far two pile {19, 16} and {20,15}
Extract these from what we have left Our resultant set
Well we have already take 20 and 15 from this set we need to take 19 ,16 from this
That leaves us with {1,2,3,4,5,6,7,8,9,10,11,12,13,14,17,18} which becomes new resultant set
Next we have two sets as answers we now need for more piles
Take the next highest number which is 18
So 35 -18 = 17 Next answer {18,17} Our Next Pile
Answer so far {19,16} ; {20,15} ; {18,17} = 3 piles sorted
Extract these from what we have left
Our resultant set was = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,17,18}
New resultant set is now = {1,2,3,4,5,6,7,8,9,10,11,12,13,14}
Do the same again
35 - 14 = 21 So thats 14 so far but searching our resultant array don't have a 21 do we ??
So take next number off the result of this sum which is 13
So our sum now looks like this
35 -14- 13 gives us 8
Still not zero yet Take the 8 off
35-14-13 -8 = 0
Collect the elements you have subtracted
Next answer will be {14,13,8}
Answer so far are {19,16} ; {20,15} ; {18,17} ; {14,13,8}
Extract these from what we have left resultant set {1,2,3,4,5,6,7,8,9,10,11,12,13,14} now becomes
Resultant set or array now = {1,2,3,4,5,6,7,9,10,11,12}
Do the same again
35 -12 = 23
Searching our resultant array don't have a 23 do we ??
Take the next highest number off and continue until your answer is zero in each case
Hence we have 35 -12 -11 -10 -2 = 0 but we have use 11 haven't we
Collect the elements you have subtracted
Answer Fifth set is {12,11,10,2}
Answer so far is
{19,16} ; {20,15} ; {18,17} ; {14,13,8} ; {12,11,10,2}
Extract this from last know resultant set {1,2,3,4,5,6,7,9,10,11,12}
New result is thus {1,3,4,5,6,7,9}
Whats left ?? Answer = {1,3,4,5,6,7,9}
Add them together you will see result is 35
Hence here are your sets
{19,16} ; {20,15} ; {18,17} ; {14,13,8} ; {12,11,10,2}; {1,3,4,5,6,7,9}
Set1 = {19,16} ;
Set2 = {20,15} ;
Set3 = {18,17} ;
Set4 = {14,13,8} ;
Set5 = {12,11,10,2} ;
Set6 = {1,3,4,5,6,7,9}
Problem solved
A question put to 9 year olds who apparently have never been taught any of this at all I don't know you tell me
Thats explains it all for you
Mark Harrington
The sum of the numbers from $1$ to $20$ is
$\left(\begin{array}{c} 20 + 1 \\ + \\ 19 + 2 \\ + \\ 18 + 3 \\ + \\ 17 + 4 \\ + \\ 16 + 5 \\ + \\ 15 + 6 \\ + \\ 14 + 7 \\ + \\ 13 + 8 \\ + \\ 12 + 9 \\ + \\ 11 + 10 \\ \end{array}\right) = \left(\begin{array}{c} 21 \\ + \\ 21 \\ + \\ 21 \\ + \\ 21 \\ + \\ 21 \\ + \\ 21 \\ + \\ 21 \\ + \\ 21 \\ + \\ 21 \\ + \\ 21 \\ \end{array}\right) =10\times 21 = 210$
So each pile must sum to $210 \div 6 = 35$
One thing that might make this problem look more difficult than it is, is that you might think that there is only one solution. There are lots of solutions. So don't worry too much about what effect the numbers you choose now will affect the choices you will have to make later.
So let's start by listing all of the pairs of numbers that add up to $35$
That's all of the pairs of numbers that add up to $35$
That leaves $1,2,3,4,5,6,7,8,9,10,11,12,13,14$ which need to be split up into three groups that add up to $35$.
So let's start by finding three of the remaining numbers that add up to $35$.
That leaves $1,2,3,4,5,6,7,9,10,11,12$ which need to be split up into two more groups that add up to $35$.
So let's look for three more of the remaining numbers that add up to $35$. Darn, there aren't any. So let's look for four of the remaining numbers that add up to $35$.
will add up to $35$
So we now have five groups of numbers that add up to $35$ and we have these numbers left
Guess what. The remaining seven number HAVE TO add up to $35$. So we are done. the complete list is