How to sort numbers from 1 to 20 into 6 piles where sum in each pile is the same?

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.

• First find out what the sum in each pile must be. Then start putting numbers on the piles. Start with the large numbers. – Daniel Fischer Oct 3 '13 at 21:36

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

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

{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

{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 $35$

• $(20, 15)$
• $(19, 16)$
• $(18, 17)$

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 $35$.

• $(14, 13, 8)$

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 $35$.

• $(12, 11, 10, 2)$

will add up to $35$

So we now have five groups of numbers that add up to $35$ and we have these numbers left

• $(1,3,4,5,6,7,9)$

Guess what. The remaining seven number HAVE TO add up to $35$. So we are done. the complete list is

• $(20, 15)$
• $(19, 16)$
• $(18, 17)$
• $(14, 13, 8)$
• $(12, 11, 10, 2)$
• $(1,3,4,5,6,7,9)$