Question

How to divide ${{1}^{2}},{{2}^{2}},....,{{64}^{2}}$ set into four subsets with equal sum and 16 numbers in each?

Answer 1

We have to divide the squares of the above into four subsets with equal sums and 16 numbers in each. We start to solve the problem by taking the first four numbers and adding the squares to them in the given set to get the required result.

Complete step-by-step solution:
We are given a set of squares from 1 to 64 and need to find four subsets with equal sums. We will start to solve the question by starting with 4 squares and adding squares to them in the given set to get the required result.
We consider the first four squares from the given set and add the next four squares to them by keeping the sum as even as possible. We get,
$\Rightarrow {{1}^{2}}+{{8}^{2}}=65$
$\Rightarrow {{2}^{2}}+{{7}^{2}}=53$
$\Rightarrow {{3}^{2}}+{{6}^{2}}=45$
$\Rightarrow {{4}^{2}}+{{5}^{2}}=41$
In the above, we have considered the first eight terms in the set.
We need to repeat the same with the next eight terms in the given set.
Applying the same, we get,
$\Rightarrow {{9}^{2}}+{{16}^{2}}=337$
$\Rightarrow {{10}^{2}}+{{15}^{2}}=325$
$\Rightarrow {{11}^{2}}+{{14}^{2}}=317$
$\Rightarrow {{12}^{2}}+{{13}^{2}}=313$
In the above, we have considered the eighth term to the sixteenth term in the given set.
From the above, we can notice that the differences between the sums are 4,7, or 12.
We combine the two sets of the sums above to free out the differences in each group of two sums. Following the same, we get,
$\Rightarrow {{1}^{2}}+{{8}^{2}}+{{10}^{2}}+{{15}^{2}}=390$
$\Rightarrow {{2}^{2}}+{{7}^{2}}+{{9}^{2}}+{{16}^{2}}=390$
$\Rightarrow {{3}^{2}}+{{6}^{2}}+{{12}^{2}}+{{13}^{2}}=358$
$\Rightarrow {{4}^{2}}+{{5}^{2}}+{{11}^{2}}+{{14}^{2}}=358$
We need to repeat the same procedure for the next sixteen squares. We get,
$\Rightarrow {{17}^{2}}+{{24}^{2}}+{{26}^{2}}+{{31}^{2}}=2502$
$\Rightarrow {{18}^{2}}+{{23}^{2}}+{{25}^{2}}+{{32}^{2}}=2502$
$\Rightarrow {{19}^{2}}+{{22}^{2}}+{{28}^{2}}+{{29}^{2}}=2470$
$\Rightarrow {{20}^{2}}+{{21}^{2}}+{{27}^{2}}+{{30}^{2}}=2470$
Now, we need to add the above two sets in reverse order to free out the differences in each group of two sums.
On adding, we get,
$\Rightarrow {{1}^{2}}+{{8}^{2}}+{{10}^{2}}+{{15}^{2}}+{{20}^{2}}+{{21}^{2}}+{{27}^{2}}+{{30}^{2}}=2860.....\left( 1 \right)$
$\Rightarrow {{2}^{2}}+{{7}^{2}}+{{9}^{2}}+{{16}^{2}}+{{19}^{2}}+{{22}^{2}}+{{28}^{2}}+{{29}^{2}}=2860.....\left( 2 \right)$
$\Rightarrow {{3}^{2}}+{{6}^{2}}+{{12}^{2}}+{{13}^{2}}+{{18}^{2}}+{{23}^{2}}+{{25}^{2}}+{{32}^{2}}=2860.....\left( 3 \right)$
$\Rightarrow {{4}^{2}}+{{5}^{2}}+{{11}^{2}}+{{14}^{2}}+{{17}^{2}}+{{24}^{2}}+{{26}^{2}}+{{31}^{2}}=2860.....\left( 4 \right)$
Adding the equations $\left( 1 \right)$ and $\left( 2 \right)$ , we get,
$\Rightarrow {{1}^{2}}+{{8}^{2}}+{{10}^{2}}+{{15}^{2}}+{{20}^{2}}+{{21}^{2}}+{{27}^{2}}+{{30}^{2}}+{{2}^{2}}+{{7}^{2}}+{{9}^{2}}+{{16}^{2}}+{{19}^{2}}+{{22}^{2}}+{{28}^{2}}+{{29}^{2}}=2860+2860$
$\Rightarrow {{1}^{2}}+{{8}^{2}}+{{10}^{2}}+{{15}^{2}}+{{20}^{2}}+{{21}^{2}}+{{27}^{2}}+{{30}^{2}}+{{2}^{2}}+{{7}^{2}}+{{9}^{2}}+{{16}^{2}}+{{19}^{2}}+{{22}^{2}}+{{28}^{2}}+{{29}^{2}}=5720$
Adding the equations $\left( 1 \right)$ and $\left( 3 \right)$ , we get,
$\Rightarrow {{1}^{2}}+{{8}^{2}}+{{10}^{2}}+{{15}^{2}}+{{20}^{2}}+{{21}^{2}}+{{27}^{2}}+{{30}^{2}}+{{3}^{2}}+{{6}^{2}}+{{12}^{2}}+{{13}^{2}}+{{18}^{2}}+{{23}^{2}}+{{25}^{2}}+{{32}^{2}}=2860+2860$
$\Rightarrow {{1}^{2}}+{{8}^{2}}+{{10}^{2}}+{{15}^{2}}+{{20}^{2}}+{{21}^{2}}+{{27}^{2}}+{{30}^{2}}+{{3}^{2}}+{{6}^{2}}+{{12}^{2}}+{{13}^{2}}+{{18}^{2}}+{{23}^{2}}+{{25}^{2}}+{{32}^{2}}=5720$
Adding the equations $\left( 1 \right)$ and $\left( 4 \right)$ , we get,
$\Rightarrow {{1}^{2}}+{{8}^{2}}+{{10}^{2}}+{{15}^{2}}+{{20}^{2}}+{{21}^{2}}+{{27}^{2}}+{{30}^{2}}+{{4}^{2}}+{{5}^{2}}+{{11}^{2}}+{{14}^{2}}+{{17}^{2}}+{{24}^{2}}+{{26}^{2}}+{{31}^{2}}=2860+2860$
$\Rightarrow {{1}^{2}}+{{8}^{2}}+{{10}^{2}}+{{15}^{2}}+{{20}^{2}}+{{21}^{2}}+{{27}^{2}}+{{30}^{2}}+{{4}^{2}}+{{5}^{2}}+{{11}^{2}}+{{14}^{2}}+{{17}^{2}}+{{24}^{2}}+{{26}^{2}}+{{31}^{2}}=5720$
Adding the equations $\left( 2 \right)$ and $\left( 3 \right)$ , we get,
$\Rightarrow {{2}^{2}}+{{7}^{2}}+{{9}^{2}}+{{16}^{2}}+{{19}^{2}}+{{22}^{2}}+{{28}^{2}}+{{29}^{2}}+{{3}^{2}}+{{6}^{2}}+{{12}^{2}}+{{13}^{2}}+{{18}^{2}}+{{23}^{2}}+{{25}^{2}}+{{32}^{2}}=2860+2860$
$\Rightarrow {{2}^{2}}+{{7}^{2}}+{{9}^{2}}+{{16}^{2}}+{{19}^{2}}+{{22}^{2}}+{{28}^{2}}+{{29}^{2}}+{{3}^{2}}+{{6}^{2}}+{{12}^{2}}+{{13}^{2}}+{{18}^{2}}+{{23}^{2}}+{{25}^{2}}+{{32}^{2}}=5720$
$\therefore$ The four subsets with equal sum and 16 numbers in each are:
\left\\{ {{1}^{2}},{{2}^{2}},{{7}^{2}},{{8}^{2}},{{9}^{2}},{{10}^{2}},{{15}^{2}},{{16}^{2}},{{19}^{2}},{{20}^{2}},{{21}^{2}},{{22}^{2}},{{27}^{2}},{{28}^{2}},{{29}^{2}},{{30}^{2}} \right\\} , \left\\{ {{1}^{2}},{{3}^{2}},{{6}^{2}},{{8}^{2}},{{10}^{2}},{{12}^{2}},{{13}^{2}},{{15}^{2}},{{18}^{2}},{{20}^{2}},{{21}^{2}},{{23}^{2}},{{25}^{2}},{{27}^{2}},{{30}^{2}},{{32}^{2}} \right\\} ,
\left\\{ {{1}^{2}},{{4}^{2}},{{5}^{2}},{{8}^{2}},{{10}^{2}},{{11}^{2}},{{14}^{2}},{{15}^{2}},{{17}^{2}},{{20}^{2}},{{21}^{2}},{{24}^{2}},{{26}^{2}},{{27}^{2}},{{30}^{2}},{{31}^{2}} \right\\} , and
\left\\{ {{2}^{2}},{{3}^{2}},{{6}^{2}},{{7}^{2}},{{9}^{2}},{{12}^{2}},{{13}^{2}},{{16}^{2}},{{18}^{2}},{{19}^{2}},{{22}^{2}},{{23}^{2}},{{25}^{2}},{{28}^{2}},{{29}^{2}},{{32}^{2}} \right\\}

Note: We should be careful while performing addition between the squares to get precise results. We must know which set of squares are to be added to free out the differences in each group of two sums.