Question

*Puzzle Time :*

A farmer has 49 cows which are numbered from 1 to 49. Cow # 1 gives 1 litre of milk per day, cow # 2 gives 2 litres of milk per day, and like this cow # 49 gives 49 litres of milk per day. The farmer has 7 sons.

Help him divide the cows among his 7 sons in such a way that each son also gets same quantity of milk given by the cows.

Clue: there are at least 2 solutions.​

Answer 1

The total amount of milk the farmer does get is,

$\longrightarrow\sf{1+2+\,\dots\,+49=1225\ L}$

And since it is equally distributed to 7 sons, amount of milk each son does get is,

$\longrightarrow\sf{\dfrac{1225}{7}=175\ L}$

Solution:-

It can be about a 7 × 7 magic square arranged with numbers from 1 to 49 as follows:

$\begin{tabular}{|c|c|c|c|c|c|c|}\cline{1-7}22&21&13&5&46&38&30\\\cline{1-7}31&23&15&14&6&47&39\\\cline{1-7}40&32&24&16&8&7&48\\\cline{1-7}49&41&33&25&17&9&1\\\cline{1-7}2&43&42&34&26&18&10\\\cline{1-7}11&3&44&36&35&27&19\\\cline{1-7}20&12&4&45&37&29&28\\\cline{1-7}\end{tabular}$

Thus we obtain two solutions for the problem:

1. Adding each row is one solution. The farmer can give the cows, numbered in each row, to each son.
2. Adding each column is another solution. The farmer can give the cows, numbered in each column, to each son.

Method of obtaining the magic square:-

• Let the space at $\sf{i^{th}}$ row and $\sf{j^{th}}$ column be $\sf{a_{ij}.}$ So here is our $\sf{7\times7}$ square.

$\begin{tabular}{|c|c|c|c|c|c|c|}\cline{1-7}&&&&&&\\\cline{1-7}&&&&&&\\\cline{1-7}&&&&&&\\\cline{1-7}&&&&&&\\\cline{1-7}&&&&&&\\\cline{1-7}&&&&&&\\\cline{1-7}&&&&&&\\\cline{1-7}\end{tabular}$

• Start inserting 1 in the center space of the final column, i.e., $\sf{a_{47}.}$

$\begin{tabular}{|c|c|c|c|c|c|c|}\cline{1-7}&&&&&&\\\cline{1-7}&&&&&&\\\cline{1-7}&&&&&&\\\cline{1-7}&&&&&&1\\\cline{1-7}&&&&&&\\\cline{1-7}&&&&&&\\\cline{1-7}&&&&&&\\\cline{1-7}\end{tabular}$

• Insert successive numbers in the order $\sf{a_{(i+1)(j+1)},}$ where $\sf{a_{ij}}$ represents the location of the predecessor.

• If $\sf{i+1=8}$ or $\sf{j+1=8}$ inclusively, then replace the 8, by 1.

→   So that 2 is placed in $\sf{a_{(4+1)(7+1)}=a_{58}\equiv a_{51},}$ 3 is placed in $\sf{a_{(5+1)(1+1)}=a_{62},}$ and so on.

$\begin{tabular}{|c|c|c|c|c|c|c|}\cline{1-7}&&&5&&&\\\cline{1-7}&&&&6&&\\\cline{1-7}&&&&&7&\\\cline{1-7}&&&&&&1\\\cline{1-7}2&&&&&&\\\cline{1-7}&3&&&&&\\\cline{1-7}&&4&&&&\\\cline{1-7}\end{tabular}$

• But if the successor has to be placed in $\sf{a_{(i+1)(j+1)}}$ which is already in use, then the successor will have to be placed in $\sf{a_{i(j-1)},}$ i.e., left to $\sf{a_{ij}.}$

→   7 is located in $\sf{a_{36}}$ so 8 has to be located in $\sf{a_{47}.}$ But since it is already used by 1, thus 8 is located in $\sf{a_{35}.}$

$\begin{tabular}{|c|c|c|c|c|c|c|}\cline{1-7}&&&5&&&\\\cline{1-7}&&&&6&&\\\cline{1-7}&&&&8&7&\\\cline{1-7}&&&&&&1\\\cline{1-7}2&&&&&&\\\cline{1-7}&3&&&&&\\\cline{1-7}&&4&&&&\\\cline{1-7}\end{tabular}$

By changing the direction of inserting the numbers, the magic square can vary, can be flipped, but not the solutions!

Answer 2

A farmer has 49 cows which are numbered from 1 to 49. Cow # 1 gives 1 litre of milk per day, cow # 2 gives 2 litres of milk per day, and like this cow # 49 gives 49 litres of milk per day. The farmer has 7 sons.

Help him divide the cows among his 7 sons in such a way that each son also gets same quantity of milk given by the cows.

Step-by-step explanation:

A farmer has 49 cows which are numbered from 1 to 49. Cow # 1 gives 1 litre of milk per day, cow # 2 gives 2 litres of milk per day, and like this cow # 49 gives 49 litres of milk per day. The farmer has 7 sons.

Help him divide the cows among his 7 sons in such a way that each son also gets same quantity of milk given by the cows.

Answer

Each row is for each son.