Logic Puzzle of Diamonds and sons

Question

I came across a math problem and I need a solution for this. An old man has 49 diamonds. Each one has a different worth as $1, $2, $3, ….. $49. He has 7 sons and he wanted to give them equal number of diamonds of equal worth. How the diamonds will be Divided?

Answer 1

So let's do this in general. Say we have $n^2$ diamonds with worths $1, 2, \ldots, n^2$, and we want to divvy them up amongst $n$ sons, equal amounts, etc.

Our total worth is $T = \frac{n^2(n^2+1)}{2}$, so each son gets $S = \frac{n(n^2+1)}{2} = \frac{n^3+n}{2}$.

Now let's look at two matrices, $M$ and $N$.

Let $M$ be such that $M_{ij} = (j - i) \pmod n$ and $N$ be such that $N_{ij} = n(j+1)$. In the case that $n = 4$, then we have $M = \left( \begin{array}{cccc} 0 & 1 & 2 & 3 \\ 3 & 0 & 1 & 2 \\ 2 & 3 & 0 & 1 \\ 1 & 2 & 3 & 0 \\ \end{array} \right)$ and $N = \left( \begin{array}{cccc} 4 & 8 & 12 & 16 \\ 4 & 8 & 12 & 16 \\ 4 & 8 & 12 & 16 \\ 4 & 8 & 12 & 16 \\ \end{array} \right)$.

Now let's look at the sum of each row. The sum of each row of $N$ is $n+2n+3n+\ldots+n^2 = n\cdot\frac{n(n+1)}{2} = \frac{n^3+n^2}{2}$. The sum of each row of $M$ is $0 + 1 + 2 + \ldots + n - 1 = \frac{n(n-1)}{2} = \frac{n^2-n}{2}$. So let's consider the matrix $N - M$. The sum of each row will then be $\frac{n^3+n^2}{2} - \frac{n^2-n}{2} =\frac{n^3+n}{2} = S$, so if we calculate $N - M$, each row will represent the worth of the diamonds to give to each son.

Addendum: I guess I technically didn't answer the question. Letting $n = 7$, we have:

$$N = \left( \begin{array}{ccccccc} 7 & 14 & 21 & 28 & 35 & 42 & 49 \\ 7 & 14 & 21 & 28 & 35 & 42 & 49 \\ 7 & 14 & 21 & 28 & 35 & 42 & 49 \\ 7 & 14 & 21 & 28 & 35 & 42 & 49 \\ 7 & 14 & 21 & 28 & 35 & 42 & 49 \\ 7 & 14 & 21 & 28 & 35 & 42 & 49 \\ 7 & 14 & 21 & 28 & 35 & 42 & 49 \\ \end{array} \right)$$ and $$M = \left( \begin{array}{ccccccc} 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 0 & 1 & 2 & 3 & 4 & 5 \\ 5 & 6 & 0 & 1 & 2 & 3 & 4 \\ 4 & 5 & 6 & 0 & 1 & 2 & 3 \\ 3 & 4 & 5 & 6 & 0 & 1 & 2 \\ 2 & 3 & 4 & 5 & 6 & 0 & 1 \\ 1 & 2 & 3 & 4 & 5 & 6 & 0 \\ \end{array} \right)$$

so

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

And note that $$(N-M)\cdot\left( \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ \end{array} \right) = \left( \begin{array}{c} 175 \\ 175 \\ 175 \\ 175 \\ 175 \\ 175 \\ 175 \\ \end{array} \right)$$

as desired.

Answer 2

Hint: If you split the diamonds up like this, you get close to the answer.

1 14 15 28 29 42 43 (172)

2 13 16 27 30 41 44 (173)

3 12 17 26 31 40 45 (174)

4 11 18 25 32 39 46 (175)

5 10 19 24 33 38 47 (176)

6 9 20 23 34 37 48 (177)

7 8 21 22 35 36 49 (178)