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Algebra Difficulty 6.2 National olympiad Prove it Belarus

Some businessmen decide to establish a firm and divide all profits in equal parts. However, some day after a good pennyworth, a head of the firm transfers a part of the funds from firm's account on his individual account. This part is three times as many as the part of each of the others if they divided the rest of the funds in equal parts. After that he leaves the firm. Next head of the firm gets the right to command of all remaining funds, and he deals with the funds just as the previous one, and so on. Finally, next to the last head of the firm transfers a part of the remaining funds from firm's account on his individual account and this part is also three times as many as he leaves to the last of the co-funder of the firm. As a result of this revenue sharing the profit of the last businessman is 210 times as little as the profit of the first head of the firm.
How many businessmen have established this firm?

Solution

Answer: 20.
Let nn be the number of co-funders and did_i be the value of the ii-th director, i=1,,ni = 1, \ldots, n. By condition,
di=3di+1+di+2++dnni. d_i = 3 \cdot \frac{d_{i+1} + d_{i+2} + \dots + d_n}{n-i}.
So,
di1=3di+di+1++dnni+1==33di+1+di+2++dnni+di+1++dnni+1==3(ni+3)(di+1+di+2++dn)(ni)(ni+1). \begin{align*} d_{i-1} &= 3 \cdot \frac{d_i + d_{i+1} + \dots + d_n}{n-i+1} = \\ &= 3 \cdot \frac{3 \cdot \frac{d_{i+1} + d_{i+2} + \dots + d_n}{n-i} + d_{i+1} + \dots + d_n}{n-i+1} = \\ &= 3 \cdot \frac{(n-i+3)(d_{i+1} + d_{i+2} + \dots + d_n)}{(n-i)(n-i+1)}. \end{align*}
Therefore, di1di=ni+3ni+1\frac{d_{i-1}}{d_i} = \frac{n-i+3}{n-i+1}, i=2,,ni = 2, \ldots, n.
Multiplying these equalities, we obtain
d1dn=d1d2d2d3dn1dn=n+1n1nn24231=(n+1)n2. \frac{d_1}{d_n} = \frac{d_1}{d_2} \cdot \frac{d_2}{d_3} \cdot \dots \cdot \frac{d_{n-1}}{d_n} = \frac{n+1}{n-1} \cdot \frac{n}{n-2} \cdot \dots \cdot \frac{4}{2} \cdot \frac{3}{1} = \frac{(n+1)n}{2}.
By condition, d1/dn=210d_1/d_n = 210, so (n+1)n=420(n+1)n = 420 which gives n=20n = 20.

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