The 3 Merchants

Number Theory Level 4

The 3 merchants named $A, B, C$ each possessed whole number of less than 100 gold coins, and this was how they traded simultaneously:

Merchant $A$ spent $\dfrac{3}{8}$ of his gold coins to buy goods from $B$ and later paid extra tax of 7 gold coins to the city.

Merchant $B$ spent $\dfrac{2}{7}$ of his gold coins to buy goods from $C$ and later paid extra tax of 5 gold coins to the city.

Merchant $C$ spent $\dfrac{1}{6}$ of his gold coins to buy goods from $A$ and later paid extra tax of 3 gold coins to the city.

In the end, if all three yielded equal balance of money, how many gold coins would each have?

1 solution

Mr X
Feb 7, 2017

Let $8a, 7b, 6c$ what the 3 merchants have at the beginning

At the end they will have the same sum s

$5a+c-7=s$ ..... (1)

$5b+3a-5=s$ ..... (2)

$5c+2b-3=s$ ..... (3)

Taking the differences between (1)-(2), (2)-(3) and (1)-(3)

$2a-5b=2-c$ ..... (4)

$3a+3b=2+5c$ ..... (5)

$5a=4c+2b+4$ ..... (6)

From (6) we conclude that a is pair $a=2a'$ Solving (4) and (5) give

$21a'=11c+8$ ..... (7)

$21b=13c-2$ ..... (8)

(8)-(7) gives

$21(b-a')=2 (c-5)$ ..... (9)

The only solution of the last equation is when $c=5, a'=b=3$ So the answer is 28