A problem from the Nine Chapters (Chapter 2, problem 38) reads as follows: Now p

Question

A problem from the Nine Chapters (Chapter 2, problem 38) reads as follows: Now pay 576 coins to purchase 78 bamboos. [They are] classified into thicker and thinner ones. Tell: how much does each of them cost? Liu Hui’s method assumes that the unit cost for each bamboo is an integer, with the thinner ones cheaper than the thick ones, and that the thick ones cost one more (per unit) than the thin ones. His method states that if we divide the number of coins by the number of bamboos, the remainder (30 in this case) is the number of thick ones, and the quotient (7 in this case) is the unit cost of the thin (cheaper) bamboos. (a) Show that this is indeed the case if we assume that the unit cost of the thick is one more than the cost of the thin ones. (b) Show that there is a second solution if we assume the thick ones cost (per unit) two more than the thin ones, and a third solution if we assume the thick ones cost 3 more than the thin ones. How many solutions are there altogether if the thin ones cost 7 coins each?

Explanation / Answer

Here we let that there are total y thicker and x thinner bamboos.

So according to given conditions, x+y= 78 ...........(i)

Now as it is said that the unit cost of thin one is 7 then clearly the unit cost of thicker one is 8.

So we have that total cost of x thicker + total cost of y thins = 576

or 7x+ 8y= 576

or 7x + 8(78-x )= 576 ( from equation (i) y= 78-x)

7x+624 - 8x = 576

or -x = 576 - 624 = - 48

or x= 48

and thus y= 78- 48= 30

that means total thicker bamboos be 30 as she already investigated. So this is indeed the case.

This is the answer of part (a)

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Now according to second condition as the unit rate of thicker one is 2 more than of thiner one, so if x=7, then y=9

so clearly 7x + 9y = 576

7x+9(78-x)= 576

or 7x+ 702-9x = 576

or -2x = 576 - 702 = -126

x= 63

So y= 78- 63 = 15

That means there is surely a second solution there.

And when x= 7and y= 10 then

second equation changes as :

7x +10(78-x)= 576

7x+780-10x= 576

or -3x = 576 - 780 = -204

or x= 204/3 = 68

and y= 78-68=10

So there is surely third solution also there.

Now it is cleared that the value of y in decreasing to with each new solution and it takes 4 mores olutions to get y=0 almost. Thus there are total 7 solutions are possible.

This is the last answer for part (b)

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