Suppose that one bowl costs more than two plates, three plates cost more than four candlesticks, and three candlesticks cost more than one bowl.
If it costs precicely $100 to purchase a plate, bowl, and candlestick, how much does each item cost?
This seems like a Diophantine problem so I will assume the prices are integer amounts.
Let B be the price of a bowl, P be the price of a plate, and C be the price of a candlestick. Then we are given B>2P, 3P>4C, 3C>B, and B+P+C=100
First I played with the inequalities to get (3/8)B>C, P>(4/9)B, (1/2)B>P, C>(1/3)B and the compound inequality 9B>18P>24C>8B.
Combining B+P+C=100 with (3/8)B>C and (1/2)B>P we get B+(3/8)B+(1/2)B>100, or B>160/3=53.333=
Similarly, combining B+P+C=100 with C>(1/3)B and P>(4/9)B we get B+(1/3)B+(4/9)B<100, or B<225/4=56.25
With B>53.333 and B<56.25 the only integers B can be are 54 and 55.
Now bring in the compound inequality 9B>18P>24C>8B. Any solution must obey this inequality.
If B=54 then 486>18P>24C>432, The multiples of 18 strictly between 486 and 432 are 450 and 468.
The multiple of 24 strictly between 486 and 432 are 456 and 480.
We need the multiple of 18 to be greater than the multiple of 24, so that leaves 18P=468>456=24C. Which makes P=26 and C=19.
Checking B+P+C=54+26+19=99, so not a solution
If B=55 then 495>18P>24C>440, The multiples of 18 strictly between 495 and 440 are 450, 468 and 486.
The multiple of 24 strictly between 495 and 440 are 456 and 480.
We need the multiple of 18 to be greater than the multiple of 24, so again 18P=468>456=24C. Which makes P=26 and C=19.
Checking B+P+C=55+26+19=100 is a solution.
Also 18P=486>456=24C, which makes P=27 and C=19.
Checking B+P+C=55+27+19=101 is not a solution.
Finally, 18P=486>480=24C, which makes P=27 and C=20.
Checking B+P+C=55+27+20=102 is not a solution.
So after all the cases are checked the only integer solution is a bowl costs 55, a plate costs 26, and a candlestick costs 19.
For completeness 55>2*26=52, 78=3*26>4*19=76, 57=3*19>55 and 55+26+19=100.
Solution
