Money for odd jobs is shared equally between x workers. Day 1 yielded $ 23.3, day 2 $ 31.65, day 3 $ 37.24. There were some cents over after each day. Day 2 gave a remainder twice as much as day 1, day 3 gave 3 times as much as the first day. How much did each person receive on day 2 ?
The soln is $ 1.37. I don't believe we have enough info to solve the problem! Amount divided by number of people = what each worker receives. Perhaps I am missing a trick ? Thanks.
So that we can work with integers, make all monies in pennies. Now let r be the remainder on day 1. Then the remainder on day 2 is 2r and on day 3 it is 3r. This gives rise to the following formulas
(1) n1 x + r = 2330
(2) n2 x + 2r = 3165
(3) n3 x +3r = 3724
As a hint in solving the problem we have the following: If you play with the formulas you can get a formula such as
y x = z
where y and z are integers. This means that x must be a divisor of z. As an example, suppose we had
a x = 51
The divisors of 51 are 1, 3, and 17. So x must be either 1, 3, 17, or 51. Now suppose we had another formula which said
b x = 34
The divisors of 34 are 1, 2, and 17. So x must be either 1, 2, 17, or 34. The only numbers in both lists are 1 and 17. Well if x were 1 there would be no remainders other than zero and I think we can eliminate that since the problem implies r > 0. Thus x is 17.
One further hint: What is twice equation 1 minus equation 2?
EDIT: BTW: Have you heard of the Chinese Remainder Theorem?
