The total for all coins counted is $4,564.50 The last coin added to the pile is a 50 cent piece There are 8 times as many 50 cent pieces as there are quarters There are 6 times as many dimes as nickles How many of each are there?

I have been helping him with this for 5 hours and have no answer yet. Can you help?


I think it is easier to work in cents rather than dollars so think of the value of the coins as 456,450 cents. Suppose that the number of quarters is Q then the value of these Q quarters is 25Q cents. The number of fifty cent pieces is then 8Q (eight times the number of quarters) so the the value of the fifty cent pieces is 50(8Q) = 400Q cents. Hence the total value of the quarters and fifty cent pieces is

25Q + 400Q = 425Q cents.

Suppose also that the number of nickles is N, then the value of these nickles is 5N cents. The number of dimes is then 6N so the value of the dimes is 10(6N) = 60N cents. Hence the total value of the nickles and dimes is

5N + 60N = 65N cents. The total value of all the coins is 456,450 cents and hence

425Q + 65N = 456,450

The first fact I notice about this equation is that all three coefficients are divisible by 5. Thus divide both sides by 5 to get

85Q + 13N = 91,290

At this point I tried some arithmetic and found that 85 divides 9,1290. In fact 9,1290 = 85 x 1074. Since N and Q are integers and 85 divides both 85Q and 9,1290 this implies that 85 divides 13N. Hence 85 divides N. Hence

N = 85 or N = 2 x 85 or N = 3 x 85 or N = 4 x 85 .....

If you substitute N = 85 in the above equation and solve for Q you will find that Q = 1061. This then gives a solution to the problem. (You should check that this is actually a solution.)

This is not the only solution however. Try N = 2 x 85 and N = 3 x 85, ... to see if you can find other solutions.

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