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The data of the problem lead to the following system of two equations:

x + 10y + 50z = 500

x + y + z = 100

We subtract the second equation from the first equation:

9y + 49z = 400

For z = 1, y = 39, x = 60

For z = 2 the system has no solution

For z = 3 the system has no solution

For z = 4 the system has no solution

For z = 5 the system has no solution

For z = 6 the system has no solution

For z = 7 the system has no solution

For z = 8 the system has no solution

For z = 9 and higher, y cannot be a positive integer, so we can ignore these.

So the only solution would be 1 half dollar, 39 dimes and 60 pennies.

That's where I end up after getting 770 Math and 650 Verbal on the SAT and 800 on the Math II achievement test (previous names of the SAT tests). *Sob...*

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Q: If you have to make exactly five dollars using exactly 100 coins and the only coins you can use are half dollars dimes and pennies how many of each coin do you have?
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