1963 AHSME Problems/Problem 23
Contents
Problem
A gives as many cents as has and as many cents as has. Similarly, then gives and as many cents as each then has. , similarly, then gives and as many cents as each then has. If each finally has cents, with how many cents does start?
Solution
Let be number of cents originally had, be number of cents originally had, and be number of cents originally had.
After gave his money away, has cents, has cents, and has cents.
After gave his money away, has cents, has cents, and has cents.
After gave his money away, has cents, has cents, and has cents.
Since all of them have cents in the end, we can write a system of equations. Note that adding the three equation yields , so . Therefore, , so . Solving for can also be done traditionally.
Thus, started out with cents, which is answer choice .
Solution 2
We know that people not giving away money on the previous turn now have twice what they previously had. Using the fact that the sum of their money is cents, we can work backward Thus at the beginning has cents.
~ Nafer
See Also
1963 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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