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Krisha Rao
- Newbie | Next Rank: 10 Posts
- Posts: 1
Perhaps the following reflects the intent of the problem above:
If z=1, then x+5y = 40, with the result that x=40-5y.What is the total number of solutions for the equation x + 5y + 10z = 50 such that x, y and z are positive integers?
Since x must be positive, y can be any integer between 1 and 7, inclusive.
Total number of options = 7.
If z=2, then x+5y = 30, with the result that x=30-5y.
Since x must be positive, y can be any integer between 1 and 5, inclusive.
Total number of options = 5.
If z=3, then x+5y = 20, with the result that x=20-5y.
Since x must be positive, y can be any integer between 1 and 3, inclusive.
Total number of options = 3.
If z=4, then x+5y = 10, with the result that x=10-5y.
Since x must be positive, y = 1.
Total number of options = 1.
Total number of solutions = 7+5+3+1 = 16.
