The total amount for any odd number of quarters ends in a 5 in the cents column. Since $6.50 ends in a zero in the cents column and our young stalwart has no nickels, she must have an even number of quarters.
The statement "...$6.50 in dimes and quarters." implies that the number both the number of dimes and the number of quarters must be greater than zero.
Since 26 quarters is exactly $6.50, 24 quarters is the most she could have, and that would have to be paired with 5 dimes. (24 X 25 = 600 and 5 X 10 = 50)
Since 65 dimes is exactly $6.50 and the fewest quarters possible is 2, the most dimes possible is 60 paired with the 2 quarters. (60 X 10 = 600 and 2 X 25 = 50)
So
24Q + 5D
22Q + 10D
20Q + 15D
and so on until you get to
6Q + 50D
4Q + 55D
2Q + 60D
Total number of combinations, either 24 divided by 2 or 60 divided by 5. If it turns out that the "both denominations greater than zero" assumption is incorrect, then add two to your total count.
John
My calculator said it, I believe it, that settles it
