Question 1132084: how many ways to make change for .80 cents?
Found 4 solutions by MathLover1, ikleyn, Alan3354, greenestamps:
Answer by MathLover1(20850)   (Show Source):
Answer by ikleyn(52803)  (Show Source):
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! Another useless "math" problem.
Nothing is learned from it, except it's not .80 cents.
It's .80 dollars, or 80 cents.
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In response to the comments, I doubt the student learned anything by having the problem done by others.
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I never "bad mouth" other tutors. If they submit a wrong answer, I might solve the problem with the correct answer.
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But, there's a lower limit to what's considered a worthwhile problem. I've seen "What is 2+2?" posted more than once and no one took it seriously. I think "making change" is only slightly more interesting.
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But my most serious complaint is doing the problems for the students. I've gotten TY notes saying, "I just want the answer." No interest in learning anything. I think doing it for them is doing a disservice.
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The general public has no interest in knots, it's true. The general public is not interested in mi/hr, km/hr, or in algebra, trigonometry or calculus, either, so why mention them?
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I won't deign to cavil the point further.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
This problem, and similar ones, are NOT useless. Solving them requires logical reasoning, organized thinking, and useful math skills.
Both of the responses from other tutors assumed the problem was to use only quarters, dimes, and nickels. There is no mention in the statement of the problem about not using pennies, or half dollars. But adding pennies and/or half dollars to the problem just makes the solution more tedious; the same problem solving skills are used if we just use quarters, dimes, and nickels. So I will address the same problem.
There is no need to list out all the combinations, as one tutor did successfully and the other tutor did unsuccessfully. And if the problem were the harder problem with pennies and/or half dollars, you would not want to try to write out all the solutions.
Instead, with only quarters, dimes, and nickels, you only need to look at each possible number of quarters and determine through logical analysis and simple arithmetic how many solutions there are with each number of quarters.
For example, if you use 2 quarters, that is 50 cents; you have 30 cents more to make using dimes and nickels. You can have any number of dimes from 0 to a maximum of 3; the nickels will make up anything that remains. So with 2 quarters there are 4 solutions.
The complete solution can then look like this:
0 quarters --> 80 cents left --> 0 to 8 dimes --> 9 solutions with 0 quarters
1 quarter --> 55 cents left --> 0 to 5 dimes --> 6 solutions with 1 quarter
2 quarters --> 30 cents left --> 0 to 3 dimes --> 4 solutions with 2 quarters
3 quarters --> 5 cents left --> 0 dimes --> 1 solution with 3 quarters
Total number of solutions: 9+6+4+1 = 20
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