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this is a tough extra credit problem... and i still don't get how to do it.
Find all integers between 100 and 1000 such that each digit is a different odd number and the sum of the cubes of the digits is the original three-digit number.
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You know that your numbers will have 3 digits, that all 3 digits must be either 1,3,5,7, or 9, and that all 3 digits must be unique.
My suggestion would be to first find the cubes of all five of the possible digits. They are 1^3 = 1, 3^3 = 27, 5^3 = 125, 7^3 = 343, and 9^3 = 729. Now just try systematically adding cubes together to see if you get a number with all 3 of the cubes you used to form it.
For example, I would start with 9. Since 729 + 343 > 1000 you know that no numbers with both 9 and 7 will work. However, 729 + 125 = 854, and you can plainly see that adding neither 1 nor 27 will give you a number that fits the parameters, so you can rule out all 3 digit numbers with a 9.
Then just keep working your way down until you've exhausted all possible numbers. There are surprisingly few that you even need to consider.
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sorry, still dont really get it. im only a 5th grader, but i understand some of it
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Ok, we'll start from the beginning then. Do you know what it means to cube a number?
If you don't, it means to multiply that number by itself 3 times. For example, 2 cubed is 2 * 2 * 2 = 8, and 3 cubed is 3 * 3 * 3 = 27.
The way I would solve the problem would first be to find the cube of every odd number. The cubes of the odd numbers are 1^3 = 1, 3^3 = 27, 5^3 = 125, 7^3 = 343, and 9^3 = 729.
After that, consider what other rules the number has to follow. It says that it has 3 different digits, so you don't have to worry about numbers like 111 or 355, since those have repeated digits.
The last step is just to try out different numbers and see if they work. When you consider that the digits can't be repeated you actually have very few numbers to test.
Wrap it in bacon
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