Math 131

Spring 2000

This is due Friday, February 25 at the beginning of class. It is worth 10 points. Papers turned in before class on Monday lose 3 points; papers will not be accepted after Monday. Write neatly or use a word-processor. Your grade will depend on the correctness of your work and how well you explain it. You may discuss your work with me but not with others in the class.

In exercise 30 on page 139, it is stated that a positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9. Thus, 18675 is divisible by 9 since 1+8+6+7+5=27 is divisible by 9. Similarly, 93626 is not divisible by since 9+3+6+2+6=26 is not divisible by 9.

The result is true for any number of digits. To simplify the proof, we will use only 3 digits.

Theorem. A 3-digit positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9.

Prove this Theorem. Note that there are actually two statements to prove: (1) if a 3-digit number is divisible by 9, then the sum of its digits is divisible by 9 and (2) if the sum of the digits of a 3-digit number is divisible by 9, then the number is divisible by 9.

Hints: The 3-digit number 462 can be written as 4· 100 + 6· 10 + 2. You should use a similar representation for the generic 3-digit number n = d₂d₁d₀ . It may help to read over exercise 40.

Although I am not asking you to do the general proof (any number of digits), the proof of the 3-digit case should convince you that it is true. In research mathematics, it is often convenient (and sometimes necessary) to try out a proof on a special case, and then hope that the full proof can be obtained by generalizing the special case.