#problem 5

Individual Tax Return Problem 5

Individual Tax Return Problem 5 in $110 only

Individual Tax Return Project (LO 2 & 3)

Tax return problem: Complete the individual tax return problem 5 on page Appendix C-7 in your textbook. This is an individual project instead of a group project. However, you are encouraged to discuss with your classmates. Make sure that you don’t simply copy the forms. You have to download the appropriate forms…

Find all (if any) functions \( f \) taking the non-negative reals such that  (a) \( f(y)f(xf(y)) = f(x + y) \) for all non-negative \( x, y \)

(b) \( f(2) = 0 \)

(c) \( f(x) \neq 0 \) for all \(0 \leq x < 2 \)

We know immediately \( f \)'s behaviour outside of \( [0, 2) \): let \( y = 2 \) then (a) says \( f(2 + x) = f(2) f(x f(2)) = 0 \cdot f(x \cdot 0) = 0 \).  ie. \( f \) is the zero function above 2. Next I claim \( f(x) \geq \frac{2}{2 - x} \) for all \( x \in [0, 2) \).  This time let \( y \leq 2 \) and \( x = 2 - y \) in (a): \( f( y)f((2-y)f(y)) = f(2) = 0 \).  But \( f(y) \neq 0 \) so it must be the case that \( (2-y)f(y) \geq 2 \rightarrow f(y) \geq \frac{2}{2-y}\).

In preparation for tomorrow's interview with App Academy (yes, it keeps getting postponed and I wish it could get over already!!), I've been working on problems from Project Euler today. I learned about Project Euler from morgtheborg. She is awesome + will be going to Dev Bootcamp in the summer and you should follow her. 

Anyways. I was solving this problem...

Problem 5: 2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

Solution: While the problem is worded differently, they are technically asking you to find the "least common multiple" for the numbers 1 to 20. While I love Math & can solve this without any problem on paper, translating math to code has always been tricky for me. I wrote multiple lines of code trying to figure out how to solve this problem till I realized that Ruby has an "lcm" method that returns the least common multiple of two numbers. Several scratching-my-head moments later, I figured out how to solve this problem in a single line. Woot. 

(1..20).inject(:lcm)