#problem 5

[temaenianpuzzle]

December whispers of treachery

[solvemyquestion]

Individual Tax Return 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…

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[vacantcatastrophe]

IMO 1986 Question 5

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 \( (2-y)f(y) \geq 2 \rightarrow f(y) \geq \frac{2}{2-y}\).

Luckily, it turns out \( f(y) \leq \frac{2}{2 - y} \) as well.  We assume not for contradiction: \( \exists y \in [0, 2) : f(y) > \frac{2}{2 - y} \).  Note that \(y \neq 0 \) since \( f(0) f(0 f(0)) = f(0 + 0) \rightarrow f(0) f(0) = f(0) \rightarrow f(0) = 1 = \frac{2}{2-0} \).  Then given such an \( y \) we can find positive \( x < 2 \) such that \( x f(y) \geq 2 \) and yet \( x + y < 2 \).  To see this let \( \alpha \) be such that \( f(y) = \alpha \frac{2}{2-y} \) and note that \( \alpha > 1 \).  Then let \( x = \frac{2 - y}{\alpha} \) and we get: \( x \cdot f(y) = (\frac{2 - y}{\alpha}) (\frac{2}{2 - y} \alpha) = 2 \). Thus we have positive \( x, y \) s.t. \( 0 = f(y) f(x f(y)) = f(x+y) > 0 \).   It remains to show that the function \(f(y) = \frac{2}{2-y} \) (on \( [0,2) \) and zero elsewhere), but this is a matter of algebra with requirement (a) and left as an exercise to the tireless reader.  \( \blacksquare \)

[harinilearnsruby]

Ruby continues to blow my mind!

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)

While lcm can be used only for two integers, using "inject" allows you to run lcm across a range of numbers. Ruby, you blow my mind!

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