Solve the equation:

\(\sqrt{a-\sqrt{a+x}}=x\)

Prove that if p is a prime number greater than 3, then the numerator of the reduce fraction is:

\(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{p-1}\)

Prove without using logarithm tables, that:

\(\frac{1}{\log_2\pi}+\frac{1}{log_5\pi}>2\)

In chess, is it possible for the knight to go from the bottom right corner to the upper left corner, but also landing on every single square in the process.

These are really hard, so if you don't know them it's fine.

I just wanted to share these cool problems with you.

GYanggg Apr 27, 2018

#1**-1 **

GYanggg: Here is a "real world" problem for you to solve:

The Red Cross in Geneva, Switzerland was offered the following deal by one of the big Swiss banks:

Deposit 1 Swiss franc in the first month, 2 Swiss francs in the second month, 3 Swiss francs in the third month......and so on for 1,000 Swiss francs in the 1000th month, the bank would pay the Red Cross 6% interest compounded monthly on ALL the money deposited. How much money would the Red Cross have accumulated in their account at the end of the 1000th month?

Guest Apr 28, 2018

#2**+3 **

Hi Gavin,

I'll give this one a go :)

\(\frac{1}{\log_2\pi}+\frac{1}{log_5\pi}>2\)

\(LHS\\ =\frac{1}{\log_2\pi}+\frac{1}{log_5\pi}\\ =1\div \frac{\log_\pi\pi}{\log_\pi2}+1\div \frac{\log_\pi\pi}{\log_\pi5}\\ =\frac{\log_\pi2}{\log_\pi\pi}+ \frac{\log_\pi5}{\log_\pi\pi}\\ =\frac{\log_\pi2+\log_\pi5}{\log_\pi\pi}\\ =\frac{\log_\pi10}{1}\\ =\log_\pi10\\ \text{Now pi^2 is approx 9.87 which is less than 10 so}\\ =\log_\pi(\pi^2*\delta) \qquad \text{where delta is a real number slightly bigger than 1}\\ =\log_\pi(\pi^2) +\log_\pi \delta\\ =2+\text{a small positive number}\\ >2\)

QED

Melody Apr 28, 2018

#4**+3 **

Here is my solutions.

4. To "light" up all the squares with a knight going from A1 to H8, the knight must travel 63 times.

Everytime a knight moves, it goes from black to white square and vise versa.

To go from white to white, the knight must move an even amount of times.

Therefore this is not possible.

GYanggg Apr 28, 2018