https://www.cs.utexas.edu/~lorenzo/corsi/cs372/06F/hw/3sol.html 
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https://www.cs.utexas.edu/~lorenzo/corsi/cs372/06F/hw/3sol.html 

Here, in problem 2:

In solution, point number 2 should be: With 4-byte entries in the page table we can reference 2^34 pages. Since each page is 2^13 B long, the maximum addressable physical memory size is 2^34 * 2^13 = 2^47 B (assuming no protection bits are used). 

REASON: 2^36/ 2^2 = 2^34, please correct me if I am wrong.

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