00:
Find a, b, c, d, e, f, and g, all positive integers greater than 1, that satisfy all equations:
a2 × b × c2 × g = 5100
a × b2 × e × f2 = 33462
a × c2 × d3 = 17150
a3 × b3 × c × d × e2 = 914760
... in the fastest way possible, using your language of choice.
01:
Download and solve the puzzle below.
02:
Solve:
789c0dccdb5283301405d04f82412ee9231c0ae8214de8ddbe594781888549dad
9c8d7dbf501ab125c539aad612b1a0a9f90882342cce4542ed3b9445ce50f5fd3
78d079e7478ab726efffde83c36ea141728bb879e883a6ce78f4eb98117fe756a
859df127292bf9024d42e628308a7467083e92ddd059a6e7d42361b03841b87c8
d0fd934de95485b0a23eb3f423d962caba52f0e52c14105d164c35b9f3506bc17
bc4d7bc75fc81e9eec38ed5d671a7a572c5f1b5d3e5423668d7985ec89cf6cf5b
5d61571e92558ad12323b9489ffe013c525446
04:
What is the 10001st prime number?
05:
Let d(n) be defined as the sum of proper divisors of n (numbers less
than n which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable
pair and each of a and b are called amicable numbers.
For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110;
therefore d(220) = 284.
The proper divisors of 284 are 1, 2, 4, 71 and 142;
therefore d(284) = 220.
-> Evaluate the sum of all the amicable numbers under 100000.
06:
By starting at the top of the triangle below and moving to adjacent
numbers on the row below, the maximum total from top to bottom is 23.
3
7 4
2 4 6
8 5 9 3
That is, 3 + 7 + 4 + 9 = 23.
Find the maximum total from top to bottom of the triangle below:
75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
07:
Solve 06 as before using the following input: