Math
posted by Krish .
How many ordered pairs of natural numbers (n, m) are there that solve the equation
2n + 3m = 100

Well we could argue about if zero counts as a natural number. If it does then
2n will always be zero or even
2 n = 0 2 4 6 8 ...... 50 ....100
3 m is zero or a multiple of 3
3 m = 0 3 6 9 12 15 ....33 .... 99
now the sum of those must be 100, which is even so delete all the odd 3ms
3 m = 0 6 12 18 24 .... 90 96
now which could add to 100
4 + 96
10 + 90
16 + 84
22 + 78
28 + 72 add six, subtract 6, see pattern
34 + 66
40 + 60
46 + 54
52 + 48
58 + 42
64 + 36
70 + 30
76 + 24
82 + 18
88 + 12
94 + 6
100 + 0 count them :)