Two integer $'a'$ and $'b'$ are randomly selected from the set $\{1, 2, \ldots\}$ (with replacement) then if the probability of $\frac{1}{5}(a^2 + b^2)$ being positive integer is $\frac{p}{q}$ (where $H.C.F(p, q) = 1$) then $q - 2p = \ldots\ldots\ldots\ldots$
Step-by-Step Solution
Key Concept: For (1/5)(a² + b²) to be a positive integer, a² + b² must be divisible by 5. This requires analyzing the last digits of perfect squares: only 0, 1, 4, 5, 6, 9 are possible. The sum a² + b² is divisible by 5 when their last digits sum to 0 or 5 (mod 10).
Find pairs $(a,b)$ where both $a^2$ and $b^2$ have the same last digit. Create a frequency table showing that last digits $0,1,4,5,6,9$ are possible for squares. Count favorable outcomes where the sum of last digits equals 0 or 5: cases include $(0,0), (0,5), (1,4), (1,9), (4,6), (5,5), (6,9)$ with multiplicities. The total number of favorable outcomes is $1 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 2 + 1 + 1 + 1 + 2 + 2 + 2 = 9$ (when considering ordered pairs from frequency 10×10 total).
Correct Answer: 7