I think that instead of (91,92) is 71,72. However, I have a doubt here.

72 and 71 is a valid pair, nonetheless, if I am not mistaken, 72*73 is also divisible by 12 and 9 at the same time. In addition, 36*37 is also divisible by 12 and 9, and the same happens with 35*36. If the last is true, it creates three new pairs, and it would imply that A > B, right?

the pairs which is divisible by 4,9
8,9
27,28
35,36
36,37
44,45
63,64
71,72
72,73
80,81
99,100

Total 10 pairs
10/100 = 1/10
C is answer

KarunMendiratta
can you please share a detailed answer to the question.

can you please explain this sir

We need \(\mathrm{x}(\mathrm{x}+1)\) to be divisible by both 12 and 9 , which means it must be divisible by \(\operatorname{lcm}(12,9)=36=4 \times 9\) .

Since x and \(\mathrm{x}+1\) are consecutive integers (coprime), I'll find the divisibility conditions \(\bmod\) 4 and \(\bmod\) 9 separately, then combine with CRT.

Divisibility by 4: Since one of \(\mathrm{x}, \mathrm{x}+1\) is even, we need that even one to actually carry a factor of 4 . Checking residues $\bmod 4$ :

- \(\mathrm{x} \equiv 0(\bmod 4): \mathrm{x}\) divisible by 4 \(\checkmark \)
- \(\mathrm{x} \equiv 3(\bmod 4): \mathrm{x}+1\) divisible by 4 \(\checkmark \)
- \(\mathrm{x} \equiv 1 or $2(\bmod 4)\) : neither term contributes a factor of \(4 \times \)

Divisibility by 9: Since x and $\mathrm{x}+1$ share no common factor of 3 , the full factor of 9 must land in just one of them:
- \(\mathrm{x} \equiv 0(\bmod 9): \mathrm{x}\) divisible by 9 \(\checkmark \)
- \(x \equiv 8(\bmod 9): x+1\) divisible by 9 \(\checkmark\)
- otherwise X


Combining via CRT (mod 36)
Pairing each valid condition $\bmod 4$ with each valid condition \(\bmod\) 9 gives four residue classes \(\bmod\) 36 :

\(\begin{tabular}{|l|l|l|}
\hline \mathbf{x} \bmod \mathbf{4} & xmod 9 & x mod 36 \\
\hline 0 & 0 & 0 \\
\hline 0 & 8 & 20 \\
\hline 3 & 0 & 27 \\
\hline 3 & 8 & 35 \\
\hline
\end{tabular}\)

Counting Valid x in [1,100]
- \(\mathbf{x} \equiv \mathbf{0}(\bmod 36): 36,72 \rightarrow \mathbf{2}\) values
- \(x \equiv 20(\bmod 36): 20,56,92 \rightarrow 3\) values
- \( \mathbf{x} \equiv \mathbf{2 7}(\bmod 36): 27,63,99 \rightarrow \mathbf{3}\) values
- \( x \equiv 35(\bmod 36): 35,71 \rightarrow 2\) values

Total: 2+3+3+2=10 values


Result


\(P=\frac{10}{100}=\frac{1}{10}\)



Quantity A = Quantity B. The two quantities are equal (answer C).