MATHS M.C.Q (1 Marks)

$n$ is natural number, and $9^{2 n}-4^{2 n}$ is the form of $a^{2 n}-b^{2 n}$ is or $\left(a^n\right)^2-\left(b^n\right)^2$ which is divisibel by $(a+b)$ and $(a-b)$ or $9+4$ and $9-4$ or $13$ and $5$ both.

$\begin{array}{c|c}2 &144\\\hline 2 & 72\\\hline 2 & 36\\\hline2 & 18\\\hline3 & 9\\\hline3 & 3 \\\hline&1 \end{array}$
$144=2^4 \times 3^2$
$\therefore$ Exponant of $2$ is $4$

We know that the co$-$prime numbers have no factor in common, or, their $\text{HCF}$ is $1.$
Thus, $p^2$ and $q^2$ have the same factors with twice of the exponents of $p$ and $q$ respectively, which again will not have any common factor.
Thus we can conclude that $p ^2$ and $q ^2$ are co$-$prime numbers.
Hence, the correct choice is $(a).$

$a=p q^2$ and $b=p^3 q$ where $a$ and $b$ are positive integers and $p, q$ are prime numbers, then $H C F=p q$.

$A$ and $b$ are two positive integers and $a=p q^2$ and $b=p^2 q$, where $p$ and $q$ are prime numbers, then $L C M=p^2 q^2$.

$n$ is any natural number and $6^n-5^n$
We know that $6^n$ ends with $6$ and $5^n$ ends with $5$
$6^n-5^n$ will end with $6-5=1$

$\begin{array}{c|c}2 &196\\\hline 2 & 98\\\hline 7 & 49\\\hline7 & 7\\\hline&1 \end{array}$
$=2 \times 2 \times 7 \times 7$
$=2^2 \times 7^2$
 Sum of exponents $=2+2=4$

Let the two odd prime numbers $p_1$ and $p_2$ be $5$ and $3 .$
Then,
$\text{p}^2_1=5^2$
$=25$
And
$\text{p}^2_2=3^2$
$=9$
Thus,
$\text{p}^2_1-\text{p}^2_2=25-9$
$=16$
16 is even number.
Take another example, with $p_1$ and $p_2$ be $11$ and $7 .$
Then,
$\text{p}^2_1=11^2$
$=121$
And
$\text{p}^2_2=7^2$
$=49$
Thus,
$\text{p}^2_1-\text{p}^2_2=121-49$
$=72$
72 is even number.
Thus, we can say that $\text{p}^2_1-\text{p}^2_2$ is even number
In general the square of odd prime number is odd. Hence the difference of square of two prime numbers is odd
Hence the correct choice is $(a).$