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.

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

The smallest rational number which should be multiplied by $\frac{1}{3}$ to get a terminating.
$\text{decimals }=\frac{3}{10}$
$\because\ \frac{1}{3}\times\frac{3}{10}=\frac{1}{10}=0.1$

Since, it is given that
$ n=2^3 \times 3^4 \times 5^4 \times 7 $
$ =2^3 \times 5^4 \times 3^4 \times 7 $
$ =2^3 \times 5^3 \times 5 \times 3^4 \times 7 $
$ =(2 \times 5)^3 \times 5 \times 3^4 \times 7 $
$ =5 \times 3^4 \times 7 \times(10)^3$
So, this means the given number n will end with 3 consecutive zeroes.

We know that the co-prime numbers have no factor in common, or, their $HCF$ 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 $(a)$.

$a = pq^2$ and $b = p^3q$ where a and b are positive integers and $p, q$ are prime numbers, then $HCF = pq$.

It is given that,
$\text{a}=\text{x}^3\text{y}^2=\text{x}\times\text{x}\times\text{x}\times\text{y}\times\text{y}$
$\text{b}=\text{xy}^3=\text{x}\times\text{y}\times\text{y}\times\text{y}$
$\text{HCF(a, b)}=\text{HCF}(\text{x}^3\text{y}^2,\text{xy}^3)=\text{x}\times\text{y}\times\text{y}=\text{xy}^2$
Hence, the correct answer is option $B$.

We know that a rational number has terminating decimal if the prime factors of its denominator are in the form $2^m× 5^n$ $\frac{16}{225}$ and $\frac{7}{250}$ has terminating decimals.

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

Rational number $=\frac{14587}{1250}=\frac{14587}{2^1\times5^4}$
$\begin{array}{c|c}2 &1250\\\hline 5 & 625\\\hline 5 & 125\\\hline5 & 25\\\hline5&5\\\hline&1 \end{array}$
$=\frac{14587}{10\times5^3}\times\frac{(2)^3}{(2)^3}$
$=\frac{14587\times8}{10\times1000}$
$=\frac{116696}{10000}=11.6696$
Hence, given rational number will terminate after four decimal places.

Let the two odd prime numbers $\mathrm{p}_1$ and $\mathrm{p}_2$ 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 $\mathrm{p}_1$ and $\mathrm{p}_2$ 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 $(a)$.