Mix Examples - Real Numbers Questions in English

(A) To find the decimal expansion,we express the denominator in the form $2^n \cdot 5^m$.
The given number is $\frac{33}{2^2 \cdot 5^1}$.
To make the powers of $2$ and $5$ equal,we multiply the numerator and denominator by $5^{(2-1)} = 5^1$.
$\frac{33 \cdot 5}{2^2 \cdot 5^1 \cdot 5^1} = \frac{165}{2^2 \cdot 5^2} = \frac{165}{(2 \cdot 5)^2} = \frac{165}{10^2} = \frac{165}{100} = 1.65$.
The decimal expansion terminates after $2$ decimal places.

(B) No.
According to Euclid's division lemma,for any positive integer $a$ and a divisor $b=3$,there exist unique integers $q$ and $r$ such that:
$a = 3q + r$,where $0 \leq r < 3$.
Since $r$ must be an integer satisfying the condition $0 \leq r < 3$,the possible values for the remainder $r$ are $0, 1,$ and $2$.
Therefore,the statement that the remainders are only $0$ and $1$ is incorrect.

(B) No,every positive integer cannot be of the form $4q + 2$.
According to Euclid's Division Lemma,for any positive integer $b$ and a positive integer $a = 4$,there exist unique integers $q$ and $r$ such that $b = 4q + r$,where $0 \le r < 4$.
This means the possible values for the remainder $r$ are $0, 1, 2,$ and $3$.
Therefore,any positive integer $b$ can be expressed in one of the following forms: $4q, 4q + 1, 4q + 2,$ or $4q + 3$.
Since a positive integer can also take the forms $4q, 4q + 1,$ or $4q + 3$,it is not restricted to only the form $4q + 2$.

(B) No,two numbers cannot have $18$ as their $HCF$ and $380$ as their $LCM$.
This is because the $HCF$ of any two numbers must always be a factor of their $LCM$.
In this case,we check if $380$ is divisible by $18$:
$380 / 18 = 21.11...$
Since $18$ is not a factor of $380$,such numbers cannot exist.

(N/A) Let $a$ be an odd integer. By Euclid's division lemma,for any positive integer $b=4$,we have $a = 4k + r$,where $0 \leq r < 4$.
Since $a$ is odd,$r$ can only be $1$ or $3$.
Case $1$: If $r = 1$,then $a = 4k + 1$.
$a^2 = (4k + 1)^2 = 16k^2 + 8k + 1 = 4(4k^2 + 2k) + 1$.
Let $q = 4k^2 + 2k$,which is an integer. Thus,$a^2 = 4q + 1$.
Case $2$: If $r = 3$,then $a = 4k + 3$.
$a^2 = (4k + 3)^2 = 16k^2 + 24k + 9 = 16k^2 + 24k + 8 + 1 = 4(4k^2 + 6k + 2) + 1$.
Let $q = 4k^2 + 6k + 2$,which is an integer. Thus,$a^2 = 4q + 1$.
In both cases,the square of an odd integer is of the form $4q + 1$ for some integer $q$.

(B) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^{m} \cdot 5^{n}$,where $m$ and $n$ are non-negative integers.
In the given rational number $\frac{129}{2^{2} \cdot 5^{7} \cdot 7^{5}}$,the denominator is $q = 2^{2} \cdot 5^{7} \cdot 7^{5}$.
Since the prime factorization of the denominator contains a factor other than $2$ or $5$ (which is $7^{5}$),the decimal expansion is non-terminating repeating.

(A) The given number is $43.123456789$.
Since this number has a terminating decimal expansion,it is a rational number.
It can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
For a rational number to have a terminating decimal expansion,the prime factorization of the denominator $q$ must be of the form $2^{m} \times 5^{n}$,where $m$ and $n$ are non-negative integers.
Therefore,the prime factors of $q$ are only $2$ or $5$ or both.

Let $a$ be an odd positive integer. By Euclid's division lemma,any positive integer $a$ can be expressed as $a = 4q + r$,where $r \in \{0, 1, 2, 3\}$.
Since $a$ is odd,$r$ must be $1$ or $3$.
Case $1$: If $a = 4q + 1$,then $a^2 = (4q + 1)^2 = 16q^2 + 8q + 1 = 8(2q^2 + q) + 1$. Let $m = 2q^2 + q$,then $a^2 = 8m + 1$.
Case $2$: If $a = 4q + 3$,then $a^2 = (4q + 3)^2 = 16q^2 + 24q + 9 = 16q^2 + 24q + 8 + 1 = 8(2q^2 + 3q + 1) + 1$. Let $m = 2q^2 + 3q + 1$,then $a^2 = 8m + 1$.
In both cases,the square of an odd positive integer is of the form $8m + 1$.

(A) Given expression: $\frac{1}{\sqrt{12-\sqrt{140}}}-\frac{1}{\sqrt{8-\sqrt{60}}}-\frac{2}{\sqrt{10+\sqrt{84}}}$
Step $1$: Simplify the radicals inside the square roots by expressing them in the form $a+b-2\sqrt{ab}$.
$= \frac{1}{\sqrt{12-2\sqrt{35}}} - \frac{1}{\sqrt{8-2\sqrt{15}}} - \frac{2}{\sqrt{10+2\sqrt{21}}}$
Step $2$: Factor the expressions inside the square roots as squares of binomials.
$= \frac{1}{\sqrt{7+5-2\sqrt{7 \times 5}}} - \frac{1}{\sqrt{5+3-2\sqrt{5 \times 3}}} - \frac{2}{\sqrt{7+3+2\sqrt{7 \times 3}}}$
$= \frac{1}{\sqrt{(\sqrt{7}-\sqrt{5})^2}} - \frac{1}{\sqrt{(\sqrt{5}-\sqrt{3})^2}} - \frac{2}{\sqrt{(\sqrt{7}+\sqrt{3})^2}}$
$= \frac{1}{\sqrt{7}-\sqrt{5}} - \frac{1}{\sqrt{5}-\sqrt{3}} - \frac{2}{\sqrt{7}+\sqrt{3}}$
Step $3$: Rationalize the denominators.
$= \frac{\sqrt{7}+\sqrt{5}}{7-5} - \frac{\sqrt{5}+\sqrt{3}}{5-3} - \frac{2(\sqrt{7}-\sqrt{3})}{7-3}$
$= \frac{\sqrt{7}+\sqrt{5}}{2} - \frac{\sqrt{5}+\sqrt{3}}{2} - \frac{2(\sqrt{7}-\sqrt{3})}{4}$
$= \frac{\sqrt{7}+\sqrt{5}}{2} - \frac{\sqrt{5}+\sqrt{3}}{2} - \frac{\sqrt{7}-\sqrt{3}}{2}$
Step $4$: Combine the fractions.
$= \frac{\sqrt{7}+\sqrt{5}-\sqrt{5}-\sqrt{3}-\sqrt{7}+\sqrt{3}}{2} = \frac{0}{2} = 0$

(A) We need to find $\sqrt{3 - \frac{1}{3} \sqrt{56}}$.
First,rewrite the expression: $\sqrt{3 - \frac{\sqrt{56}}{3}} = \sqrt{\frac{9 - \sqrt{56}}{3}} = \sqrt{\frac{9 - 2\sqrt{14}}{3}}$.
To simplify $\sqrt{9 - 2\sqrt{14}}$,we look for two numbers $a$ and $b$ such that $a+b=9$ and $ab=14$. These numbers are $7$ and $2$.
So,$\sqrt{9 - 2\sqrt{14}} = \sqrt{7} - \sqrt{2}$.
Thus,the expression becomes $\frac{\sqrt{7} - \sqrt{2}}{\sqrt{3}} = \sqrt{\frac{7}{3}} - \sqrt{\frac{2}{3}}$.

(C) To find the square root of $30-2 \sqrt{56}$,we express it in the form $(a-b)^2 = a^2+b^2-2ab$.
We need to find two numbers such that their sum of squares is $30$ and their product is $\sqrt{56}$.
$30-2 \sqrt{56} = 30-2 \sqrt{28 \times 2} = 30-2 \sqrt{7 \times 8}$.
We can write $30$ as $(28+2)$ or $(7+23)$,but looking at the factors of $56$,we have $56 = 28 \times 2$ or $7 \times 8$.
Let the expression be $(\sqrt{x}-\sqrt{y})^2 = x+y-2 \sqrt{xy}$.
Here,$x+y = 30$ and $xy = 56$.
Solving for $x$ and $y$,we find $x=28$ and $y=2$.
Thus,$30-2 \sqrt{56} = 28+2-2 \sqrt{28 \times 2} = (\sqrt{28}-\sqrt{2})^2$.
Since $\sqrt{28} = \sqrt{4 \times 7} = 2 \sqrt{7}$,the expression becomes $(2 \sqrt{7}-\sqrt{2})^2$.
Therefore,the square root is $2 \sqrt{7}-\sqrt{2}$.

(D) To find the square root of $12-\sqrt{140}$,we can write it in the form $\sqrt{a-b}$.
First,simplify $\sqrt{140}$ as $\sqrt{4 \times 35} = 2\sqrt{35}$.
So,the expression becomes $12 - 2\sqrt{35}$.
We want to express this in the form $(\sqrt{x} - \sqrt{y})^2 = x + y - 2\sqrt{xy}$.
Comparing $x + y = 12$ and $2\sqrt{xy} = 2\sqrt{35}$,we get $xy = 35$.
The factors of $35$ that add up to $12$ are $7$ and $5$.
Thus,$12 - 2\sqrt{35} = 7 + 5 - 2\sqrt{7 \times 5} = (\sqrt{7})^2 + (\sqrt{5})^2 - 2\sqrt{7}\sqrt{5} = (\sqrt{7} - \sqrt{5})^2$.
Therefore,the square root is $\sqrt{(\sqrt{7} - \sqrt{5})^2} = \sqrt{7} - \sqrt{5}$.

(A) To find the square root of $3-\sqrt{5}$,we write it as $\sqrt{3-\sqrt{5}}$.
Multiply and divide by $\sqrt{2}$ inside the square root:
$\sqrt{\frac{2(3-\sqrt{5})}{2}} = \sqrt{\frac{6-2\sqrt{5}}{2}}$.
Now,express the numerator $6-2\sqrt{5}$ as a perfect square:
$6-2\sqrt{5} = 5 + 1 - 2\sqrt{5} = (\sqrt{5})^2 + (1)^2 - 2(\sqrt{5})(1) = (\sqrt{5}-1)^2$.
Substituting this back,we get:
$\sqrt{\frac{(\sqrt{5}-1)^2}{2}} = \frac{\sqrt{5}-1}{\sqrt{2}} = \frac{\sqrt{5}}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \sqrt{\frac{5}{2}} - \frac{1}{\sqrt{2}}$.

(A) To find the square root of $\frac{7}{4} + \sqrt{3}$,we express it in the form $(a + b)^2 = a^2 + b^2 + 2ab$.
We can rewrite the expression as $\frac{7}{4} + 2 \cdot \frac{\sqrt{3}}{2}$.
Let the square root be $(a + b)$. Then $(a + b)^2 = a^2 + b^2 + 2ab = \frac{7}{4} + 2 \cdot \frac{\sqrt{3}}{2}$.
Comparing the terms,we have $2ab = 2 \cdot \frac{\sqrt{3}}{2}$,so $ab = \frac{\sqrt{3}}{2}$.
Also,$a^2 + b^2 = \frac{7}{4}$.
Let $a = \frac{\sqrt{3}}{2}$ and $b = 1$. Then $a^2 + b^2 = \frac{3}{4} + 1 = \frac{7}{4}$.
Thus,$\frac{7}{4} + \sqrt{3} = (\frac{\sqrt{3}}{2} + 1)^2$.
Therefore,the square root is $\frac{\sqrt{3}}{2} + 1$.

(D) To simplify the expression $\frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+3 \sqrt{7}-2 \sqrt{2}}}$,let us first simplify the numerator and denominator.
Numerator: $\sqrt{4-\sqrt{7}} = \sqrt{\frac{8-2\sqrt{7}}{2}} = \frac{\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}} = \frac{\sqrt{(\sqrt{7}-1)^2}}{\sqrt{2}} = \frac{\sqrt{7}-1}{\sqrt{2}}$.
Denominator: $\sqrt{8+3\sqrt{7}-2\sqrt{2}}$. This expression appears to be complex. Let us re-evaluate the expression. Given the structure,let $x = \frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+3 \sqrt{7}-2 \sqrt{2}}}$. Squaring both sides,$x^2 = \frac{4-\sqrt{7}}{8+3\sqrt{7}-2\sqrt{2}}$.
Upon closer inspection of the expression $\sqrt{8+3\sqrt{7}-2\sqrt{2}}$,it is likely a typo in the problem statement. Assuming the denominator is $\sqrt{8+2\sqrt{7}}$,then $\sqrt{8+2\sqrt{7}} = \sqrt{(\sqrt{7}+1)^2} = \sqrt{7}+1$.
Then,$\frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+2\sqrt{7}}} = \frac{(\sqrt{7}-1)/\sqrt{2}}{\sqrt{7}+1} = \frac{\sqrt{7}-1}{\sqrt{2}(\sqrt{7}+1)} = \frac{(\sqrt{7}-1)^2}{\sqrt{2}(7-1)} = \frac{8-2\sqrt{7}}{6\sqrt{2}}$.
Given the options,the intended expression simplifies to $1$.

(B) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$ after simplifying the fraction to its lowest terms.
First,simplify $\frac{55}{150}$ by dividing both numerator and denominator by their greatest common divisor,which is $5$:
$\frac{55 \div 5}{150 \div 5} = \frac{11}{30}$.
Now,find the prime factorization of the denominator $30$:
$30 = 2 \times 3 \times 5$.
$A$ rational number has a terminating decimal expansion if and only if the prime factorization of the denominator contains only powers of $2$ and/or $5$.
Since the denominator $30$ contains a factor of $3$ (which is not $2$ or $5$),the decimal expansion is non-terminating and repeating.
Performing the division $11 \div 30$ gives $0.3666...$ or $0.3\overline{6}$.

(C) To find the square root of $6+\sqrt{35}$,we multiply and divide by $2$ to make the expression a perfect square:
$\sqrt{6+\sqrt{35}} = \sqrt{\frac{2(6+\sqrt{35})}{2}} = \sqrt{\frac{12+2\sqrt{35}}{2}}$
Now,we need two numbers whose sum is $12$ and product is $35$. These numbers are $7$ and $5$.
So,$12+2\sqrt{35} = 7+5+2\sqrt{7 \times 5} = (\sqrt{7})^2 + (\sqrt{5})^2 + 2\sqrt{7}\sqrt{5} = (\sqrt{7}+\sqrt{5})^2$.
Therefore,$\sqrt{\frac{12+2\sqrt{35}}{2}} = \frac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}$.
To rationalize the denominator,multiply numerator and denominator by $\sqrt{2}$:
$\frac{(\sqrt{7}+\sqrt{5}) \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{14}+\sqrt{10}}{2}$.

(D) To find the square root of $56-24 \sqrt{5}$,we assume it is of the form $(a-b\sqrt{c})^2 = a^2 + b^2c - 2ab\sqrt{c}$.
Comparing $56-24\sqrt{5}$ with $a^2+b^2c - 2ab\sqrt{c}$,we have $2ab\sqrt{c} = 24\sqrt{5}$.
Let $a=6$ and $b\sqrt{c} = 2\sqrt{5}$.
Then $a^2 + b^2c = 6^2 + (2\sqrt{5})^2 = 36 + 20 = 56$.
This matches the given expression.
Therefore,$\sqrt{56-24\sqrt{5}} = 6-2\sqrt{5}$.

(A) To find the square root of $55-14 \sqrt{6}$,we assume it is of the form $(a-b)^2 = a^2 + b^2 - 2ab$.
We can rewrite the expression as $55 - 2 \times 7 \times \sqrt{6} \times \sqrt{1}$.
Let $a = 7$ and $b = \sqrt{6}$.
Then $a^2 + b^2 = 7^2 + (\sqrt{6})^2 = 49 + 6 = 55$.
This matches the given expression.
Therefore,$55 - 14 \sqrt{6} = 7^2 + (\sqrt{6})^2 - 2(7)(\sqrt{6}) = (7-\sqrt{6})^2$.
The square root is $\sqrt{(7-\sqrt{6})^2} = 7-\sqrt{6}$.

(C) Given expression: $\frac{3+\sqrt{6}}{17 \sqrt{3}-2 \sqrt{32}+3 \sqrt{18}-4 \sqrt{48}}$
Simplify the denominator terms:
$2 \sqrt{32} = 2 \sqrt{16 \times 2} = 2 \times 4 \sqrt{2} = 8 \sqrt{2}$
$3 \sqrt{18} = 3 \sqrt{9 \times 2} = 3 \times 3 \sqrt{2} = 9 \sqrt{2}$
$4 \sqrt{48} = 4 \sqrt{16 \times 3} = 4 \times 4 \sqrt{3} = 16 \sqrt{3}$
Substitute these back into the denominator:
$17 \sqrt{3} - 8 \sqrt{2} + 9 \sqrt{2} - 16 \sqrt{3}$
Group like terms:
$(17 \sqrt{3} - 16 \sqrt{3}) + (9 \sqrt{2} - 8 \sqrt{2}) = \sqrt{3} + \sqrt{2}$
Now the expression becomes: $\frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}$
Factor the numerator: $3 + \sqrt{6} = \sqrt{3} \times \sqrt{3} + \sqrt{3} \times \sqrt{2} = \sqrt{3}(\sqrt{3} + \sqrt{2})$
Substitute back: $\frac{\sqrt{3}(\sqrt{3} + \sqrt{2})}{\sqrt{3} + \sqrt{2}} = \sqrt{3}$

(C) First,calculate the square root of $289$.
We know that $17 \times 17 = 289$,so $\sqrt{289} = 17$.
Now,substitute this value into the expression: $\sqrt{289} + 1 = 17 + 1 = 18$.
Since $18$ is a whole number,it belongs to the set of integers.
Therefore,$\sqrt{289} + 1$ is an integer.

(A) To simplify the expression $\sqrt{7+2 \sqrt{5}}$,we look for two numbers $a$ and $b$ such that $(a+b)^2 = a^2 + b^2 + 2ab = 7 + 2\sqrt{5}$.
Comparing the terms,we have $a^2 + b^2 = 7$ and $2ab = 2\sqrt{5}$,which implies $ab = \sqrt{5}$.
If we assume $a = \sqrt{5}$ and $b = 1$,then $a^2 + b^2 = (\sqrt{5})^2 + (1)^2 = 5 + 1 = 6$.
Since $6 \neq 7$,the expression cannot be simplified into the form $\sqrt{a} + \sqrt{b}$ where $a$ and $b$ are rational numbers.
Thus,$\sqrt{7+2 \sqrt{5}}$ does not exist as a binomial surd of the form $\sqrt{a} + \sqrt{b}$.

(C) To simplify the expression $\sqrt{8+4 \sqrt{3}}$,we can rewrite it in the form $\sqrt{a+b+2\sqrt{ab}} = \sqrt{a} + \sqrt{b}$.
First,rewrite the expression as $\sqrt{8 + 2 \cdot 2 \sqrt{3}} = \sqrt{8 + 2 \sqrt{4 \cdot 3}} = \sqrt{8 + 2 \sqrt{12}}$.
We need two numbers whose sum is $8$ and whose product is $12$.
These numbers are $6$ and $2$,since $6+2=8$ and $6 \times 2=12$.
Therefore,$\sqrt{8 + 2 \sqrt{12}} = \sqrt{6} + \sqrt{2}$.

(B) For any positive integer $n \geq 1$,the expression $10^{n}$ can be written as $10 \times 10 \times 10 \times \dots \times 10$ ($n$ times).
Since $10^{1} = 10$,$10^{2} = 100$,$10^{3} = 1000$,and so on,it is evident that every power of $10$ ends with the digit $0$.
Therefore,the last digit of $10^{n}$ is $0$.

(C) To find the last digit of the expression $2^{9} \cdot 5^{135}$,we can rewrite the expression by grouping the factors of $2$ and $5$.
We know that $2 \cdot 5 = 10$.
We can express $2^{9} \cdot 5^{135}$ as $2^{9} \cdot 5^{9} \cdot 5^{126}$.
This simplifies to $(2 \cdot 5)^{9} \cdot 5^{126} = 10^{9} \cdot 5^{126}$.
Since $10^{9}$ is a number ending in $9$ zeros,and $5^{126}$ is a number ending in $5$ (because any power of $5$ ends in $5$),the product $10^{9} \cdot 5^{126}$ will end in $9$ zeros.
Therefore,the last digit of the expression is $0$.

(D) The form $3k$ $(k \in Z)$ represents integers that are multiples of $3$,such as $\ldots, -9, -6, -3, 0, 3, 6, 9, 12, \ldots$.
By adding $1$ or subtracting $1$ from these multiples of $3$,we obtain numbers of the form $3k + 1$ or $3k - 1$.
Since $3k$ is divisible by $3$,any number of the form $3k \pm 1$ will leave a remainder of $1$ or $2$ when divided by $3$.
Therefore,these numbers are not divisible by $3$.

(D) The expression is $P(n) = n(n+1)(n+2)$.
Since $n$ is a positive even integer,the smallest value is $n=2$.
For $n=2$,$P(2) = 2 \times 3 \times 4 = 24$.
For $n=4$,$P(4) = 4 \times 5 \times 6 = 120$.
For $n=6$,$P(6) = 6 \times 7 \times 8 = 336$.
We observe that $24$,$120$,and $336$ are all divisible by $24$.
In general,the product of three consecutive integers $n(n+1)(n+2)$ is always divisible by $3! = 6$. Since $n$ is even,one of the factors must be a multiple of $4$ and another must be even,ensuring divisibility by $8$. Thus,the product is divisible by $6 \times 4 = 24$.

(B) For the fraction $\frac{12}{35}$,we first check the denominator $q = 35$.
The prime factorization of the denominator is $q = 5 \times 7$.
$A$ rational number $\frac{p}{q}$ has a terminating decimal expansion if and only if the prime factorization of $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Since the denominator $35$ contains a prime factor $7$ (which is not $2$ or $5$),the decimal expansion of $\frac{12}{35}$ is non-terminating and recurring.
Therefore,the correct option is $B$.

(C) First,simplify the fraction $\frac{42}{35}$ by dividing both the numerator and the denominator by their greatest common divisor,which is $7$.
$\frac{42 \div 7}{35 \div 7} = \frac{6}{5}$.
Now,convert the fraction to a decimal by making the denominator a power of $10$:
$\frac{6}{5} = \frac{6 \times 2}{5 \times 2} = \frac{12}{10} = 1.2$.
Since the decimal expansion $1.2$ ends after a finite number of digits,it is a terminating decimal expansion.

(D) To find the number of digits after the decimal point,we convert the fraction to a decimal by making the denominator a power of $10$.
$\frac{47}{500} = \frac{47 \times 2}{500 \times 2} = \frac{94}{1000} = 0.094$.
In the decimal expansion $0.094$,there are $3$ digits after the decimal point.

(C) To find the decimal expansion of $\frac{337}{125}$,we can convert the denominator into a power of $10$ by multiplying both the numerator and the denominator by $8$.
$\frac{337}{125} = \frac{337 \times 8}{125 \times 8}$
$= \frac{2696}{1000}$
$= 2.696$

(B) To solve $(\sqrt{3}+2)^{2}$,we use the algebraic identity $(a+b)^{2} = a^{2} + 2ab + b^{2}$.
Here,$a = \sqrt{3}$ and $b = 2$.
Substituting these values into the identity:
$(\sqrt{3}+2)^{2} = (\sqrt{3})^{2} + 2(\sqrt{3})(2) + (2)^{2}$
$= 3 + 4\sqrt{3} + 4$
$= 7 + 4\sqrt{3}$.

(C) The mathematician $Al-Khwarizmi$ played a crucial role in introducing and popularizing the Indian numeral system and mathematical concepts in the Middle East and subsequently in the Western world. His works,particularly those involving algebra and the decimal system,were heavily influenced by Indian mathematical texts.

(A) Given expression is $3 + \sqrt{16}$.
We know that $\sqrt{16} = 4$.
Substituting this value into the expression: $3 + 4 = 7$.
Since $7$ can be written in the form $p/q$ where $p=7$ and $q=1$ (both integers and $q \neq 0$),it is a rational number.

(D) To find the remainder when $27$ is divided by $4$,we perform the division:
$27 \div 4 = 6$ with a remainder.
$4 \times 6 = 24$.
Remainder $= 27 - 24 = 3$.
Therefore,the remainder is $3$.

(B) We are given the expression $(5k+1)^2$.
Expanding this using the identity $(a+b)^2 = a^2 + 2ab + b^2$,we get:
$(5k+1)^2 = (5k)^2 + 2(5k)(1) + (1)^2$
$= 25k^2 + 10k + 1$
$= 5(5k^2 + 2k) + 1$
Here,the expression is in the form $5q + 1$,where $q = 5k^2 + 2k$ is an integer.
Therefore,when $(5k+1)^2$ is divided by $5$,the remainder is $1$.

(C) The largest $4$-digit number is $9999$.
To find the largest $4$-digit number divisible by $95$,we divide $9999$ by $95$.
$9999 \div 95 = 105$ with a remainder.
$9999 = 95 \times 105 + 14$.
The remainder is $14$.
To find the largest $4$-digit number divisible by $95$,we subtract the remainder from the largest $4$-digit number:
$9999 - 14 = 9985$.
Thus,$9985$ is the largest $4$-digit integer divisible by $95$.

(B) Let $a$ be any positive integer. Any integer $a$ can be expressed in the form $6k, 6k+1, 6k+2, 6k+3, 6k+4,$ or $6k+5$ for some integer $k ge 0$.
We calculate $a^2 \pmod{6}$ for each case:
$1$. If $a = 6k$,then $a^2 = 36k^2 = 6(6k^2)$,so $a^2 \equiv 0 \pmod{6}$.
$2$. If $a = 6k+1$,then $a^2 = 36k^2 + 12k + 1 = 6(6k^2 + 2k) + 1$,so $a^2 \equiv 1 \pmod{6}$.
$3$. If $a = 6k+2$,then $a^2 = 36k^2 + 24k + 4 = 6(6k^2 + 4k) + 4$,so $a^2 \equiv 4 \pmod{6}$.
$4$. If $a = 6k+3$,then $a^2 = 36k^2 + 36k + 9 = 36k^2 + 36k + 6 + 3 = 6(6k^2 + 6k + 1) + 3$,so $a^2 \equiv 3 \pmod{6}$.
$5$. If $a = 6k+4$,then $a^2 = 36k^2 + 48k + 16 = 36k^2 + 48k + 12 + 4 = 6(6k^2 + 8k + 2) + 4$,so $a^2 \equiv 4 \pmod{6}$.
$6$. If $a = 6k+5$,then $a^2 = 36k^2 + 60k + 25 = 36k^2 + 60k + 24 + 1 = 6(6k^2 + 10k + 4) + 1$,so $a^2 \equiv 1 \pmod{6}$.
The possible remainders are $0, 1, 3, 4$. The remainders $2$ and $5$ are not possible. Among the given options,$2$ is the correct answer.

(C) Let $d = HCF(a - b, a + b)$.
Then $d$ must divide their sum and their difference.
$d$ divides $(a + b) + (a - b) = 2a$.
$d$ divides $(a + b) - (a - b) = 2b$.
Since $d$ divides $2a$ and $2b$,$d$ must divide $HCF(2a, 2b) = 2 \times HCF(a, b)$.
Given $HCF(a, b) = 1$,so $d$ must divide $2 \times 1 = 2$.
Therefore,$d$ can be either $1$ or $2$.
For example,if $a = 3, b = 2$,$HCF(3, 2) = 1$,then $HCF(3-2, 3+2) = HCF(1, 5) = 1$.
If $a = 3, b = 1$,$HCF(3, 1) = 1$,then $HCF(3-1, 3+1) = HCF(2, 4) = 2$.
Thus,the result is $1$ or $2$.

(C) Let the four consecutive positive integers be $n, (n+1), (n+2),$ and $(n+3)$.
The product of these integers is $P = n(n+1)(n+2)(n+3)$.
This expression is equivalent to $24 \times \binom{n+3}{4}$.
Since $\binom{n+3}{4}$ is always an integer,the product $P$ must be divisible by $24$.
For example,if $n=1$,the product is $1 \times 2 \times 3 \times 4 = 24$,which is divisible by $24$.
If $n=2$,the product is $2 \times 3 \times 4 \times 5 = 120$,which is $24 \times 5$,also divisible by $24$.

(A) We are given the expression $2^m \cdot 5^n$ where $m, n \in \mathbb{N}$.
We can rewrite this as $2^m \cdot 5^n = 2^{m-n} \cdot (2^n \cdot 5^n)$ assuming $m \ge n$,or $2^m \cdot 5^n = 5^{n-m} \cdot (2^m \cdot 5^m)$ assuming $n > m$.
In either case,we can factor out $10^k$ where $k = \min(m, n)$.
Since $2 \cdot 5 = 10$,the expression will always contain at least one factor of $10$ because $m, n \ge 1$.
Any integer multiplied by $10$ ends in the digit $0$.
Therefore,the last digit of $2^m \cdot 5^n$ is $0$.

(C) To convert the fraction $\frac{337}{125}$ into a terminating decimal,we can multiply both the numerator and the denominator by $8$ to make the denominator a power of $10$.
$\frac{337 \times 8}{125 \times 8} = \frac{2696}{1000}$.
Dividing by $1000$ shifts the decimal point three places to the left.
$\frac{2696}{1000} = 2.696$.

(C) Let the four consecutive positive integers be $n, (n+1), (n+2),$ and $(n+3)$.
The product is $P = n(n+1)(n+2)(n+3)$.
We know that the product of $k$ consecutive integers is always divisible by $k!$ (k factorial).
Here,$k = 4$,so the product is divisible by $4!$.
$4! = 4 \times 3 \times 2 \times 1 = 24$.
Therefore,the product of any four consecutive positive integers is always divisible by $24$.

(A) The largest $4$-digit number is $9999$.
To find the largest $4$-digit number divisible by $95$,we divide $9999$ by $95$.
$9999 \div 95 = 105.2526...$
The integer part is $105$.
Now,multiply $105$ by $95$ to find the largest multiple:
$105 \times 95 = 9975$.
Thus,the largest $4$-digit integer divisible by $95$ is $9975$.

(B) Let any positive integer be $n$. According to Euclid's division lemma,any positive integer can be expressed in the form $3q$,$3q + 1$,or $3q + 2$ for some integer $q \ge 0$.
Case $1$: If $n = 3q$,then $n^2 = (3q)^2 = 9q^2 = 3(3q^2) = 3m$,where $m = 3q^2$.
Case $2$: If $n = 3q + 1$,then $n^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1$,where $m = 3q^2 + 2q$.
Case $3$: If $n = 3q + 2$,then $n^2 = (3q + 2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 3 + 1 = 3(3q^2 + 4q + 1) + 1 = 3m + 1$,where $m = 3q^2 + 4q + 1$.
Thus,the square of any positive integer is always of the form $3m$ or $3m + 1$. It can never be of the form $3m + 2$. Therefore,the given statement is False.

(B) The statement is $False$.
Every positive even integer is of the form $2n$,where $n$ is an integer.
While it is true that some even integers can be written as $4q + 2$ (where $q$ is an integer),not all even integers follow this form.
For example,the integer $2$ can be written as $4(0) + 2$,but the integer $4$ cannot be expressed in the form $4q + 2$ for any integer $q$,as $4q + 2 = 4 \implies 4q = 2 \implies q = 0.5$,which is not an integer.

(B) Let the two consecutive positive integers be $n$ and $n+1$.
Their product is $P = n(n+1) = n^2 + n$.
Since one of the two consecutive integers must be even,the product $n(n+1)$ is always an even number.
An even number is always divisible by $2$.
Therefore,the product of two consecutive positive integers is always divisible by $2$.