Mix Examples - Real Numbers Questions in English

(N/A) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$.
If $q = 2^n \times 5^m$,where $n$ and $m$ are non-negative integers,the decimal expansion is terminating.
Here,the denominator is $343 = 7^3$.
Since the prime factorization of the denominator contains a factor other than $2$ or $5$ (specifically,$7$),the rational number $\frac{17}{343}$ does not have a terminating decimal expansion; it has a non-terminating repeating decimal expansion.

(A) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$ in its simplest form.
First,simplify the fraction: $\frac{9}{30} = \frac{3 \times 3}{3 \times 10} = \frac{3}{10}$.
The denominator is $10 = 2^1 \times 5^1$.
Since the prime factorization of the denominator is of the form $2^n \times 5^m$ (where $n, m$ are non-negative integers),the rational number has a terminating decimal expansion.
To find the decimal expansion: $\frac{3}{10} = 0.3$.

(N/A) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Here,$q = 256 = 2^8$.
Since the denominator is of the form $2^n \times 5^m$ (where $n=8, m=0$),the rational number $\frac{19}{256}$ has a terminating decimal expansion.
To find the decimal expansion,we multiply the numerator and denominator by $5^8$:
$\frac{19}{2^8} \times \frac{5^8}{5^8} = \frac{19 \times 390625}{10^8} = \frac{7421875}{100000000} = 0.07421875$.

(B) Let the square root be $\sqrt{a} + \sqrt{b}$.
By squaring both sides,we get $a + b + 2\sqrt{ab} = 8 + \sqrt{63}$.
Comparing the rational and irrational parts,we have $a + b = 8$ and $2\sqrt{ab} = \sqrt{63}$.
Squaring the second equation,$4ab = 63$,so $ab = \frac{63}{4}$.
Now,consider a quadratic equation with roots $a$ and $b$: $t^2 - (a+b)t + ab = 0$,which is $t^2 - 8t + \frac{63}{4} = 0$.
Multiplying by $4$,we get $4t^2 - 32t + 63 = 0$.
Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $t = \frac{32 \pm \sqrt{1024 - 1008}}{8} = \frac{32 \pm 4}{8}$.
Thus,$t_1 = \frac{36}{8} = \frac{9}{2}$ and $t_2 = \frac{28}{8} = \frac{7}{2}$.
The square root is $\sqrt{\frac{9}{2}} + \sqrt{\frac{7}{2}} = \frac{3}{\sqrt{2}} + \frac{\sqrt{7}}{\sqrt{2}} = \frac{3 + \sqrt{7}}{\sqrt{2}}$.
Rationalizing the denominator by multiplying the numerator and denominator by $\sqrt{2}$,we get $\frac{(3 + \sqrt{7})\sqrt{2}}{2} = \frac{3\sqrt{2} + \sqrt{14}}{2}$.

(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 $\sqrt{(7-\sqrt{6})^2} = 7-\sqrt{6}$.

(B) To simplify the expression,we rationalize each term individually.
First term: $\frac{2 \sqrt{2}}{\sqrt{5}+\sqrt{3}} = \frac{2 \sqrt{2}(\sqrt{5}-\sqrt{3})}{5-3} = \frac{2(\sqrt{10}-\sqrt{6})}{2} = \sqrt{10}-\sqrt{6}$.
Second term: $\frac{3 \sqrt{3}}{\sqrt{5}+\sqrt{2}} = \frac{3 \sqrt{3}(\sqrt{5}-\sqrt{2})}{5-2} = \frac{3(\sqrt{15}-\sqrt{6})}{3} = \sqrt{15}-\sqrt{6}$.
Third term: $\frac{\sqrt{5}}{\sqrt{3}+\sqrt{2}} = \frac{\sqrt{5}(\sqrt{3}-\sqrt{2})}{3-2} = \sqrt{15}-\sqrt{10}$.
Now,substitute these back into the original expression:
$(\sqrt{10}-\sqrt{6}) - (\sqrt{15}-\sqrt{6}) + (\sqrt{15}-\sqrt{10})$
$= \sqrt{10} - \sqrt{6} - \sqrt{15} + \sqrt{6} + \sqrt{15} - \sqrt{10}$
$= 0$.

(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: $\frac{\sqrt{3}(\sqrt{3} + \sqrt{2})}{\sqrt{3} + \sqrt{2}} = \sqrt{3}$

(A) To find the least number of chocolates that are equal for both brands,we need to calculate the Least Common Multiple $(LCM)$ of $24$ and $15$.
Prime factorization of $24 = 2^3 \times 3^1$.
Prime factorization of $15 = 3^1 \times 5^1$.
$LCM(24, 15) = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120$.
To find the number of boxes for the first brand ($24$ chocolates per pack): $120 / 24 = 5$ boxes.
To find the number of boxes for the second brand ($15$ chocolates per pack): $120 / 15 = 8$ boxes.
Therefore,one needs to buy $5$ boxes of the first kind and $8$ boxes of the second kind.

(A) To find the largest number that divides $615$ and $963$ leaving a remainder of $6$ in each case,we first subtract the remainder from both numbers.
$615 - 6 = 609$
$963 - 6 = 957$
Now,we need to find the Highest Common Factor $(HCF)$ of $609$ and $957$.
Prime factorization of $609 = 3 \times 7 \times 29$
Prime factorization of $957 = 3 \times 11 \times 29$
The common factors are $3$ and $29$.
$HCF = 3 \times 29 = 87$.
Therefore,the largest number is $87$.

(B) To find the largest number that divides $626$,$3127$,and $15628$ leaving remainders $1$,$2$,and $3$ respectively,we subtract the remainders from the respective numbers:
$626 - 1 = 625$
$3127 - 2 = 3125$
$15628 - 3 = 15625$
Now,we need to find the Highest Common Factor $(HCF)$ of $625$,$3125$,and $15625$.
Prime factorization of $625 = 5^4$
Prime factorization of $3125 = 5^5$
Prime factorization of $15625 = 5^6$
The $HCF$ is the product of the lowest power of common prime factors,which is $5^4 = 625$.
Therefore,the largest number is $625$.

(C) To find the longest rod that can measure the three dimensions ($15\,m$,$12\,m$,and $21\,m$) exactly,we need to calculate the Highest Common Factor $(HCF)$ of these three numbers.
Step $1$: Find the prime factorization of each dimension:
$15 = 3 \times 5$
$12 = 2^2 \times 3$
$21 = 3 \times 7$
Step $2$: Identify the common prime factors.
The only common prime factor among $15$,$12$,and $21$ is $3$.
Step $3$: The $HCF$ is the product of the lowest powers of common prime factors.
$HCF(15, 12, 21) = 3$.
Therefore,the longest rod that can measure the dimensions exactly is $3\,m$.

(D) To find the smallest five-digit number divisible by $15$,$55$,and $99$,we first find the Least Common Multiple $(LCM)$ of these numbers.
Step $1$: Prime factorization of the numbers:
$15 = 3 \times 5$
$55 = 5 \times 11$
$99 = 3^2 \times 11$
Step $2$: Calculate the $LCM$ by taking the highest power of each prime factor present:
$LCM(15, 55, 99) = 3^2 \times 5 \times 11 = 9 \times 5 \times 11 = 495$.
Step $3$: The smallest five-digit number is $10000$.
Divide $10000$ by $495$:
$10000 \div 495 = 20.202...$
Step $4$: To find the smallest five-digit multiple,we take the next integer quotient,which is $21$.
$495 \times 21 = 10395$.
Thus,the smallest five-digit number divisible by $15$,$55$,and $99$ is $10395$.

(A) To find when the bells toll together,we need to calculate the Least Common Multiple $(LCM)$ of $8$,$12$,$15$,and $18$.
Prime factorization:
$8 = 2^3$
$12 = 2^2 \times 3$
$15 = 3 \times 5$
$18 = 2 \times 3^2$
$LCM = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$ seconds.
This means the bells toll together every $360$ seconds,which is $360 / 60 = 6$ minutes.
In one hour ($60$ minutes),the number of times they toll together is $60 / 6 = 10$ times.
Since we exclude the toll at the start,the total count remains $10$.

(B) Let two consecutive natural numbers be $n$ and $n+1$,where $n \in \mathbb{N}$.
Any common divisor of $n$ and $n+1$ must also divide their difference.
The difference is $(n+1) - n = 1$.
The only positive divisor of $1$ is $1$.
Therefore,the greatest common divisor $(\text{g.c.d.})$ of $n$ and $n+1$ is $1$.

(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.

(D) We know that for any two positive integers $a$ and $b$,the relationship between their $\text{g.c.d.}$ (Greatest Common Divisor) and $\text{l.c.m.}$ (Least Common Multiple) is given by:
$\text{g.c.d.}(a, b) \times \text{l.c.m.}(a, b) = a \times b$
Given that $\text{g.c.d.} = 18$ and the product of the two numbers $(a \times b) = 6480$.
Substituting these values into the formula:
$18 \times \text{l.c.m.} = 6480$
$\text{l.c.m.} = \frac{6480}{18}$
$\text{l.c.m.} = 360$
Therefore,the $\text{l.c.m.}$ of the two numbers is $360$.

(A) By definition,a prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.
Since $k_{1}$ and $k_{2}$ are two distinct prime integers,they share no common factors other than $1$.
For any two numbers $a$ and $b$,the relationship between their $\text{H.C.F.}$ and $\text{L.C.M.}$ is given by the product of the numbers: $a \times b = \text{H.C.F.}(a, b) \times \text{L.C.M.}(a, b)$.
Since $k_{1}$ and $k_{2}$ are distinct primes,their $\text{H.C.F.}(k_{1}, k_{2}) = 1$.
Therefore,$k_{1} \times k_{2} = 1 \times \text{L.C.M.}(k_{1}, k_{2})$.
Thus,$\text{L.C.M.}(k_{1}, k_{2}) = k_{1} k_{2}$.

(B) number is rational if it can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Any terminating decimal or non-terminating repeating decimal is a rational number.
The given number $2.031\overline{2}$ is a non-terminating repeating decimal because the digit $2$ repeats infinitely.
Therefore,$2.031\overline{2}$ is a rational number.

(C) To find the $\text{g.c.d.}$ (Greatest Common Divisor) of $24$ and $63$,we perform prime factorization of both numbers:
$24 = 2^3 \times 3^1$
$63 = 3^2 \times 7^1$
The $\text{g.c.d.}$ is the product of the smallest power of each common prime factor.
The common prime factor is $3$.
The smallest power of $3$ present in both is $3^1 = 3$.
Therefore,$\text{g.c.d.}$ $(24, 63) = 3$.

(D) To find the $\text{l.c.m.}$ of $36$ and $94$,we first find their prime factorizations:
$36 = 2^2 \times 3^2$
$94 = 2 \times 47$
The $\text{l.c.m.}$ is the product of the highest power of each prime factor present in the numbers:
$\text{l.c.m.} = 2^2 \times 3^2 \times 47$
$\text{l.c.m.} = 4 \times 9 \times 47$
$\text{l.c.m.} = 36 \times 47 = 1692$

(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 $\sqrt{a} + \sqrt{b}$.

(B) To find the number of decimal places after which the expansion of a rational number $\frac{p}{q}$ terminates,we express the denominator in the form $2^n \times 5^m$.
Given the expression: $\frac{43}{2^{4} \times 5^{3}}$.
To make the powers of $2$ and $5$ equal,we multiply the numerator and denominator by $5^1$:
$\frac{43 \times 5^1}{2^{4} \times 5^{3} \times 5^1} = \frac{43 \times 5}{2^{4} \times 5^{4}} = \frac{215}{(2 \times 5)^{4}} = \frac{215}{10^4} = \frac{215}{10000} = 0.0215$.
The decimal expansion terminates after $4$ digits.

(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,$\sqrt{8 + 2 \sqrt{12}} = \sqrt{6} + \sqrt{2}$.

(D) The greatest common divisor (g.c.d.) of $12$ and $40$ is calculated as follows:
Prime factorization of $12 = 2^2 \times 3^1$.
Prime factorization of $40 = 2^3 \times 5^1$.
The g.c.d. is the product of the smallest powers of common prime factors: $2^2 = 4$.
Given the equation: $\text{g.c.d.}(12, 40) = 40 + 4x$.
Substituting the value: $4 = 40 + 4x$.
Subtracting $40$ from both sides: $4 - 40 = 4x$.
$-36 = 4x$.
Dividing by $4$: $x = -9$.

(A) To find the decimal form of the fraction $\frac{6}{15}$,we first simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor,which is $3$.
$\frac{6 \div 3}{15 \div 3} = \frac{2}{5}$.
Now,to convert $\frac{2}{5}$ into a decimal,we can multiply both the numerator and the denominator by $2$ to make the denominator $10$:
$\frac{2 \times 2}{5 \times 2} = \frac{4}{10} = 0.4$.
Therefore,the decimal form of $\frac{6}{15}$ is $0.4$.

(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 $0$.

(A) We know that for any two positive integers $a$ and $b$,the relationship between their greatest common divisor $(GCD)$ and least common multiple $(LCM)$ is given by the formula:
$\text{GCD}(a, b) \times \text{LCM}(a, b) = a \times b$
Given $a = 336$ and $b = 52$.
First,we find the prime factorization of $336$ and $52$:
$336 = 2^4 \times 3 \times 7$
$52 = 2^2 \times 13$
The $\text{GCD}(336, 52) = 2^2 = 4$.
Now,using the formula:
$4 \times \text{LCM}(336, 52) = 336 \times 52$
$\text{LCM}(336, 52) = \frac{336 \times 52}{4}$
$\text{LCM}(336, 52) = 84 \times 52 = 4368$
Thus,the correct option is $A$.

(D) To find the remainder when $(3k+2)^2$ is divided by $3$,we expand the expression:
$(3k+2)^2 = (3k)^2 + 2(3k)(2) + 2^2$
$= 9k^2 + 12k + 4$
We can rewrite this as:
$= 9k^2 + 12k + 3 + 1$
$= 3(3k^2 + 4k + 1) + 1$
Since $3(3k^2 + 4k + 1)$ is clearly divisible by $3$,the remainder when the expression is divided by $3$ is $1$.

(A) To find the rationalising factor of an expression of the form $a - \sqrt{b}$,we multiply it by its conjugate $a + \sqrt{b}$ such that the product is a rational number.
Given the expression $-2 - \sqrt{5}$.
Let the rationalising factor be $x$.
We know that $(-2 - \sqrt{5}) \times (-2 + \sqrt{5}) = (-2)^2 - (\sqrt{5})^2$.
$= 4 - 5 = -1$.
Since $-1$ is a rational number,the expression $-2 + \sqrt{5}$ is the rationalising factor.
Alternatively,we can also multiply by $2 + \sqrt{5}$:
$(-2 - \sqrt{5}) \times (2 + \sqrt{5}) = -(2 + \sqrt{5}) \times (2 + \sqrt{5}) = -(4 + 5 + 4\sqrt{5}) = -(9 + 4\sqrt{5})$,which is not rational.
Therefore,the correct rationalising factor is $-2 + \sqrt{5}$.

(B) The given number $0.123123123...$ can be written as $0.\overline{123}$.
Since the decimal expansion is non-terminating and repeating,it can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Let $x = 0.123123...$ (Equation $1$).
Multiplying both sides by $1000$,we get $1000x = 123.123123...$ (Equation $2$).
Subtracting Equation $1$ from Equation $2$:
$1000x - x = 123.123123... - 0.123123...$
$999x = 123$
$x = \frac{123}{999} = \frac{41}{333}$.
Since the number can be expressed as a ratio of two integers,it is a rational number.

(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) rationalising factor is a number that,when multiplied by a given irrational number,results in a rational number.
For the expression $5 \sqrt{2}$,the irrational part is $\sqrt{2}$.
To make $\sqrt{2}$ rational,we multiply it by $\sqrt{2}$ because $\sqrt{2} \times \sqrt{2} = 2$,which is a rational number.
Therefore,the rationalising factor of $5 \sqrt{2}$ is $\sqrt{2}$.

(A) surd is defined as an irrational root of a rational number,such as $\sqrt[n]{x}$,where $x$ is a rational number and the result is irrational.
$\pi$ is a transcendental number. It is irrational,but it cannot be expressed as the $n$-th root of any rational number. Therefore,$\pi$ is an irrational number that is not a surd.
$\sqrt{16} = 4$,which is a rational number.
$\sqrt{8} = 2\sqrt{2}$,which is a surd.
$3\sqrt{27} = 3 \times 3\sqrt{3} = 9\sqrt{3}$,which is a surd.
Thus,the correct option is $A$.

(A) Bezout's identity states that for any two integers $a$ and $b$ (not both zero),there exist integers $x$ and $y$ such that $ax + by = \text{gcd}(a, b)$,where $\text{gcd}(a, b)$ is the greatest common divisor of $a$ and $b$.

(B) An integer is defined as odd if it is not divisible by $2$.
By Euclid's division lemma,any integer $a$ can be expressed as $a = bq + r$,where $0 \le r < b$.
For divisibility by $2$,we take $b = 2$.
Thus,$a = 2k + r$,where $r$ can be $0$ or $1$.
If $r = 0$,then $a = 2k$,which is an even integer.
If $r = 1$,then $a = 2k + 1$,which is an odd integer.
Therefore,every odd integer is of the form $2k + 1$ for some integer $k$.

(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 $6 \times 4 = 24$.

(D) To find the $L.C.M.$ of $115$ and $25$,we first find their prime factorizations:
$115 = 5 \times 23$
$25 = 5 \times 5 = 5^2$
The $L.C.M.$ is the product of the highest power of each prime factor present in the numbers:
$L.C.M.(115, 25) = 5^2 \times 23$
$L.C.M.(115, 25) = 25 \times 23 = 575$

(C) To find the Greatest Common Divisor ($G$.$C$.$D$.) of $28$,$35$,and $91$,we first find the prime factorization of each number:
$28 = 2^2 \times 7$
$35 = 5 \times 7$
$91 = 7 \times 13$
The $G$.$C$.$D$. is the product of the smallest power of each common prime factor present in all the numbers.
Here,the only common prime factor is $7$,and its smallest power is $7^1$.
Therefore,$G$.$C$.$D$. $(28, 35, 91) = 7$.

(B) For any two positive integers $a$ and $b$,the fundamental theorem of arithmetic establishes a relationship between their $G$.$C$.$D$. and $L$.$C$.$M$.
Specifically,the product of the $G$.$C$.$D$. $(a, b)$ and the $L$.$C$.$M$. $(a, b)$ is equal to the product of the two numbers themselves.
Therefore,$\text{G.C.D.}(a, b) \times \text{L.C.M.}(a, b) = a \times b = ab$.

(A) To find the $LCM$ of $15, 21,$ and $35$,we first find the prime factorization of each number:
$15 = 3 \times 5$
$21 = 3 \times 7$
$35 = 5 \times 7$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM(15, 21, 35) = 3^1 \times 5^1 \times 7^1 = 3 \times 5 \times 7 = 105$
Therefore,the correct option is $A$.

(C) To find the $LCM$ of $40, 60,$ and $80$,we first find the prime factorization of each number:
$40 = 2^3 \times 5^1$
$60 = 2^2 \times 3^1 \times 5^1$
$80 = 2^4 \times 5^1$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM = 2^4 \times 3^1 \times 5^1$
$LCM = 16 \times 3 \times 5$
$LCM = 16 \times 15 = 240$

(D) To find the smallest number that leaves a remainder of $5$ when divided by $20, 30,$ and $40,$ we first find the Least Common Multiple $(LCM)$ of the divisors.
$LCM(20, 30, 40)$:
$20 = 2^2 \times 5$
$30 = 2 \times 3 \times 5$
$40 = 2^3 \times 5$
$LCM = 2^3 \times 3 \times 5 = 8 \times 3 \times 5 = 120$.
The number must be of the form $(LCM \times k) + \text{remainder}$.
Here,$LCM = 120$ and $\text{remainder} = 5$.
So,the number $= 120k + 5$.
For the smallest positive number greater than $5$,we take $k = 1$.
Number $= 120(1) + 5 = 125$.

(C) To find the smallest positive number divisible by $24, 36$,and $48$,we need to calculate the Least Common Multiple $(LCM)$ of these numbers.
Step $1$: Find the prime factorization of each number:
$24 = 2^3 \times 3^1$
$36 = 2^2 \times 3^2$
$48 = 2^4 \times 3^1$
Step $2$: The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM = 2^4 \times 3^2$
$LCM = 16 \times 9 = 144$
Therefore,the required smallest number is $144$.

(D) The integers from $2$ to $6$ are $2, 3, 4, 5,$ and $6$.
To find the smallest positive number divisible by all these integers,we need to calculate the Least Common Multiple $(LCM)$ of $2, 3, 4, 5,$ and $6$.
Prime factorization of each number:
$2 = 2^1$
$3 = 3^1$
$4 = 2^2$
$5 = 5^1$
$6 = 2^1 \times 3^1$
Taking the highest power of each prime factor present:
$LCM = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$.
Therefore,$60$ is the smallest positive number divisible by $2, 3, 4, 5,$ and $6$.

(B) To find the smallest positive number divisible by every integer from $2$ to $10$,we need to calculate the Least Common Multiple $(LCM)$ of the numbers $2, 3, 4, 5, 6, 7, 8, 9, 10$.
First,write the prime factorization of each number:
$2 = 2^1$
$3 = 3^1$
$4 = 2^2$
$5 = 5^1$
$6 = 2 \times 3$
$7 = 7^1$
$8 = 2^3$
$9 = 3^2$
$10 = 2 \times 5$
The $LCM$ is found by taking the highest power of each prime factor present in the numbers:
$LCM = 2^3 \times 3^2 \times 5^1 \times 7^1$
$LCM = 8 \times 9 \times 5 \times 7$
$LCM = 72 \times 35 = 2520$.
Therefore,the smallest positive number divisible by all integers from $2$ to $10$ is $2520$.

(C) According to the fundamental property of numbers,for any two positive integers $a$ and $b$,the product of their $\text{g.c.d.}$ and $\text{l.c.m.}$ is equal to the product of the numbers themselves.
Formula: $\text{g.c.d.}(a, b) \times \text{l.c.m.}(a, b) = a \times b$.
Here,$a = 18$ and $b = 24$.
Therefore,$\text{g.c.d.}(18, 24) \times \text{l.c.m.}(18, 24) = 18 \times 24$.
Calculating the product: $18 \times 24 = 432$.

(B) We know the fundamental relationship between the greatest common divisor (g.c.d.) and the least common multiple (l.c.m.) of two positive integers $a$ and $b$ is given by the formula:
$\text{l.c.m.} (a, b) \times \text{g.c.d.} (a, b) = a \times b$
Given that $\text{g.c.d.} (a, b) = b$,we can substitute this into the formula:
$\text{l.c.m.} (a, b) \times b = a \times b$
Dividing both sides by $b$ (since $b \in N$,$b \neq 0$):
$\text{l.c.m.} (a, b) = \frac{a \times b}{b} = a$
Therefore,the correct option is $B$.