HCF and LCM Questions in English

(B) Let the two numbers be $39x$ and $39y$,where $x$ and $y$ are coprime (i.e.,$gcd(x, y) = 1$).
Given that the sum of the numbers is $468$,we have $39x + 39y = 468$.
Dividing by $39$,we get $x + y = \frac{468}{39} = 12$.
We need to find pairs of coprime numbers $(x, y)$ such that $x + y = 12$ and $x < y$ (to avoid counting the same pair twice).
The possible pairs $(x, y)$ that sum to $12$ are:
$(1, 11)$ where $gcd(1, 11) = 1$ (Valid)
$(2, 10)$ where $gcd(2, 10) = 2$ (Invalid)
$(3, 9)$ where $gcd(3, 9) = 3$ (Invalid)
$(4, 8)$ where $gcd(4, 8) = 4$ (Invalid)
$(5, 7)$ where $gcd(5, 7) = 1$ (Valid)
$(6, 6)$ where $gcd(6, 6) = 6$ (Invalid)
Thus,there are $2$ such pairs: $(1, 11)$ and $(5, 7)$.

(D) The least number which is divisible by $6, 8$ and $12$ is the $LCM$ of $6, 8$ and $12$,which is $24$.
We need to find the numbers between $500$ and $700$ that are multiples of $24$.
Dividing $500$ by $24$: $500 = 24 \times 20 + 20$. The first multiple of $24$ greater than $500$ is $24 \times 21 = 504$.
Dividing $700$ by $24$: $700 = 24 \times 29 + 4$. The last multiple of $24$ less than $700$ is $24 \times 29 = 696$.
These numbers form an arithmetic progression: $504, 528, 552, \dots, 696$.
Here,the first term $a = 504$,the last term $l = 696$,and the common difference $d = 24$.
The number of terms $n$ is given by $l = a + (n - 1)d$,so $696 = 504 + (n - 1)24$.
$192 = (n - 1)24 \implies n - 1 = 8 \implies n = 9$.
The sum $S_n = \frac{n}{2}(a + l) = \frac{9}{2}(504 + 696) = \frac{9}{2}(1200) = 9 \times 600 = 5400$.

(D) Let the two parts be $15x$ and $15y$,where $x$ and $y$ are coprime integers.
Their sum is $15x + 15y = 1000$.
Dividing by $15$,we get $x + y = 1000 / 15 = 200 / 3$.
Since $x$ and $y$ must be integers,their sum $x + y$ must also be an integer.
However,$200 / 3$ is not an integer $(66.66...)$.
Therefore,it is not possible to divide $1000$ into two parts such that their $HCF$ is $15$.
Hence,the correct answer is $(d)$.

(D) Number of boys $= 442$; Number of girls $= 374$.
To find the largest possible equal number of students in each class,we calculate the $HCF$ of $442$ and $374$.
Prime factorization of $442 = 2 \times 13 \times 17$.
Prime factorization of $374 = 2 \times 11 \times 17$.
$HCF(442, 374) = 2 \times 17 = 34$.
So,each class contains $34$ students.
Number of classes of boys $= 442 \div 34 = 13$.
Number of classes of girls $= 374 \div 34 = 11$.
Total number of classes $= 13 + 11 = 24$.

(B) The greatest number of $4$ digits is $9999$.
Dividing $9999$ by $196$,we get $9999 = 196 \times 51 + 3$. The remainder is $3$.
Thus,the greatest $4$-digit number divisible by $196$ is $9999 - 3 = 9996$.
The least number of $5$ digits is $10000$.
Dividing $10000$ by $196$,we get $10000 = 196 \times 51 + 4$. The remainder is $4$.
To find the least $5$-digit number divisible by $196$,we add $(196 - 4)$ to $10000$.
Thus,the required $5$-digit number is $10000 + (196 - 4) = 10000 + 192 = 10192$.

(B) Let the two numbers be $9a$ and $9b$,where $a$ and $b$ are co-prime numbers.
Given that the product of the numbers is $4212$.
So,$(9a) \times (9b) = 4212$.
$81ab = 4212$.
$ab = \frac{4212}{81} = 52$.
Now,we find pairs of co-prime factors of $52$. The factors of $52$ are $(1, 52)$ and $(4, 13)$.
Case $1$: If $a=1, b=52$,then the numbers are $9 \times 1 = 9$ and $9 \times 52 = 468$.
Case $2$: If $a=4, b=13$,then the numbers are $9 \times 4 = 36$ and $9 \times 13 = 117$.
Comparing with the given options,the pair $(36, 117)$ is correct.

(B) Given that $HCF = 36$ and $LCM = 756$.
We know that the product of two numbers is equal to the product of their $HCF$ and $LCM$.
Let the two numbers be $36a$ and $36b$,where $a$ and $b$ are coprime numbers.
Then,$HCF \times LCM = 36a \times 36b = 36 \times 756$.
Dividing both sides by $36^2$,we get $a \times b = \frac{756}{36} = 21$.
The possible pairs of coprime factors of $21$ are $(1, 21)$ and $(3, 7)$.
Case $1$: If $(a, b) = (1, 21)$,the numbers are $(36 \times 1, 36 \times 21) = (36, 756)$.
Case $2$: If $(a, b) = (3, 7)$,the numbers are $(36 \times 3, 36 \times 7) = (108, 252)$.
Comparing with the given options,the pair $(108, 252)$ is present.

(B) To find the maximum length that can measure the given lengths exactly,we need to calculate the $HCF$ (Highest Common Factor) of the lengths.
First,convert all lengths into centimeters $(cm)$:
$6 \, m = 600 \, cm$
$4 \, m \, 75 \, cm = 475 \, cm$
$10 \, m \, 25 \, cm = 1025 \, cm$
Now,find the $HCF$ of $600$,$475$,and $1025$:
Prime factorization of $600 = 2^3 \times 3 \times 5^2$
Prime factorization of $475 = 5^2 \times 19$
Prime factorization of $1025 = 5^2 \times 41$
The common factor with the lowest exponent is $5^2 = 25$.
Therefore,the maximum possible length is $25 \, cm$.

(B) Let the second number be $x$.
The product of two numbers is equal to the product of their $HCF$ and $LCM$.
$x \times 144 = 36 \times 1008$
$x = \frac{36 \times 1008}{144}$
Since $144 = 36 \times 4$,we have:
$x = \frac{36 \times 1008}{36 \times 4} = \frac{1008}{4} = 252$
Therefore,the other number is $252$.

(C) Since the bells ring at intervals of $6$ seconds,$7$ seconds,and $8$ seconds respectively,the next time they will ring together is after an interval equal to the $LCM$ of $6, 7,$ and $8$ seconds.
To find the $LCM$ of $6, 7,$ and $8$:
$6 = 2 \times 3$
$7 = 7 \times 1$
$8 = 2^3$
$LCM = 2^3 \times 3 \times 7 = 8 \times 3 \times 7 = 24 \times 7 = 168$.
Therefore,the required time is $168$ seconds.

(A) To find the least number of square tiles,we must first determine the maximum size of a square tile that can perfectly fit the dimensions of the hall.
The maximum side length of the square tile is the $HCF$ of the length and breadth of the hall.
Length $= 17 \, m \, 55 \, cm = 1755 \, cm$.
Breadth $= 8 \, m \, 10 \, cm = 810 \, cm$.
Now,find the $HCF$ of $1755$ and $810$:
$1755 = 135 \times 13$
$810 = 135 \times 6$
So,$HCF = 135 \, cm$.
The number of tiles required $= \frac{\text{Area of the hall}}{\text{Area of one tile}} = \frac{1755 \times 810}{135 \times 135}$.
$= \frac{1755}{135} \times \frac{810}{135} = 13 \times 6 = 78$.

(D) To find the smallest whole number exactly divisible by a set of numbers,we need to calculate the Least Common Multiple $(LCM)$ of those numbers.
Step $1$: Find the prime factorization of each number:
$6 = 2 \times 3$
$8 = 2^3$
$15 = 3 \times 5$
$18 = 2 \times 3^2$
$24 = 2^3 \times 3$
Step $2$: Identify the highest power of each prime factor present in the factorizations:
Prime factors involved are $2, 3,$ and $5$.
Highest power of $2 = 2^3 = 8$
Highest power of $3 = 3^2 = 9$
Highest power of $5 = 5^1 = 5$
Step $3$: Multiply these highest powers to find the $LCM$:
$LCM = 8 \times 9 \times 5 = 72 \times 5 = 360$.
Therefore,the smallest whole number divisible by $6, 8, 15, 18,$ and $24$ is $360$.

(C) To find the highest common factor $(HCF)$ of $36$ and $84$,we perform prime factorization for both numbers:
$36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$
$84 = 2 \times 2 \times 3 \times 7 = 2^2 \times 3^1 \times 7^1$
The $HCF$ is the product of the lowest powers of common prime factors:
$HCF = 2^2 \times 3^1 = 4 \times 3 = 12$
Therefore,the highest common factor of $36$ and $84$ is $12$.

(C) To find the Least Common Multiple $(LCM)$ of $24, 36,$ and $40$,we use the prime factorization method or the division method.
Using the division method:
$\begin{array}{c|ccc} 2 & 24, & 36, & 40 \\ \hline 2 & 12, & 18, & 20 \\ \hline 2 & 6, & 9, & 10 \\ \hline 3 & 3, & 9, & 5 \\ \hline & 1, & 3, & 5 \end{array}$
$\therefore LCM = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 360$
Thus,the $LCM$ of $24, 36,$ and $40$ is $360$.

(D) Let the two numbers be $3x$ and $4x$,where $x$ is the common factor.
Given that the $HCF$ of the numbers is $4$,we have $x = 4$.
Therefore,the numbers are $3 \times 4 = 12$ and $4 \times 4 = 16$.
The $LCM$ of two numbers is given by the product of the ratio terms and the $HCF$: $LCM = 3 \times 4 \times 4 = 48$.

(C) Let the two numbers be $37x$ and $37y$,where $x$ and $y$ are coprime numbers.
Given that the product of the numbers is $4107$.
So,$(37x) \times (37y) = 4107$.
$1369 \times (xy) = 4107$.
$xy = \frac{4107}{1369} = 3$.
Since $x$ and $y$ are coprime,the possible pairs for $(x, y)$ are $(1, 3)$.
Therefore,the numbers are $37 \times 1 = 37$ and $37 \times 3 = 111$.
The greater number is $111$.

(C) Given: $HCF = 21$ and $LCM = 84$.
Let the two numbers be $21x$ and $21y$,where $x$ and $y$ are coprime.
The ratio of the numbers is given as $1:4$,so $\frac{21x}{21y} = \frac{1}{4}$,which implies $\frac{x}{y} = \frac{1}{4}$.
We know the property: Product of two numbers $= HCF \times LCM$.
So,$(21x) \times (21y) = 21 \times 84$.
$441(xy) = 1764$.
$xy = \frac{1764}{441} = 4$.
Since $\frac{x}{y} = \frac{1}{4}$,we have $y = 4x$. Substituting this into $xy = 4$,we get $x(4x) = 4$,so $4x^2 = 4$,which means $x^2 = 1$,so $x = 1$.
Then $y = 4(1) = 4$.
The numbers are $21(1) = 21$ and $21(4) = 84$.
The larger number is $84$.

(D) The $HCF$ (Highest Common Factor) of two numbers always divides their $LCM$ (Least Common Multiple) exactly.
Given that the $HCF$ is $8$,the $LCM$ must be a multiple of $8$.
Checking the options:
$24 \div 8 = 3$ (Divisible)
$48 \div 8 = 6$ (Divisible)
$56 \div 8 = 7$ (Divisible)
$60 \div 8 = 7.5$ (Not divisible)
Therefore,$60$ can never be the $LCM$ of two numbers whose $HCF$ is $8$.

(C) Let the two numbers be $29x$ and $29y$,where $x$ and $y$ are coprime.
Given that $LCM = 29 \times x \times y = 4147$.
Therefore,$xy = \frac{4147}{29} = 143$.
The factors of $143$ are $(1, 143)$ and $(11, 13)$.
If we choose $(1, 143)$,the numbers are $29 \times 1 = 29$ and $29 \times 143 = 4147$. However,the problem states that both numbers must be greater than $29$.
If we choose $(11, 13)$,the numbers are $29 \times 11 = 319$ and $29 \times 13 = 377$.
Both $319$ and $377$ are greater than $29$.
Therefore,the sum of the numbers is $319 + 377 = 696$.

(B) First,find the $LCM$ of $5, 6, 7$ and $8$.
$\begin{array}{c|cccc} 2 & 5, & 6, & 7, & 8 \\ \hline & 5, & 3, & 7, & 4 \end{array}$
$LCM = 2 \times 5 \times 3 \times 7 \times 4 = 840$.
The number must be of the form $840k + 3$,where $k$ is a positive integer.
We need to find the smallest $k$ such that $(840k + 3)$ is divisible by $9$.
$840k + 3 = (837k + 3k) + 3 = 9(93k) + 3k + 3$.
For $(840k + 3)$ to be divisible by $9$,$(3k + 3)$ must be divisible by $9$.
If $k = 1$,$3(1) + 3 = 6$ (not divisible by $9$).
If $k = 2$,$3(2) + 3 = 9$ (divisible by $9$).
Therefore,the required number is $840(2) + 3 = 1680 + 3 = 1683$.

(D) First,find the $LCM$ of $6, 9, 15,$ and $18$.
$\begin{array}{c|cccc} 2 & 6, & 9, & 15, & 18 \\ \hline 3 & 3, & 9, & 15, & 9 \\ \hline 3 & 1, & 3, & 5, & 3 \\ \hline & 1, & 1, & 5, & 1 \end{array}$
$LCM = 2 \times 3 \times 3 \times 5 = 90$.
Let the number be of the form $90k + 4$,where $k$ is a positive integer.
We are given that the number is divisible by $7$. Therefore,$90k + 4 \equiv 0 \pmod{7}$.
Simplifying $90 \pmod{7}$: $90 = 7 \times 12 + 6$,so $90 \equiv 6 \equiv -1 \pmod{7}$.
Thus,$-k + 4 \equiv 0 \pmod{7}$,which means $k \equiv 4 \pmod{7}$.
The least positive integer $k$ is $4$.
Substituting $k = 4$ into the expression $90k + 4$:
Required number $= 90(4) + 4 = 360 + 4 = 364$.

(D) To find the least number that leaves a remainder of $8$ when divided by $12, 15, 20$ and $54$, we first find the Least Common Multiple $(LCM)$ of these numbers.
$\begin{array}{r|rrrr} 2 & 12, & 15, & 20, & 54 \\ \hline 2 & 6, & 15, & 10, & 27 \\ \hline 3 & 3, & 15, & 5, & 27 \\ \hline 3 & 1, & 5, & 5, & 9 \\ \hline 5 & 1, & 1, & 1, & 3 \end{array}$
$LCM = 2 \times 2 \times 3 \times 3 \times 5 \times 3 = 540$.
The required number is $LCM + \text{remainder} = 540 + 8 = 548$.

(B) To find the smallest number which when diminished by $7$ is divisible by $12, 16, 18, 21,$ and $28,$ we first need to find the Least Common Multiple $(LCM)$ of these numbers.
Step $1$: Find the prime factorization of each number:
$12 = 2^2 \times 3$
$16 = 2^4$
$18 = 2 \times 3^2$
$21 = 3 \times 7$
$28 = 2^2 \times 7$
Step $2$: The $LCM$ is the product of the highest powers of all prime factors present:
$LCM = 2^4 \times 3^2 \times 7 = 16 \times 9 \times 7 = 1008$
Step $3$: The number which when diminished by $7$ gives $1008$ is $1008 + 7 = 1015$.
Therefore,the required number is $1015$.

(B) To find the least number which when doubled is divisible by $12, 18, 21$ and $30$,we first find the Least Common Multiple $(LCM)$ of these numbers.
$LCM$ of $12, 18, 21, 30$:

$2$	$12, 18, 21, 30$
$3$	$6, 9, 21, 15$
	$2, 3, 7, 5$

$LCM = 2 \times 3 \times 2 \times 3 \times 7 \times 5 = 1260$.
Let the required number be $x$. According to the problem,$2x$ must be divisible by $1260$.
Therefore,$2x = 1260$.
$x = \frac{1260}{2} = 630$.
Thus,the required number is $630$.

(B) To find the greatest number that divides $1657$ and $2037$ leaving remainders $6$ and $5$ respectively,we need to find the $HCF$ of $(1657 - 6)$ and $(2037 - 5)$.
Step $1$: Subtract the remainders from the numbers.
$1657 - 6 = 1651$
$2037 - 5 = 2032$
Step $2$: Find the $HCF$ of $1651$ and $2032$ using the division method.
$2032 = 1651 \times 1 + 381$
$1651 = 381 \times 4 + 127$
$381 = 127 \times 3 + 0$
Since the remainder is $0$,the $HCF$ is $127$.
Therefore,the greatest number is $127$.

(A) To find the greatest number $N$ that divides $a, b,$ and $c$ leaving the same remainder,we calculate the $HCF$ of the differences of the numbers: $(b-a), (c-b),$ and $(c-a)$.
Here,$a = 1305, b = 4665, c = 6905$.
Differences are:
$4665 - 1305 = 3360$
$6905 - 4665 = 2240$
$6905 - 1305 = 5600$
Now,find the $HCF$ of $3360, 2240,$ and $5600$.
$3360 = 1120 \times 3$
$2240 = 1120 \times 2$
$5600 = 1120 \times 5$
Thus,$N = HCF(3360, 2240, 5600) = 1120$.
The sum of the digits of $N = 1 + 1 + 2 + 0 = 4$.

(A) To find the greatest number that divides $43, 91,$ and $183$ leaving the same remainder,we calculate the differences between the numbers:
$91 - 43 = 48$
$183 - 91 = 92$
$183 - 43 = 140$
Now,we find the $HCF$ of these differences: $48, 92,$ and $140$.
Prime factorization of $48 = 2^4 \times 3$
Prime factorization of $92 = 2^2 \times 23$
Prime factorization of $140 = 2^2 \times 5 \times 7$
The common factor with the lowest power is $2^2 = 4$.
Therefore,the greatest number is $4$.

(A) To find the maximum number of students,we need to calculate the Highest Common Factor $(HCF)$ of $1001$ and $910$.
Using the division method:
$1001 = 910 \times 1 + 91$
$910 = 91 \times 10 + 0$
Since the remainder is $0$,the $HCF$ of $1001$ and $910$ is $91$.
Therefore,the maximum number of students is $91$.

(C) To find the greatest possible length that can measure the given lengths exactly,we need to find the Highest Common Factor $(HCF)$ of the given values.
First,convert all lengths into centimeters $(cm)$:
$7 \ m = 700 \ cm$
$3 \ m \ 85 \ cm = 385 \ cm$
$12 \ m \ 95 \ cm = 1295 \ cm$
Now,find the $HCF$ of $700$,$385$,and $1295$.
Using the division method:
$HCF$ of $700$ and $385$:
$700 = 385 \times 1 + 315$
$385 = 315 \times 1 + 70$
$315 = 70 \times 4 + 35$
$70 = 35 \times 2 + 0$
So,$HCF$ of $700$ and $385$ is $35$.
Now,find the $HCF$ of $35$ and $1295$:
$1295 = 35 \times 37 + 0$
Thus,the $HCF$ of $700$,$385$,and $1295$ is $35$.
The greatest possible length is $35 \ cm$.

(A) Let the two numbers be $a$ and $b$.
We know that the product of two numbers is equal to the product of their $HCF$ and $LCM$,so $a \times b = HCF \times LCM$.
Given $HCF = 11$ and $LCM = 385$.
Therefore,$a \times b = 11 \times 385 = 11 \times (5 \times 7 \times 11) = (11 \times 5) \times (11 \times 7) = 55 \times 77$.
Since both numbers must be multiples of the $HCF$ $(11)$,the possible factors are $11 \times 1, 11 \times 5, 11 \times 7, 11 \times 35$.
The two numbers are $55$ and $77$.
We are given that one number lies between $75$ and $125$.
Since $77$ is between $75$ and $125$,the required number is $77$.

(C) Let the two numbers be $a$ and $b$.
Given that the $HCF$ of the two numbers is $23$.
Any two numbers can be expressed as $a = HCF \times x$ and $b = HCF \times y$,where $x$ and $y$ are coprime factors.
The $LCM$ of these two numbers is given by $LCM = HCF \times x \times y$.
We are given that the other two factors of the $LCM$ are $13$ and $14$,so $x = 13$ and $y = 14$.
The two numbers are:
$a = 23 \times 13 = 299$
$b = 23 \times 14 = 322$
Comparing the two numbers,the larger number is $322$.

(B) Two numbers are said to be co-prime if their $HCF$ (Highest Common Factor) is $1$.
We know the relationship between two numbers,their $HCF$,and their $LCM$ is given by:
$\text{Product of two numbers} = HCF \times LCM$
Given that the product of the two co-prime numbers is $117$ and their $HCF$ is $1$,we have:
$117 = 1 \times LCM$
Therefore,$LCM = 117$.

(C) Let the two numbers be $x$ and $y$.
Given that the sum of the numbers is $x + y = 55$ $....(1)$
We are given $HCF = 5$ and $LCM = 120$.
We know the property that for any two numbers,the product of the numbers is equal to the product of their $HCF$ and $LCM$:
$x \times y = HCF \times LCM$
$x \times y = 5 \times 120 = 600$ $....(2)$
We need to find the sum of the reciprocals of the numbers,which is $\frac{1}{x} + \frac{1}{y}$.
$\frac{1}{x} + \frac{1}{y} = \frac{x + y}{x \times y}$
Substituting the values from equation $(1)$ and $(2)$:
$\frac{1}{x} + \frac{1}{y} = \frac{55}{600}$
Simplifying the fraction by dividing both numerator and denominator by $5$:
$\frac{55 \div 5}{600 \div 5} = \frac{11}{120}$

(A) Let the two numbers be $x$ and $y$.
Given that the sum of the numbers is $x + y = 100$ $....(1)$
We know that for any two numbers,the product of the numbers is equal to the product of their $LCM$ and $HCF$.
$x \times y = LCM \times HCF$
$x \times y = 495 \times 5 = 2475$ $....(2)$
We need to find the difference,$|x - y|$.
Using the algebraic identity $(x - y)^2 = (x + y)^2 - 4xy$:
$(x - y)^2 = (100)^2 - 4(2475)$
$(x - y)^2 = 10000 - 9900$
$(x - y)^2 = 100$
Taking the square root on both sides,we get:
$x - y = \sqrt{100} = 10$
Therefore,the difference between the two numbers is $10$.

(C) Given: $HCF = 11$,$LCM = 7700$,and one number $x = 275$.
We know the relationship: $\text{Product of two numbers} = HCF \times LCM$.
Let the other number be $y$.
So,$275 \times y = 11 \times 7700$.
$y = \frac{11 \times 7700}{275}$.
Dividing $275$ by $11$,we get $25$.
$y = \frac{7700}{25}$.
$y = 308$.
Therefore,the other number is $308$.

(B) Let the two numbers be $x$ and $y$.
Given that the product $x \times y = 2028$ and the $HCF = 13$.
Since $HCF$ is a factor of both numbers,we can write $x = 13a$ and $y = 13b$,where $a$ and $b$ are coprime numbers (i.e.,$gcd(a, b) = 1$).
Substituting these into the product equation: $(13a) \times (13b) = 2028$.
$169 \times ab = 2028$.
$ab = \frac{2028}{169} = 12$.
Now,we find pairs of factors $(a, b)$ such that their product is $12$ and they are coprime:
$1$. $(1, 12)$ where $gcd(1, 12) = 1$.
$2$. $(3, 4)$ where $gcd(3, 4) = 1$.
Note: $(2, 6)$ is not considered because $gcd(2, 6) = 2 \neq 1$.
Thus,there are $2$ such pairs of numbers.

(A) Let the two numbers be $33a$ and $33b$,where $a$ and $b$ are coprime numbers (i.e.,$gcd(a, b) = 1$).
Given that the sum of the numbers is $528$,we have $33a + 33b = 528$.
Dividing by $33$,we get $a + b = \frac{528}{33} = 16$.
We need to find pairs $(a, b)$ such that $a + b = 16$ and $gcd(a, b) = 1$ (since $HCF$ is $33$):
Possible pairs $(a, b)$ where $a < b$ are $(1, 15), (3, 13), (5, 11), (7, 9)$.
Note: $(2, 14), (4, 12), (6, 10), (8, 8)$ are excluded because they have a common factor other than $1$.
Thus,there are $4$ such pairs.

(D) Let the three numbers be $x, 2x,$ and $3x$ where $x$ is a common factor.
Since the $HCF$ of these numbers is $12$,the common factor $x$ must be equal to $12$.
Substituting the value of $x$ into the expressions:
First number $= 1 \times 12 = 12$
Second number $= 2 \times 12 = 24$
Third number $= 3 \times 12 = 36$
Therefore,the numbers are $12, 24,$ and $36$.
Hence,the correct option is $(d)$.

(C) To find the $LCM$ of $22, 54, 108, 135,$ and $198$,we perform prime factorization:
$22 = 2 \times 11$
$54 = 2 \times 3^3$
$108 = 2^2 \times 3^3$
$135 = 3^3 \times 5$
$198 = 2 \times 3^2 \times 11$
The $LCM$ is the product of the highest powers of all prime factors present:
$LCM = 2^2 \times 3^3 \times 5 \times 11 = 4 \times 27 \times 5 \times 11 = 5940$
Alternatively,using the division method:

$2$	$22, 54, 108, 135, 198$
$2$	$11, 27, 54, 135, 99$
$3$	$11, 27, 27, 135, 99$
$3$	$11, 9, 9, 45, 33$
$3$	$11, 3, 3, 15, 11$
$5$	$11, 1, 1, 5, 11$
$11$	$11, 1, 1, 1, 11$
	$1, 1, 1, 1, 1$

$LCM = 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 11 = 5940$.

(A) First,express each number in its prime factorization form:
$4 \times 27 \times 3125 = 2^2 \times 3^3 \times 5^5$
$8 \times 9 \times 25 \times 7 = 2^3 \times 3^2 \times 5^2 \times 7^1$
$16 \times 81 \times 5 \times 11 \times 49 = 2^4 \times 3^4 \times 5^1 \times 11^1 \times 7^2$
To find the $HCF$,take the lowest power of each common prime factor present in all three numbers:
Common prime factors are $2, 3,$ and $5$.
Lowest power of $2$ is $2^2 = 4$.
Lowest power of $3$ is $3^2 = 9$.
Lowest power of $5$ is $5^1 = 5$.
Therefore,$HCF = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$.

(B) To simplify the fraction $\frac{1095}{1168}$,we need to find the Greatest Common Divisor $(GCD)$ of $1095$ and $1168$.
Using the Euclidean algorithm:
$1168 = 1 \times 1095 + 73$
$1095 = 15 \times 73 + 0$
Thus,the $GCD$ is $73$.
Now,divide both the numerator and the denominator by $73$:
$\frac{1095 \div 73}{1168 \div 73} = \frac{15}{16}$.

(C) Let the three co-prime numbers be $x, y,$ and $z$.
Given that the product of the first two is $xy = 551$ and the product of the last two is $yz = 1073$.
Since $y$ is a common factor of both products,we find the Highest Common Factor $(HCF)$ of $551$ and $1073$.
Using the division method:
$1073 = 551 \times 1 + 522$
$551 = 522 \times 1 + 29$
$522 = 29 \times 18 + 0$
Thus,the $HCF$ is $29$,which means $y = 29$.
Now,find $x$ and $z$:
$x = 551 / 29 = 19$
$z = 1073 / 29 = 37$
The three numbers are $19, 29,$ and $37$. These are co-prime to each other.
The sum of the three numbers is $19 + 29 + 37 = 85$.

(C) Given that the $LCM$ of two numbers is $48$ and the ratio of the numbers is $2:3$.
Let the two numbers be $2x$ and $3x$,where $x$ is a common multiplier.
The $LCM$ of $2x$ and $3x$ is $2 \times 3 \times x = 6x$.
According to the problem,$6x = 48$.
Solving for $x$,we get $x = 48 / 6 = 8$.
Therefore,the numbers are $2 \times 8 = 16$ and $3 \times 8 = 24$.
The sum of the numbers is $16 + 24 = 40$.

(A) Let the three numbers be $3x, 4x,$ and $5x$, where $x$ is the $HCF$ of the numbers.
The $LCM$ of these numbers is given by the product of the common factor $x$ and the $LCM$ of the remaining ratios $(3, 4, 5)$.
$LCM = x \times LCM(3, 4, 5) = x \times 60 = 60x$.
Given that the $LCM$ is $2400$, we have:
$60x = 2400$.
Solving for $x$:
$x = 2400 / 60 = 40$.
Since $x$ represents the $HCF$, the $HCF$ of the numbers is $40$.

(A) Given that the $LCM$ of two prime numbers $x$ and $y$ is $161$.
Since $x$ and $y$ are prime numbers,their $HCF$ is $1$.
We know that for any two numbers,$LCM \times HCF = \text{Product of the numbers}$.
Therefore,$x \times y = 161$.
By prime factorization of $161$,we get $161 = 7 \times 23$.
Since $x > y$,we assign $x = 23$ and $y = 7$.
Now,calculating the value of $3y - x$:
$3(7) - 23 = 21 - 23 = -2$.

(B) First, observe the difference between each divisor and its corresponding remainder:
$48 - 38 = 10$
$60 - 50 = 10$
$72 - 62 = 10$
$108 - 98 = 10$
$140 - 130 = 10$
Since the difference is constant $(10)$, the required number is $(LCM \text{ of } 48, 60, 72, 108, 140) - 10$.
Prime factorization:
$48 = 2^4 \times 3$
$60 = 2^2 \times 3 \times 5$
$72 = 2^3 \times 3^2$
$108 = 2^2 \times 3^3$
$140 = 2^2 \times 5 \times 7$
$LCM = 2^4 \times 3^3 \times 5 \times 7 = 16 \times 27 \times 35 = 15120$.
Therefore, the required number $= 15120 - 10 = 15110$.

(C) First,find the $LCM$ of $4, 6, 10,$ and $15$.
$4 = 2^2$
$6 = 2 \times 3$
$10 = 2 \times 5$
$15 = 3 \times 5$
$LCM = 2^2 \times 3 \times 5 = 60$.
The smallest six-digit number is $100000$.
Divide $100000$ by $60$:
$100000 = 60 \times 1666 + 40$.
To find the smallest six-digit number exactly divisible by $60$,we add $(60 - 40) = 20$ to $100000$.
So,the number is $100000 + 20 = 100020$.
Since the number must leave a remainder of $2$ in each case,we add $2$ to this result:
$N = 100020 + 2 = 100022$.
The sum of the digits of $N$ is $1 + 0 + 0 + 0 + 2 + 2 = 5$.

(C) The smallest five-digit number is $10000$.
First,find the $LCM$ of $12, 16, 18$ and $27$:
$12 = 2^2 \times 3$
$16 = 2^4$
$18 = 2 \times 3^2$
$27 = 3^3$
$LCM = 2^4 \times 3^3 = 16 \times 27 = 432$.
Now,divide $10000$ by $432$ to find the remainder:
$10000 \div 432 = 23$ with a remainder of $64$.
To find the smallest five-digit number exactly divisible by $432$,we subtract the remainder from the divisor and add it to the smallest five-digit number:
Required number $= 10000 - 64 + 432 = 10368$.

(C) The largest four-digit number is $9999$.
To find the number exactly divisible by $14, 24, 27$,and $32$,we first calculate their $LCM$:
$14 = 2 \times 7$
$24 = 2^{3} \times 3$
$27 = 3^{3}$
$32 = 2^{5}$
$LCM = 2^{5} \times 3^{3} \times 7 = 32 \times 27 \times 7 = 6048$.
Now,we divide the largest four-digit number by the $LCM$:
$9999 \div 6048 = 1$ with a remainder of $3951$.
To find the largest four-digit number divisible by $6048$,we subtract the remainder from $9999$:
$9999 - 3951 = 6048$.
Therefore,the largest four-digit number exactly divisible by $14, 24, 27$,and $32$ is $6048$.

(B) To find the $LCM$ of numbers given in prime factorization form,we take the highest power of each prime factor present in any of the numbers.
The given numbers are:
$1) \ 2^{2} \times 3^{2} \times 5^{1} \times 11^{1}$
$2) \ 2^{4} \times 3^{4} \times 5^{2} \times 7^{1}$
$3) \ 2^{5} \times 3^{3} \times 5^{2} \times 7^{2} \times 11^{1}$
Identifying the highest powers of each prime factor:
- For $2$: The highest power is $2^{5}$.
- For $3$: The highest power is $3^{4}$.
- For $5$: The highest power is $5^{2}$.
- For $7$: The highest power is $7^{2}$.
- For $11$: The highest power is $11^{1}$.
Therefore,the $LCM = 2^{5} \times 3^{4} \times 5^{2} \times 7^{2} \times 11^{1}$.