HCF and LCM Questions in English

(A) To find the next time the devices beep together,we need to calculate the Least Common Multiple $(LCM)$ of their individual beep intervals: $48$,$72$,and $108$.
Prime factorization:
$48 = 2^4 \times 3^1$
$72 = 2^3 \times 3^2$
$108 = 2^2 \times 3^3$
$LCM(48, 72, 108) = 2^4 \times 3^3 = 16 \times 27 = 432\, seconds$.
Now,convert $432\, seconds$ into minutes and seconds:
$432 \div 60 = 7\, minutes$ and $12\, seconds$.
Since they last beeped together at $10:00\, a.m.$,they will next beep together at $10:00\, a.m. + 7\, minutes\, 12\, seconds = 10:07:12\, a.m.$

(C) To find the greatest length of the pipe pieces of the same size that can be cut from both pipes without any remainder,we need to calculate the Highest Common Factor $(HCF)$ of the two lengths.
Given lengths are $1.5\, m$ and $1.2\, m$.
To simplify,we can express these in terms of decimeters or multiply by $10$: $15$ and $12$.
The factors of $15$ are $1, 3, 5, 15$.
The factors of $12$ are $1, 2, 3, 4, 6, 12$.
The greatest common factor of $15$ and $12$ is $3$.
Therefore,the $HCF$ of $1.5$ and $1.2$ is $0.3\, m$.

(B) To find the number that is divisible by $7, 11,$ and $14$,we first need to find their Least Common Multiple $(LCM)$.
Step $1$: Find $\operatorname{LCM}(7, 11, 14)$.
$7 = 7 \times 1$
$11 = 11 \times 1$
$14 = 2 \times 7$
$\operatorname{LCM} = 2 \times 7 \times 11 = 154$.
Step $2$: Divide $1812$ by $154$ to find the remainder.
$1812 \div 154 = 11$ with a remainder.
$154 \times 11 = 1694$.
$1812 - 1694 = 118$.
So,the remainder is $118$.
Step $3$: Calculate the value to be added.
To make $1812$ divisible by $154$,we need to add the difference between the divisor $(154)$ and the remainder $(118)$.
Value to be added $= 154 - 118 = 36$.

(C) To find numbers divisible by $2, 3,$ and $7$,we must find the Least Common Multiple $(LCM)$ of these numbers.
Since $2, 3,$ and $7$ are prime numbers,their $LCM = 2 \times 3 \times 7 = 42$.
We need to find the count of multiples of $42$ in the range $[300, 700]$.
First,find the number of multiples of $42$ up to $700$: $\lfloor 700 / 42 \rfloor = 16$.
Next,find the number of multiples of $42$ up to $299$: $\lfloor 299 / 42 \rfloor = 7$.
The number of multiples between $300$ and $700$ is $16 - 7 = 9$.

(A) Since $x$ and $y$ are prime numbers,their $LCM$ is simply their product,$x \times y = 161$.
We factorize $161$ into prime factors: $161 = 7 \times 23$.
Given $x > y$,we assign $x = 23$ and $y = 7$.
Now,substitute these values into the expression $(3y - x)$:
$(3 \times 7) - 23 = 21 - 23 = -2$.

(C) Let the two numbers be $12x$ and $12y$,where $x$ and $y$ are coprime.
Given that the product of the numbers is $2160$,we have:
$12x \times 12y = 2160$
$144xy = 2160$
$xy = \frac{2160}{144} = 15$
Since $x$ and $y$ are coprime and their product is $15$,the possible pairs for $(x, y)$ are $(1, 15)$ or $(3, 5)$.
If $(x, y) = (1, 15)$,the numbers are $12(1) = 12$ and $12(15) = 180$. However,$180$ is not a $2$-digit number.
If $(x, y) = (3, 5)$,the numbers are $12(3) = 36$ and $12(5) = 60$. Both are $2$-digit numbers.
Thus,the numbers are $(36, 60)$.

(B) To find the greatest number that divides $390, 495,$ and $300$ without leaving a remainder,we need to calculate the Highest Common Factor $(HCF)$ of these numbers.
Step $1$: Prime factorization of the numbers:
$390 = 2 \times 3 \times 5 \times 13$
$495 = 3^2 \times 5 \times 11$
$300 = 2^2 \times 3 \times 5^2$
Step $2$: Identify the common prime factors with the lowest exponent:
The common prime factors are $3$ and $5$.
The lowest power of $3$ is $3^1 = 3$.
The lowest power of $5$ is $5^1 = 5$.
Step $3$: Calculate the $HCF$:
$HCF = 3 \times 5 = 15$.
Therefore,the greatest number that divides all three numbers is $15$.

(C) To find the largest possible number of students in each class such that the number of boys and girls in each class is equal,we need to calculate the $H.C.F.$ of $391$ and $323$.
First,find the prime factorization of $391$ and $323$:
$391 = 17 \times 23$
$323 = 17 \times 19$
The $H.C.F.$ of $391$ and $323$ is $17$. This means each class contains $17$ students.
Now,calculate the total number of classes:
Number of boys' classes = $391 / 17 = 23$
Number of girls' classes = $323 / 17 = 19$
Total number of classes = $23 + 19 = 42$.

(C) To find the $HCF$,we factorize both expressions.
First expression: $x^{8}-1 = (x^{4}-1)(x^{4}+1) = (x^{2}-1)(x^{2}+1)(x^{4}+1) = (x-1)(x+1)(x^{2}+1)(x^{4}+1)$.
Second expression: $x^{4}+2x^{3}-2x-1 = (x^{4}-1) + 2x(x^{2}-1) = (x^{2}-1)(x^{2}+1) + 2x(x^{2}-1) = (x^{2}-1)(x^{2}+2x+1) = (x-1)(x+1)(x+1)^{2} = (x-1)(x+1)^{3}$.
The common factors are $(x-1)$ and $(x+1)$.
Therefore,the $HCF = (x-1)(x+1) = x^{2}-1$.

(A) Let the three numbers be $1x, 2x,$ and $3x$,where $x$ is the common factor.
Since the $HCF$ of these numbers is $12$,the common factor $x$ must be equal to $12$.
Therefore,the numbers are:
$1 \times 12 = 12$
$2 \times 12 = 24$
$3 \times 12 = 36$
Thus,the numbers are $12, 24, 36$.

(D) The required number is given by the formula: $\text{Required Number} = (\text{L.C.M. of } 15, 20, 36, 48) + 3$.
First,we find the $L.C.M.$ of $15, 20, 36$ and $48$ using the division method:
$\begin{array}{c|cccc} 2 & 15, & 20, & 36, & 48 \\ \hline 2 & 15, & 10, & 18, & 24 \\ \hline 3 & 15, & 5, & 9, & 12 \\ \hline 5 & 5, & 1, & 3, & 4 \end{array}$
$\therefore L.C.M. = 2 \times 2 \times 3 \times 5 \times 1 \times 1 \times 3 \times 4 = 720$.
Now,add the remainder $3$ to the $L.C.M.$:
$\text{Required Number} = 720 + 3 = 723$.

(A) We know that for any two expressions, the product of the expressions is equal to the product of their $H.C.F.$ and $L.C.M.$
First expression $\times$ Second expression $= H.C.F. \times L.C.M.$
Given:
$L.C.M. = (x^{2}+6x+8)(x+1) = (x+2)(x+4)(x+1)$
$H.C.F. = (x+1)$
First expression $= x^{2}+3x+2 = (x+2)(x+1)$
Let the second expression be $P(x)$.
$(x+2)(x+1) \times P(x) = (x+2)(x+4)(x+1) \times (x+1)$
$P(x) = \frac{(x+2)(x+4)(x+1)(x+1)}{(x+2)(x+1)}$
$P(x) = (x+4)(x+1)$
$P(x) = x^{2}+x+4x+4 = x^{2}+5x+4$

(B) First,find the prime factorization of $625$:
$\begin{array}{l|l} 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline & 5 \end{array}$
Thus,$625 = 5 \times 5 \times 5 \times 5 = 5^3 \times 5$.
To make the quotient a perfect cube,we must divide $625$ by the factor that is not part of the triplet,which is $5$.
Therefore,$625 \div 5 = 125$,and $125 = 5^3$,which is a perfect cube.
The smallest number is $5$.

(D) To find the greatest number that exactly divides $200$ and $320$,we need to calculate the Highest Common Factor $(H.C.F.)$ of $200$ and $320$.
Using the division method:
$320 = 200 \times 1 + 120$
$200 = 120 \times 1 + 80$
$120 = 80 \times 1 + 40$
$80 = 40 \times 2 + 0$
Since the remainder is $0$,the $H.C.F.$ is $40$.
Therefore,the greatest number is $40$.

(B) Step $1$: Find the $L.C.M.$ of $10, 15$ and $20$.
$10 = 2 \times 5$
$15 = 3 \times 5$
$20 = 2^2 \times 5$
$L.C.M. = 2^2 \times 3 \times 5 = 60$.
Step $2$: Identify the greatest $4-$digit number,which is $9999$.
Step $3$: Divide $9999$ by $60$ to find the remainder.
$9999 \div 60 = 166$ with a remainder of $39$.
Step $4$: Subtract the remainder from the greatest $4-$digit number to get the largest number divisible by $60$.
$9999 - 39 = 9960$.
Therefore,the greatest $4-$digit number exactly divisible by $10, 15$ and $20$ is $9960$.

(D) To find the number of students that can be arranged in rows of $6, 8, 12$,or $16$ without any student being left behind,we need to find the Least Common Multiple $(L.C.M.)$ of these numbers.
$L.C.M.$ of $6, 8, 12, 16$:

$2$	$6, 8, 12, 16$
$2$	$3, 4, 6, 8$
$2$	$3, 2, 3, 4$
$2$	$3, 1, 3, 2$
$3$	$3, 1, 3, 1$
	$1, 1, 1, 1$

$L.C.M. = 2 \times 2 \times 2 \times 2 \times 3 = 48$.
The number of students must be a multiple of $48$. Among the given options,$96$ is a multiple of $48$ $(48 \times 2 = 96)$.

(D) We know that for any two numbers or expressions $x$ and $y$,the product of their $H.C.F.$ $(A)$ and $L.C.M.$ $(B)$ is equal to the product of the numbers themselves.
So,$A \times B = x \times y$.
We are given $A + B = x + y$.
Consider the identity $(A+B)^3 = A^3 + B^3 + 3AB(A+B)$.
Substituting the given values,we get $(x+y)^3 = A^3 + B^3 + 3xy(x+y)$.
Rearranging for $A^3 + B^3$,we have $A^3 + B^3 = (x+y)^3 - 3xy(x+y)$.
Factoring out $(x+y)$,we get $A^3 + B^3 = (x+y)((x+y)^2 - 3xy)$.
Expanding the term inside the bracket: $A^3 + B^3 = (x+y)(x^2 + 2xy + y^2 - 3xy)$.
$A^3 + B^3 = (x+y)(x^2 - xy + y^2)$.
Using the algebraic identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$,we conclude that $A^3 + B^3 = x^3 + y^3$.

(C) To find the greatest number that divides $411, 684,$ and $821$ leaving remainders $3, 4,$ and $5$ respectively,we first subtract the remainders from the respective numbers:
$411 - 3 = 408$
$684 - 4 = 680$
$821 - 5 = 816$
Now,we need to find the $H.C.F.$ of $408, 680,$ and $816$.
First,find the $H.C.F.$ of $408$ and $680$:
$680 = 408 \times 1 + 272$
$408 = 272 \times 1 + 136$
$272 = 136 \times 2 + 0$
So,the $H.C.F.$ of $408$ and $680$ is $136$.
Next,find the $H.C.F.$ of $136$ and $816$:
$816 = 136 \times 6 + 0$
Since $136$ divides $816$ completely,the $H.C.F.$ of $408, 680,$ and $816$ is $136$.
Therefore,the required number is $136$.

(D) We know that for any two numbers $x$ and $y$,the product of the numbers is equal to the product of their $H.C.F.$ and $L.C.M.$
$x \times y = H.C.F. \times L.C.M.$
Given that $H.C.F. = 3$ and $L.C.M. = 315$,we have:
$x \times y = 3 \times 315 = 945$
We need to find the value of $\frac{1}{x} + \frac{1}{y}$.
$\frac{1}{x} + \frac{1}{y} = \frac{x + y}{x y}$
Given $x + y = 36$,substitute the values:
$\frac{1}{x} + \frac{1}{y} = \frac{36}{3 \times 315}$
Simplify the expression:
$\frac{1}{x} + \frac{1}{y} = \frac{12}{315} = \frac{4}{35}$

(B) To compare the surds,we find the $L.C.M.$ of the indices $(3, 2, 6, 12)$,which is $12$.
Convert each surd to an equivalent surd with index $12$:
$1. \sqrt[3]{4} = 4^{1/3} = 4^{4/12} = \sqrt[12]{4^4} = \sqrt[12]{256}$
$2. \sqrt{3} = 3^{1/2} = 3^{6/12} = \sqrt[12]{3^6} = \sqrt[12]{729}$
$3. \sqrt[6]{25} = 25^{1/6} = 25^{2/12} = \sqrt[12]{25^2} = \sqrt[12]{625}$
$4. \sqrt[12]{289} = \sqrt[12]{289}$
Comparing the radicands: $729 > 625 > 289 > 256$.
Therefore,the greatest value is $\sqrt{3}$ and the least value is $\sqrt[3]{4}$.

(B) Let the four consecutive prime numbers be $a, b, c,$ and $d$ in ascending order,such that $a < b < c < d$.
Given that the product of the first three is $a \times b \times c = 385$ and the product of the last three is $b \times c \times d = 1001$.
To find the common factors,we calculate the Highest Common Factor $(HCF)$ of $385$ and $1001$.
$385 = 5 \times 7 \times 11$
$1001 = 7 \times 11 \times 13$
The common factors are $7$ and $11$,so $b \times c = 7 \times 11 = 77$.
Now,using the product of the last three numbers: $b \times c \times d = 1001$.
Substituting $b \times c = 77$,we get $77 \times d = 1001$.
Therefore,$d = \frac{1001}{77} = 13$.
The four consecutive prime numbers are $5, 7, 11, 13$. The largest prime number is $13$.

(B) To find the $H.C.F.$ of fractions,we use the formula:
$H.C.F. = \frac{H.C.F. \text{ of numerators}}{L.C.M. \text{ of denominators}}$
Given fractions: $\frac{2}{3}, \frac{4}{5}, \frac{6}{7}$
Step $1$: Find the $H.C.F.$ of the numerators $(2, 4, 6)$.
The factors of $2$ are $1, 2$.
The factors of $4$ are $1, 2, 4$.
The factors of $6$ are $1, 2, 3, 6$.
The greatest common factor is $2$.
Step $2$: Find the $L.C.M.$ of the denominators $(3, 5, 7)$.
Since $3, 5,$ and $7$ are all prime numbers,their $L.C.M.$ is their product:
$3 \times 5 \times 7 = 105$.
Step $3$: Combine the results.
$H.C.F. = \frac{2}{105}$.

(A) To find when the bells ring together,we need to calculate the $L.C.M.$ of the given intervals: $3, 6, 9, 12, 15$ seconds.
Prime factorization:
$3 = 3^1$
$6 = 2^1 \times 3^1$
$9 = 3^2$
$12 = 2^2 \times 3^1$
$15 = 3^1 \times 5^1$
$L.C.M. = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 180$ seconds.
This means the bells ring together every $180$ seconds,which is equal to $3$ minutes.
In $36$ minutes,the number of times they ring together is $\frac{36}{3} = 12$ times.
Since they start ringing together at the $0^{th}$ minute,we add $1$ to the result.
Total times = $12 + 1 = 13$ times.

(C) To find the time when they meet again at the starting point, we need to calculate the Least Common Multiple $(LCM)$ of the time taken by each person to complete one round.
The times are $18\, \text{seconds}$, $22\, \text{seconds}$, and $30\, \text{seconds}$.
Prime factorization:
$18 = 2 \times 3^2$
$22 = 2 \times 11$
$30 = 2 \times 3 \times 5$
$LCM(18, 22, 30) = 2^1 \times 3^2 \times 5^1 \times 11^1 = 2 \times 9 \times 5 \times 11 = 990\, \text{seconds}$.
Now, convert $990\, \text{seconds}$ into minutes:
$990 / 60 = 16.5\, \text{minutes}$.
$0.5\, \text{minutes} = 0.5 \times 60 = 30\, \text{seconds}$.
Therefore, they will meet again at the starting point after $16\, \text{minutes and } 30\, \text{seconds}$.

(C) Let the two numbers be $12x$ and $12y$,where $x$ and $y$ are coprime (i.e.,their $H.C.F.$ is $1$).
The $L.C.M.$ of these two numbers is given by $12xy$.
According to the problem,$12xy = 924$.
Dividing both sides by $12$,we get $xy = \frac{924}{12} = 77$.
Now,we need to find pairs of coprime factors $(x, y)$ such that their product is $77$.
The factors of $77$ are $1, 7, 11, 77$.
The possible pairs $(x, y)$ are $(1, 77)$ and $(7, 11)$.
Since both pairs consist of coprime numbers,there are $2$ such pairs.

(C) First,find the $L.C.M.$ of $5, 6, 7, 8$.
$5 = 5$
$6 = 2 \times 3$
$7 = 7$
$8 = 2^3$
$L.C.M. = 5 \times 3 \times 7 \times 8 = 840$.
Any number that leaves a remainder of $3$ when divided by $5, 6, 7, 8$ is of the form $(840x + 3)$,where $x$ is a natural number.
We are given that this number is divisible by $9$.
So,$(840x + 3) \div 9$ must be an integer.
$840x + 3 = (837x + 3x + 3) = 9(93x) + 3(x + 1)$.
For this to be divisible by $9$,$(x + 1)$ must be a multiple of $3$.
For the least number,let $x + 1 = 3$,which gives $x = 2$.
Substituting $x = 2$ in $(840x + 3)$:
Required number $= 840(2) + 3 = 1680 + 3 = 1683$.

(D) Let the two numbers be $10x$ and $10y$,where $x$ and $y$ are coprime to each other.
Given that the $H.C.F.$ is $10$,the numbers must be multiples of $10$.
The $L.C.M.$ of $10x$ and $10y$ is $10xy$.
We are given $10xy = 120$,which implies $xy = 12$.
The possible pairs of $(x, y)$ such that they are coprime are $(1, 12)$ and $(3, 4)$.
Case $1$: If the pair is $(1, 12)$,the numbers are $10(1) = 10$ and $10(12) = 120$. Their sum is $10 + 120 = 130$.
Case $2$: If the pair is $(3, 4)$,the numbers are $10(3) = 30$ and $10(4) = 40$. Their sum is $30 + 40 = 70$.
Comparing with the given options,$70$ is present.

(B) To find when the traffic lights will change simultaneously again,we need to calculate the Least Common Multiple $(LCM)$ of the time intervals: $24$,$36$,and $54$ seconds.
Prime factorization:
$24 = 2^3 \times 3^1$
$36 = 2^2 \times 3^2$
$54 = 2^1 \times 3^3$
$LCM = 2^3 \times 3^3 = 8 \times 27 = 216$ seconds.
Convert $216$ seconds into minutes:
$216 \div 60 = 3$ minutes and $36$ seconds.
Adding this duration to the initial time of $10:15:00$ am:
$10:15:00 + 0:03:36 = 10:18:36$ am.
Therefore,the traffic lights will again change simultaneously at $10:18:36$ am.

(B) To find the time when they will meet again at the starting point,we need to calculate the Least Common Multiple $(LCM)$ of the times taken by each person to complete one round.
The times taken are $18$,$24$,and $32$ seconds.
Prime factorization:
$18 = 2 \times 3^2$
$24 = 2^3 \times 3$
$32 = 2^5$
$LCM$ is the product of the highest powers of all prime factors involved:
$LCM(18, 24, 32) = 2^5 \times 3^2 = 32 \times 9 = 288$.
Therefore,they will meet again at the starting point after $288 \text{ seconds}$.

(D) To find when they will meet again at the starting point,we need to calculate the Least Common Multiple $(LCM)$ of the time taken by each person to complete one round.
The times are $54 \text{ seconds}$,$42 \text{ seconds}$,and $63 \text{ seconds}$.
Prime factorization:
$54 = 2 \times 3^3$
$42 = 2 \times 3 \times 7$
$63 = 3^2 \times 7$
$LCM = 2 \times 3^3 \times 7 = 2 \times 27 \times 7 = 378 \text{ seconds}$.
To convert seconds into minutes,divide by $60$:
$\frac{378}{60} = 6.3 \text{ minutes}$.
Rounding to the nearest whole number,they will meet after approximately $6 \text{ minutes}$.

(B) To find the number that is exactly divisible by $20, 28, 32$ and $35$,we must first calculate their Least Common Multiple $(L.C.M.)$.
Prime factorization of the numbers:
$20 = 2^2 \times 5$
$28 = 2^2 \times 7$
$32 = 2^5$
$35 = 5 \times 7$
$L.C.M. = 2^5 \times 5 \times 7 = 32 \times 35 = 1120$.
We need to find a number $x$ such that $(5834 - x)$ is exactly divisible by $1120$. To make $(5834 - x)$ the largest possible multiple of $1120$ that is less than $5834$,we divide $5834$ by $1120$:
$5834 \div 1120 = 5$ with a remainder of $234$.
So,$5834 = (1120 \times 5) + 234$.
To get a number exactly divisible by $1120$,we must subtract the remainder $234$ from $5834$:
$5834 - 234 = 5600$.
However,the question asks for the number which when subtracted from $5834$ gives a result divisible by $1120$. If we subtract $234$ from $5834$,we get $5600$,which is divisible by $1120$. Thus,the number to be subtracted is $234$.
Wait,re-evaluating the question: "The greatest number,which when subtracted from $5834$,gives a number exactly divisible by each of $20, 28, 32$ and $35$". This implies we want the result $(5834 - x)$ to be a multiple of $1120$. The multiples of $1120$ are $1120, 2240, 3360, 4480, 5600$.
To make $x$ the greatest,$(5834 - x)$ must be the smallest multiple,which is $1120$.
$5834 - x = 1120 \implies x = 5834 - 1120 = 4714$.

(D) The relationship between two numbers and their $H.C.F.$ and $L.C.M.$ is given by the formula:
Product of two numbers $= H.C.F. \times L.C.M.$
Let the other number be $x$.
Given: $H.C.F. = 8$,$L.C.M. = 48$,and one number $= 24$.
Substituting the values in the formula:
$24 \times x = 8 \times 48$
$x = \frac{8 \times 48}{24}$
$x = 8 \times 2 = 16$
Therefore,the other number is $16$.

(C) Let the two numbers be $3x$ and $4x$,where $x$ is the common factor.
The prime factorization of the numbers is:
$3x = 3 \times x$
$4x = 2^2 \times x$
The $L.C.M.$ of $3x$ and $4x$ is $3 \times 2^2 \times x = 12x$.
Given that the $L.C.M.$ is $84$,we set up the equation:
$12x = 84$
Solving for $x$:
$x = 84 / 12 = 7$
Now,calculate the two numbers:
First number $= 3x = 3 \times 7 = 21$
Second number $= 4x = 4 \times 7 = 28$
Comparing the two numbers,$21$ and $28$,the greater number is $28$.

(C) Let the two numbers be $5x$ and $6x$,where $x$ is the common factor.
The $H.C.F.$ of $5x$ and $6x$ is $x$.
Given that the $H.C.F.$ is $4$,we have $x = 4$.
Therefore,the numbers are $5 \times 4 = 20$ and $6 \times 4 = 24$.
The $L.C.M.$ of two numbers is given by the product of the ratio parts and the $H.C.F.$:
$L.C.M. = 5 \times 6 \times 4 = 120$.
Alternatively,using the formula: $L.C.M. = \frac{\text{Product of numbers}}{H.C.F.} = \frac{20 \times 24}{4} = 20 \times 6 = 120$.

(C) Observe the difference between the divisor and the remainder in each case:
$4 - 2 = 2$  
$5 - 3 = 2$  
$6 - 4 = 2$
Since the difference is constant $(2)$, the required number is of the form
$\text{LCM of } 4, 5, 6 - 2$.
First, find the LCM of $4, 5$ and $6$:
$4 = 2^2$  
$5 = 5^1$  
$6 = 2 \times 3$
$\text{LCM} = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60$
Therefore, the least such number is:
$60 - 2 = 58$

(C) Let the greatest number be $x$ and the common remainder be $k$.
Then,we have:
$43 = nx + k$ $...(1)$
$91 = mx + k$ $...(2)$
$183 = lx + k$ $...(3)$
Subtracting the equations:
$(2) - (1) \Rightarrow 91 - 43 = (m - n)x \Rightarrow 48 = (m - n)x$
$(3) - (2) \Rightarrow 183 - 91 = (l - m)x \Rightarrow 92 = (l - m)x$
$(3) - (1) \Rightarrow 183 - 43 = (l - n)x \Rightarrow 140 = (l - n)x$
The greatest number $x$ is the $H.C.F.$ of the differences $48, 92,$ and $140$.
Prime factorization:
$48 = 2^4 \times 3$
$92 = 2^2 \times 23$
$140 = 2^2 \times 5 \times 7$
$H.C.F. = 2^2 = 4$.
Thus,the greatest number is $4$.