HCF and LCM by Prime Factorisation Questions in English

(N/A) Since $12576 > 4052,$ we apply the division lemma to $12576$ and $4052$ to get:
$12576 = 4052 \times 3 + 420$
Since the remainder $420 \neq 0,$ we apply the division lemma to $4052$ and $420$ to get:
$4052 = 420 \times 9 + 272$
Since the remainder $272 \neq 0,$ we apply the division lemma to $420$ and $272$ to get:
$420 = 272 \times 1 + 148$
Since the remainder $148 \neq 0,$ we apply the division lemma to $272$ and $148$ to get:
$272 = 148 \times 1 + 124$
Since the remainder $124 \neq 0,$ we apply the division lemma to $148$ and $124$ to get:
$148 = 124 \times 1 + 24$
Since the remainder $24 \neq 0,$ we apply the division lemma to $124$ and $24$ to get:
$124 = 24 \times 5 + 4$
Since the remainder $4 \neq 0,$ we apply the division lemma to $24$ and $4$ to get:
$24 = 4 \times 6 + 0$
The remainder is now $0,$ so the procedure stops. The divisor at this stage is $4.$ Therefore,the $HCF$ of $12576$ and $4052$ is $4.$

(D) To minimize the area of the tray,we need to maximize the number of barfis in each stack. This is equivalent to finding the Highest Common Factor $(HCF)$ of $420$ and $130$.
Using Euclid's division algorithm:
$420 = 130 \times 3 + 30$
$130 = 30 \times 4 + 10$
$30 = 10 \times 3 + 0$
The $HCF$ of $420$ and $130$ is $10$.
Therefore,the sweetseller can place $10$ barfis in each stack to minimize the area occupied on the tray.

(A) To find the $HCF$ of $135$ and $225$ using Euclid's division algorithm:
Since $225 > 135$,we apply the division lemma to $225$ and $135$ to obtain:
$225 = 135 \times 1 + 90$
Since the remainder $90 \neq 0$,we apply the division lemma to $135$ and $90$ to obtain:
$135 = 90 \times 1 + 45$
Now,we consider the new divisor $90$ and the new remainder $45$,and apply the division lemma to obtain:
$90 = 45 \times 2 + 0$
Since the remainder is $0$,the process stops.
Since the divisor at this stage is $45$,the $HCF$ of $135$ and $225$ is $45$.

(B) To find the $HCF$ of $196$ and $38220$ using Euclid's division algorithm:
Since $38220 > 196$,we apply Euclid's division lemma to $38220$ and $196$:
$38220 = 196 \times 195 + 0$
Since the remainder is $0$,the process stops.
The divisor at this stage is $196$.
Therefore,the $HCF$ of $196$ and $38220$ is $196$.

(C) Given numbers are $867$ and $255$.
Since $867 > 255$,we apply Euclid's division lemma to $867$ and $255$ to obtain:
$867 = 255 \times 3 + 102$
Since the remainder $102 \neq 0$,we apply the division lemma to $255$ and $102$ to obtain:
$255 = 102 \times 2 + 51$
Since the remainder $51 \neq 0$,we apply the division lemma to $102$ and $51$ to obtain:
$102 = 51 \times 2 + 0$
Since the remainder is $0$,the process stops.
The divisor at this stage is $51$.
Therefore,the $HCF$ of $867$ and $255$ is $51$.

(A) The maximum number of columns in which they can march is given by the $HCF$ of $616$ and $32$.
We use Euclid's division algorithm to find the $HCF$:
$616 = 32 \times 19 + 8$
$32 = 8 \times 4 + 0$
Since the remainder is $0$,the $HCF$ of $616$ and $32$ is $8$.
Therefore,the maximum number of columns in which they can march is $8$.

(N/A) First,we find the prime factorisation of the given numbers:
$6 = 2^1 \times 3^1$
$20 = 2^2 \times 5^1$
To find the $HCF$,we take the product of the smallest power of each common prime factor:
$HCF(6, 20) = 2^1 = 2$
To find the $LCM$,we take the product of the greatest power of each prime factor involved in the numbers:
$LCM(6, 20) = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$
Thus,the $HCF$ is $2$ and the $LCM$ is $60$.

(B) The prime factorisation of $96$ and $404$ is as follows:
$96 = 2^5 \times 3$
$404 = 2^2 \times 101$
The $HCF$ is the product of the smallest power of each common prime factor in the numbers:
$HCF(96, 404) = 2^2 = 4$
Using the relationship $LCM(a, b) \times HCF(a, b) = a \times b$:
$LCM(96, 404) = \frac{96 \times 404}{HCF(96, 404)}$
$LCM(96, 404) = \frac{96 \times 404}{4}$
$LCM(96, 404) = 96 \times 101 = 9696$

(C) We have the prime factorisation of the numbers:
$6 = 2^1 \times 3^1$
$72 = 2^3 \times 3^2$
$120 = 2^3 \times 3^1 \times 5^1$
To find the $HCF$,we take the product of the smallest power of each common prime factor:
$HCF(6, 72, 120) = 2^1 \times 3^1 = 6$
To find the $LCM$,we take the product of the greatest power of each prime factor involved in the numbers:
$LCM(6, 72, 120) = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360$
Thus,the $HCF$ is $6$ and the $LCM$ is $360$.

(A) First,find the prime factorization of the given numbers:
$26 = 2 \times 13$
$91 = 7 \times 13$
The $HCF$ is the product of the smallest power of each common prime factor:
$HCF = 13$
The $LCM$ is the product of the greatest power of each prime factor involved:
$LCM = 2 \times 7 \times 13 = 182$
Now,verify the relationship:
Product of the two numbers $= 26 \times 91 = 2366$
$HCF \times LCM = 13 \times 182 = 2366$
Since $2366 = 2366$,the relationship $LCM \times HCF =$ product of the two numbers is verified.

(B) First,find the prime factorization of the given numbers:
$510 = 2 \times 3 \times 5 \times 17$
$92 = 2^2 \times 23 = 2 \times 2 \times 23$
To find the $HCF$,take the product of the smallest power of each common prime factor:
$HCF = 2^1 = 2$
To find the $LCM$,take the product of the highest power of each prime factor involved:
$LCM = 2^2 \times 3 \times 5 \times 17 \times 23 = 4 \times 3 \times 5 \times 17 \times 23 = 23460$
Now,verify the relationship:
Product of the two numbers $= 510 \times 92 = 46920$
$HCF \times LCM = 2 \times 23460 = 46920$
Since $46920 = 46920$,the relationship $LCM \times HCF =$ product of the two numbers is verified.

(C) First,find the prime factorization of the numbers:
$336 = 2^4 \times 3 \times 7$
$54 = 2 \times 3^3$
$HCF$ is the product of the smallest power of each common prime factor:
$HCF = 2^1 \times 3^1 = 6$
$LCM$ is the product of the greatest power of each prime factor involved:
$LCM = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024$
Now,verify the relationship:
Product of the two numbers $= 336 \times 54 = 18144$
$LCM \times HCF = 3024 \times 6 = 18144$
Since $18144 = 18144$,the relationship is verified.

(D) To find the $LCM$ and $HCF$ of $12, 15$ and $21$ using the prime factorisation method:
Step $1$: Find the prime factors of each number.
$12 = 2^{2} \times 3$
$15 = 3 \times 5$
$21 = 3 \times 7$
Step $2$: The $HCF$ is the product of the smallest power of each common prime factor.
The only common prime factor is $3$,and its smallest power is $3^{1}$.
Therefore,$HCF = 3$.
Step $3$: The $LCM$ is the product of the greatest power of each prime factor involved.
The prime factors involved are $2, 3, 5$ and $7$.
The greatest powers are $2^{2}, 3^{1}, 5^{1}$ and $7^{1}$.
$LCM = 2^{2} \times 3^{1} \times 5^{1} \times 7^{1} = 4 \times 3 \times 5 \times 7 = 420$.

(A) To find the $LCM$ and $HCF$ of $17, 23$ and $29$ using the prime factorisation method:
Step $1$: Write the prime factorisation of each number.
$17 = 1 \times 17$
$23 = 1 \times 23$
$29 = 1 \times 29$
Step $2$: Find the $HCF$.
The $HCF$ is the product of the smallest power of each common prime factor in the numbers.
Since there are no common prime factors other than $1$,$HCF = 1$.
Step $3$: Find the $LCM$.
The $LCM$ is the product of the greatest power of each prime factor involved in the numbers.
$LCM = 17 \times 23 \times 29 = 11339$.
Thus,the $HCF$ is $1$ and the $LCM$ is $11339$.

(B) To find the $LCM$ and $HCF$ of $8, 9$ and $25$ using the prime factorisation method:
Step $1$: Write the prime factorisation of each number:
$8 = 2 \times 2 \times 2 = 2^3$
$9 = 3 \times 3 = 3^2$
$25 = 5 \times 5 = 5^2$
Step $2$: Find the $HCF$ (Highest Common Factor):
The $HCF$ is the product of the smallest power of each common prime factor. Since there are no common prime factors other than $1$,the $HCF = 1$.
Step $3$: Find the $LCM$ (Least Common Multiple):
The $LCM$ is the product of the greatest power of each prime factor involved:
$LCM = 2^3 \times 3^2 \times 5^2$
$LCM = 8 \times 9 \times 25$
$LCM = 72 \times 25 = 1800$
Thus,the $LCM$ is $1800$ and the $HCF$ is $1$.

(C) We are given that $HCF (306, 657) = 9$.
We know the fundamental relationship between two numbers,their $HCF$,and their $LCM$:
$LCM (a, b) \times HCF (a, b) = a \times b$
Substituting the given values:
$LCM (306, 657) \times 9 = 306 \times 657$
To find the $LCM$,divide the product of the numbers by their $HCF$:
$LCM (306, 657) = \frac{306 \times 657}{9}$
First,divide $306$ by $9$:
$306 \div 9 = 34$
Now,multiply the result by $657$:
$LCM (306, 657) = 34 \times 657 = 22338$
Therefore,the $LCM (306, 657)$ is $22338$.

(B) To find the time when Sonia and Ravi will meet again at the starting point,we need to find the Least Common Multiple $(LCM)$ of the time taken by them to complete one round.
Sonia takes $18$ minutes and Ravi takes $12$ minutes.
Prime factorization of $18 = 2 \times 3^2$.
Prime factorization of $12 = 2^2 \times 3$.
$LCM(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36$.
Therefore,they will meet again at the starting point after $36$ minutes.

(B) To find the $HCF$ of $65$ and $117$ using Euclid's division algorithm:
$117 = 65 \times 1 + 52$
$65 = 52 \times 1 + 13$
$52 = 13 \times 4 + 0$
Therefore,the $HCF(65, 117) = 13$.
Given that the $HCF$ is expressible as $65m - 117$,we set up the equation:
$65m - 117 = 13$
$65m = 13 + 117$
$65m = 130$
$m = \frac{130}{65}$
$m = 2$

(C) Since $5$ and $8$ are the remainders when $70$ and $125$ are divided by the required number,respectively.
Subtracting these remainders from the given numbers,we get:
$70 - 5 = 65$
$125 - 8 = 117$
The required number is the $HCF$ of $65$ and $117$.
Using Euclid's division algorithm:
$117 = 65 \times 1 + 52$
$65 = 52 \times 1 + 13$
$52 = 13 \times 4 + 0$
Since the remainder is $0$,the $HCF$ is $13$.
Therefore,$13$ is the largest number that divides $70$ and $125$ leaving remainders $5$ and $8$ respectively.

(D) Given that,$a = x^3 y^2 = x \times x \times x \times y \times y$.
And $b = x y^3 = x \times y \times y \times y$.
The $HCF$ of two numbers is the product of the smallest power of each common prime factor involved in the numbers.
Here,the common prime factors are $x$ and $y$.
The smallest power of $x$ is $x^1$ and the smallest power of $y$ is $y^2$.
Therefore,$HCF(a, b) = x^1 \times y^2 = x y^2$.

(A) Given that,$p = a b^{2}$ and $q = a^{3} b$,where $a$ and $b$ are prime numbers.
To find the $LCM(p, q)$,we take the highest power of each prime factor present in the expressions.
The prime factors involved are $a$ and $b$.
The highest power of $a$ is $a^{3}$ (from $q$).
The highest power of $b$ is $b^{2}$ (from $p$).
Therefore,$LCM(p, q) = a^{3} \times b^{2} = a^{3} b^{2}$.

(C) To find the least number divisible by all numbers from $1$ to $10$,we need to calculate the Least Common Multiple $(LCM)$ of these numbers.
Prime factorization of each number from $1$ to $10$:
$1 = 1$
$2 = 2^1$
$3 = 3^1$
$4 = 2^2$
$5 = 5^1$
$6 = 2^1 \times 3^1$
$7 = 7^1$
$8 = 2^3$
$9 = 3^2$
$10 = 2^1 \times 5^1$
The $LCM$ is found by taking the highest power of each prime factor present in the numbers:
$LCM = 2^3 \times 3^2 \times 5^1 \times 7^1$
$LCM = 8 \times 9 \times 5 \times 7$
$LCM = 72 \times 35 = 2520$
Thus,the least number divisible by all numbers from $1$ to $10$ is $2520$.

(A) The $HCF$ (Highest Common Factor) is the largest number that divides both given numbers.
Given that $525$ and $3000$ are divisible by $3, 5, 15, 25,$ and $75$,the largest among these is $75$.
To justify using Euclid's Division Lemma:
$3000 = 525 \times 5 + 375$
$525 = 375 \times 1 + 150$
$375 = 150 \times 2 + 75$
$150 = 75 \times 2 + 0$
Since the remainder is $0$ when the divisor is $75$,the $HCF(525, 3000) = 75$.

(B) To check if two numbers are co-prime,their $HCF$ must be $1$. We use Euclid's division algorithm:
$396 = 231 \times 1 + 165$
$231 = 165 \times 1 + 66$
$165 = 66 \times 2 + 33$
$66 = 33 \times 2 + 0$
Since the $HCF$ is $33$ (which is not $1$),the numbers $231$ and $396$ are not co-prime.

(B) To find the $HCF$ of $847$ and $2160$ using Euclid's division algorithm,we perform the following steps:
$2160 = 847 \times 2 + 466$
$847 = 466 \times 1 + 381$
$466 = 381 \times 1 + 85$
$381 = 85 \times 4 + 41$
$85 = 41 \times 2 + 3$
$41 = 3 \times 13 + 2$
$3 = 2 \times 1 + 1$
$2 = 1 \times 2 + 0$
Since the remainder is $0$,the $HCF$ is the divisor at this stage,which is $1$.
Two numbers are said to be co-prime if their $HCF$ is $1$. Since the $HCF$ of $847$ and $2160$ is $1$,they are co-prime.

(D) Let $a = 693$,$b = 567$,and $c = 441$.
By Euclid's division algorithm,$a = bq + r$,where $0 \le r < b$.
First,we find the $HCF$ of $693$ and $567$:
$693 = 567 \times 1 + 126$
$567 = 126 \times 4 + 63$
$126 = 63 \times 2 + 0$
Since the remainder is $0$,the $HCF(693, 567) = 63$.
Now,we find the $HCF$ of the result $(63)$ and the remaining number $(441)$:
$441 = 63 \times 7 + 0$
Since the remainder is $0$,the $HCF(63, 441) = 63$.
Therefore,the $HCF$ of $441, 567, 693$ is $63$.

(A) Since $1, 2$ and $3$ are the remainders of $1251, 9377$ and $15628$ respectively,we subtract these remainders from the numbers to get numbers that are exactly divisible by the required number.
$1251 - 1 = 1250$
$9377 - 2 = 9375$
$15628 - 3 = 15625$
The required number is the $HCF$ of $1250, 9375$ and $15625$.
Using Euclid's division algorithm,$a = bq + r$:
First,find $HCF$ of $15625$ and $9375$:
$15625 = 9375 \times 1 + 6250$
$9375 = 6250 \times 1 + 3125$
$6250 = 3125 \times 2 + 0$
So,$HCF(15625, 9375) = 3125$.
Now,find $HCF$ of $3125$ and $1250$:
$3125 = 1250 \times 2 + 625$
$1250 = 625 \times 2 + 0$
So,$HCF(3125, 1250) = 625$.
Thus,the largest number that divides $1251, 9377$ and $15628$ leaving remainders $1, 2$ and $3$ respectively is $625$.

(D) To find the minimum distance that each should walk so that each can cover the same distance in complete steps,we need to calculate the Least Common Multiple $(LCM)$ of the step measurements: $40\, cm$,$42\, cm$,and $45\, cm$.
First,we find the prime factorization of each number:
$40 = 2^3 \times 5$
$42 = 2 \times 3 \times 7$
$45 = 3^2 \times 5$
The $LCM$ is found by taking the highest power of each prime factor present in the numbers:
$LCM(40, 42, 45) = 2^3 \times 3^2 \times 5^1 \times 7^1$
$LCM = 8 \times 9 \times 5 \times 7$
$LCM = 72 \times 35$
$LCM = 2520$
Therefore,the minimum distance each should walk is $2520\, cm$.

(B) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check 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.
First,simplify the fraction $\frac{77}{210}$ by dividing both numerator and denominator by their greatest common divisor,which is $7$:
$\frac{77 \div 7}{210 \div 7} = \frac{11}{30}$.
Now,find the prime factorization of the denominator $30$:
$30 = 2 \times 3 \times 5$.
Since the prime factorization of the denominator contains a factor $3$ (other than $2$ and $5$),the rational number $\frac{11}{30}$ has a non-terminating repeating decimal expansion.

(D) The denominator of the rational number $\frac{257}{5000}$ is $5000$.
First,we find the prime factorization of $5000$:
$5000 = 5 \times 1000 = 5 \times 10^3 = 5 \times (2 \times 5)^3 = 5 \times 2^3 \times 5^3 = 2^3 \times 5^4$.
This is in the form $2^m \times 5^n$,where $m = 3$ and $n = 4$ are non-negative integers.
To find the decimal expansion without actual division,we make the powers of $2$ and $5$ equal:
$\frac{257}{5000} = \frac{257}{2^3 \times 5^4}$.
To make the powers equal to $4$,we multiply the numerator and denominator by $2^1$:
$\frac{257 \times 2}{2^3 \times 5^4 \times 2^1} = \frac{514}{2^4 \times 5^4} = \frac{514}{(2 \times 5)^4} = \frac{514}{10^4}$.
$\frac{514}{10000} = 0.0514$.
Thus,the decimal expansion is $0.0514$.

(B) Here,$210 > 55$.
Applying Euclid's division lemma,$a = bq + r$:
$210 = 55 \times 3 + 45$
$55 = 45 \times 1 + 10$
$45 = 10 \times 4 + 5$
$10 = 5 \times 2 + 0$
Since the remainder is $0$,the divisor at this stage is the $g.c.d.$
Therefore,the $g.c.d.(210, 55) = 5$.

(A) Let $d$ be the number of items in each stack.
Since each stack must contain the same number of items,$d$ must be a common divisor of $420$ and $130$.
To occupy the least area of the tray,the number of stacks must be minimized,which implies that the number of items in each stack $(d)$ must be the greatest common divisor $(GCD)$ of $420$ and $130$.
Using Euclid's division algorithm:
$420 = 130 \times 3 + 30$
$130 = 30 \times 4 + 10$
$30 = 10 \times 3 + 0$
The last non-zero divisor is $10$.
Therefore,$GCD(420, 130) = 10$.
Thus,the maximum number of burfis that can be placed in each stack is $10$.

(A) To ensure that each stack has the same height and the number of stacks is minimized (which minimizes the surface area),we need to find the Greatest Common Divisor $(GCD)$ of the number of English books $(56)$ and Gujarati books $(72)$.
Step $1$: Find the prime factorization of $56$ and $72$.
$56 = 2^3 \times 7$
$72 = 2^3 \times 3^2$
Step $2$: The $GCD$ is the product of the lowest powers of common prime factors.
$GCD(56, 72) = 2^3 = 8$.
Therefore,the maximum number of books that can be placed in each stack is $8$.

(A) To find the $g.c.d.$ of $12576$ and $4052$ using Euclid's division algorithm,we follow these steps:
Step $1$: Since $12576 > 4052$,we apply Euclid's division lemma to $12576$ and $4052$:
$12576 = 4052 \times 3 + 420$
Step $2$: Since the remainder $420 \neq 0$,we apply the lemma to $4052$ and $420$:
$4052 = 420 \times 9 + 272$
Step $3$: Since the remainder $272 \neq 0$,we apply the lemma to $420$ and $272$:
$420 = 272 \times 1 + 148$
Step $4$: Since the remainder $148 \neq 0$,we apply the lemma to $272$ and $148$:
$272 = 148 \times 1 + 124$
Step $5$: Since the remainder $124 \neq 0$,we apply the lemma to $148$ and $124$:
$148 = 124 \times 1 + 24$
Step $6$: Since the remainder $24 \neq 0$,we apply the lemma to $124$ and $24$:
$124 = 24 \times 5 + 4$
Step $7$: Since the remainder $4 \neq 0$,we apply the lemma to $24$ and $4$:
$24 = 4 \times 6 + 0$
Since the remainder is now $0$,the divisor at this stage is the $g.c.d.$
Therefore,the $g.c.d.$ of $12576$ and $4052$ is $4$.

(C) To find the $g.c.d.$ of $117$ and $65$ using Euclid's division algorithm,we follow these steps:
Step $1$: Since $117 > 65$,we apply Euclid's division lemma to $117$ and $65$:
$117 = 65 \times 1 + 52$
Step $2$: Since the remainder $52 \neq 0$,we apply the lemma to $65$ and $52$:
$65 = 52 \times 1 + 13$
Step $3$: Since the remainder $13 \neq 0$,we apply the lemma to $52$ and $13$:
$52 = 13 \times 4 + 0$
Since the remainder is now $0$,the divisor at this stage is the $g.c.d.$
Therefore,the $g.c.d.$ of $117$ and $65$ is $13$.

(D) To find the $g.c.d.$ of $240$ and $6552$ using Euclid's division algorithm,we follow these steps:
Step $1$: Apply Euclid's division lemma to $6552$ and $240$:
$6552 = 240 \times 27 + 72$
Step $2$: Apply the lemma to $240$ and $72$:
$240 = 72 \times 3 + 24$
Step $3$: Apply the lemma to $72$ and $24$:
$72 = 24 \times 3 + 0$
Since the remainder is $0$,the divisor at this stage is the $g.c.d.$
Therefore,the $g.c.d.$ of $240$ and $6552$ is $24$.

(A) To find the $g.c.d.$ of $155$ and $1385$ using Euclid's division algorithm,we follow these steps:
Step $1$: Apply Euclid's division lemma to $1385$ and $155$:
$1385 = 155 \times 8 + 145$
Step $2$: Apply the lemma to $155$ and $145$:
$155 = 145 \times 1 + 10$
Step $3$: Apply the lemma to $145$ and $10$:
$145 = 10 \times 14 + 5$
Step $4$: Apply the lemma to $10$ and $5$:
$10 = 5 \times 2 + 0$
Since the remainder is $0$,the divisor at this stage is the $g.c.d.$
Therefore,the $g.c.d.$ of $155$ and $1385$ is $5$.

(B) To find the $g.c.d.$ of $75$ and $243$ using Euclid's division algorithm,we apply the lemma $a = bq + r$ where $0 \le r < b$.
Step $1$: Divide $243$ by $75$.
$243 = 75 \times 3 + 18$
Step $2$: Divide $75$ by $18$.
$75 = 18 \times 4 + 3$
Step $3$: Divide $18$ by $3$.
$18 = 3 \times 6 + 0$
Since the remainder is $0$,the divisor at this stage is the $g.c.d.$
Therefore,the $g.c.d.$ of $75$ and $243$ is $3$.

(C) To find the largest number that divides $245$ and $1029$ leaving a remainder of $5$ in each case,we first subtract the remainder from both numbers:
$245 - 5 = 240$
$1029 - 5 = 1024$
Now,we need to find the Highest Common Factor $(HCF)$ of $240$ and $1024$.
Prime factorization of $240 = 2^4 \times 3 \times 5$
Prime factorization of $1024 = 2^{10}$
The $HCF$ is the product of the lowest power of common prime factors:
$HCF(240, 1024) = 2^4 = 16$.
Therefore,the largest number is $16$.

(D) To find the largest number that divides $2053$ and $967$ leaving remainders $5$ and $7$ respectively,we first subtract the remainders from the given numbers.
$2053 - 5 = 2048$
$967 - 7 = 960$
Now,we need to find the Highest Common Factor $(HCF)$ of $2048$ and $960$.
Using the prime factorization method:
$2048 = 2^{11}$
$960 = 2^6 \times 3 \times 5$
The $HCF$ is the product of the lowest powers of common prime factors:
$HCF = 2^6 = 64$
Therefore,the largest number is $64$.

(A) Given numbers are $a = 96$ and $b = 404$.
First,find the prime factorization of $96$ and $404$:
$96 = 2^5 \times 3$
$404 = 2^2 \times 101$
The $g.c.d.(96, 404)$ is the product of the smallest power of each common prime factor:
$g.c.d.(96, 404) = 2^2 = 4$.
Using the formula $g.c.d.(a, b) \times l.c.m.(a, b) = a \times b$:
$4 \times l.c.m.(96, 404) = 96 \times 404$
$l.c.m.(96, 404) = \frac{96 \times 404}{4}$
$l.c.m.(96, 404) = 96 \times 101 = 9696$.

(B) Given the property: $g.c.d. (a, b) \times l.c.m. (a, b) = a \times b$.
First,find the prime factorization of $26$ and $91$:
$26 = 2 \times 13$
$91 = 7 \times 13$
Therefore,$g.c.d. (26, 91) = 13$.
Using the formula: $13 \times l.c.m. (26, 91) = 26 \times 91$.
$l.c.m. (26, 91) = \frac{26 \times 91}{13}$.
$l.c.m. (26, 91) = 2 \times 91 = 182$.

(A) Let $d$ be the number of books in each stack.
Since all stacks must have the same height and books are of the same thickness,$d$ must be a common divisor of $96$,$240$,and $336$.
To minimize the number of stacks,we need to maximize the number of books per stack,$d$. Thus,$d = \text{HCF}(96, 240, 336)$.
Prime factorization:
$96 = 2^5 \times 3$
$240 = 2^4 \times 3 \times 5$
$336 = 2^4 \times 3 \times 7$
$\text{HCF}(96, 240, 336) = 2^4 \times 3 = 16 \times 3 = 48$.
So,each stack contains $48$ books.
Number of stacks of English books $= \frac{96}{48} = 2$.
Number of stacks of Hindi books $= \frac{240}{48} = 5$.
Number of stacks of Mathematics books $= \frac{336}{48} = 7$.
Therefore,the number of stacks are $2, 5,$ and $7$ respectively.

(N/A) Step $1$: Prime factorization of $96$:
$96 = 2^5 \times 3^1$
Step $2$: Prime factorization of $404$:
$404 = 2^2 \times 101^1$
Step $3$: The $\text{g.c.d.}$ is the product of the smallest power of each common prime factor:
$\text{g.c.d.}(96, 404) = 2^2 = 4$
Step $4$: The $\text{l.c.m.}$ is the product of the greatest power of each prime factor involved:
$\text{l.c.m.}(96, 404) = 2^5 \times 3^1 \times 101^1 = 32 \times 3 \times 101 = 9696$
Therefore,the $\text{g.c.d.}$ is $4$ and the $\text{l.c.m.}$ is $9696$.

(N/A) Step $1$: Prime factorization of the numbers:
$144 = 2^4 \times 3^2$
$180 = 2^2 \times 3^2 \times 5^1$
$192 = 2^6 \times 3^1$
Step $2$: To find the $\text{g.c.d.}$,take the product of the smallest power of each common prime factor:
$\text{g.c.d.} = 2^2 \times 3^1 = 4 \times 3 = 12$
Step $3$: To find the $\text{l.c.m.}$,take the product of the highest power of each prime factor involved:
$\text{l.c.m.} = 2^6 \times 3^2 \times 5^1 = 64 \times 9 \times 5 = 2880$

(N/A) To find the $\text{l.c.m.}$ and $\text{g.c.d.}$ using the fundamental theorem of arithmetic,we first find the prime factorization of each number:
$84 = 2^2 \times 3^1 \times 7^1$
$90 = 2^1 \times 3^2 \times 5^1$
$120 = 2^3 \times 3^1 \times 5^1$
$\text{g.c.d.}$ is the product of the smallest power of each common prime factor:
$\text{g.c.d.}(84, 90, 120) = 2^1 \times 3^1 = 6$
$\text{l.c.m.}$ is the product of the highest power of each prime factor present in the numbers:
$\text{l.c.m.}(84, 90, 120) = 2^3 \times 3^2 \times 5^1 \times 7^1 = 8 \times 9 \times 5 \times 7 = 2520$

(A) The given numbers are $17$,$23$,and $29$.
Since $17$,$23$,and $29$ are all prime numbers,their prime factorizations are:
$17 = 17^1 \times 1^1$
$23 = 23^1 \times 1^1$
$29 = 29^1 \times 1^1$
$\text{GCD}$ (Greatest Common Divisor) is the product of the smallest power of each common prime factor. Here,the only common factor is $1$.
$\text{GCD}(17, 23, 29) = 1$
$\text{LCM}$ (Least Common Multiple) is the product of the highest power of each prime factor involved in the numbers.
$\text{LCM}(17, 23, 29) = 17 \times 23 \times 29$
$17 \times 23 = 391$
$391 \times 29 = 11339$
Therefore,$\text{LCM} = 11339$ and $\text{GCD} = 1$.

(D) Given that $\text{g.c.d.}(306, 657) = 9$ (as the prime factorization of $306 = 2 \times 3^2 \times 17$ and $657 = 3^2 \times 73$,so $\text{g.c.d.} = 3^2 = 9$).
Using the formula: $\text{g.c.d.}(a, b) \times \text{l.c.m.}(a, b) = a \times b$.
Substitute the values: $9 \times \text{l.c.m.}(306, 657) = 306 \times 657$.
$\text{l.c.m.}(306, 657) = \frac{306 \times 657}{9}$.
$\text{l.c.m.}(306, 657) = 34 \times 657 = 22338$.

(A) Let the $\text{g.c.d.}$ of the two numbers be $x$.
Given that the $\text{l.c.m.}$ is $14$ times the $\text{g.c.d.}$,so $\text{l.c.m.} = 14x$.
The sum of $\text{l.c.m.}$ and $\text{g.c.d.}$ is $600$,so $14x + x = 600$.
$15x = 600 \implies x = 40$.
Thus,$\text{g.c.d.} = 40$ and $\text{l.c.m.} = 14 \times 40 = 560$.
We know that for two numbers $a$ and $b$,$\text{l.c.m.} \times \text{g.c.d.} = a \times b$.
Given $a = 280$,we have $560 \times 40 = 280 \times b$.
$b = \frac{560 \times 40}{280} = 2 \times 40 = 80$.
Therefore,the other number is $80$.

(B) We know that for any two positive integers $a$ and $b$,the relationship between their $\text{g.c.d.}$ and $\text{l.c.m.}$ is given by the formula:
$\text{g.c.d.}(a, b) \times \text{l.c.m.}(a, b) = a \times b$
Given that the $\text{g.c.d.}$ is $16$ and the product of the two numbers $(a \times b)$ is $3072$,we can substitute these values into the formula:
$16 \times \text{l.c.m.} = 3072$
To find the $\text{l.c.m.}$,divide $3072$ by $16$:
$\text{l.c.m.} = \frac{3072}{16}$
$\text{l.c.m.} = 192$
Therefore,the $\text{l.c.m.}$ of the two numbers is $192$.