HCF and LCM by Prime Factorisation Questions in English

(B) To find the Greatest Common Divisor $(GCD)$ of $20 a^{2} b$ and $30 a b^{2}$:
$1$. Prime factorization of $20 a^{2} b = 2^{2} \times 5 \times a^{2} \times b$.
$2$. Prime factorization of $30 a b^{2} = 2 \times 3 \times 5 \times a \times b^{2}$.
$3$. The $GCD$ is the product of the lowest power of each common prime factor:
- Common factors are $2, 5, a, b$.
- Lowest powers are $2^{1}, 5^{1}, a^{1}, b^{1}$.
- $GCD$ $= 2 \times 5 \times a \times b = 10 ab$.
Since $10 ab \neq 10 a^{2} b^{2}$,the statement is False.

(B) The factors of $5$ are $1, 5$.
The factors of $15$ are $1, 3, 5, 15$.
The common factors are $1$ and $5$.
The greatest common factor $(GCD)$ is $5$.
Since the statement claims the $GCD$ is $10$,the statement is False.

(A) For any two positive integers $a$ and $b$,the relationship between their Highest Common Factor $(HCF)$ and Least Common Multiple $(LCM)$ is given by the formula: $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$.
Since $a$ and $b$ are consecutive positive integers,they are coprime,meaning their only common factor is $1$. Thus,$\text{HCF}(a, b) = 1$.
Substituting this into the formula,we get $1 \times \text{LCM}(a, b) = a \times b$,which simplifies to $\text{LCM}(a, b) = a \times b$.
Therefore,$\text{HCF}(a, b) \times \text{LCM}(a, b) = 1 \times (a \times b) = a \times b$.

(A) To find the $HCF$ of $25$ and $52$,we find their prime factors:
$25 = 5^2$
$52 = 2^2 \times 13$
Since there are no common prime factors other than $1$,the Highest Common Factor $(HCF)$ is $1$.
Therefore,the statement is True.

(A) To find the smallest integer divisible by both $28$ and $63$,we need to calculate the Least Common Multiple $(LCM)$ of $28$ and $63$.
First,find the prime factorization of each number:
$28 = 2^2 \times 7^1$
$63 = 3^2 \times 7^1$
The $LCM$ is found by taking the highest power of each prime factor present in the numbers:
$LCM = 2^2 \times 3^2 \times 7^1$
$LCM = 4 \times 9 \times 7$
$LCM = 36 \times 7 = 252$
Therefore,the smallest integer divisible by both $28$ and $63$ is $252$.

(A) To find the $LCM$ of $36$ and $100$,we first find their prime factorizations:
$36 = 2^2 \times 3^2$
$100 = 2^2 \times 5^2$
The $LCM$ is the product of the highest powers of all prime factors involved:
$LCM = 2^2 \times 3^2 \times 5^2$
$LCM = 4 \times 9 \times 25$
$LCM = 36 \times 25 = 900$
Therefore,the $LCM$ of $36$ and $100$ is $900$.

(A) To find the $HCF$ of $65$ and $117$,we first find their prime factorizations:
$65 = 5 \times 13$
$117 = 9 \times 13 = 3^2 \times 13$
The common prime factor with the lowest power is $13$.
Therefore,the $HCF$ of $65$ and $117$ is $13$.

(A) To find the $LCM$ of $220$ and $60$,we first find the prime factorization of each number:
$220 = 2^2 \times 5^1 \times 11^1$
$60 = 2^2 \times 3^1 \times 5^1$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM(220, 60) = 2^2 \times 3^1 \times 5^1 \times 11^1$
$LCM(220, 60) = 4 \times 3 \times 5 \times 11$
$LCM(220, 60) = 12 \times 55 = 660$
Therefore,the $LCM$ of $220$ and $60$ is $660$.

(A) To find the $HCF(156, 455)$,we use the prime factorization method:
$156 = 2^2 \times 3 \times 13$
$455 = 5 \times 7 \times 13$
The common prime factor is $13$.
Therefore,the $HCF(156, 455) = 13$.

(Q.1-B, Q.2-A) Step $1$: Find the $\text{LCM}(7, 49)$.
Since $49$ is a multiple of $7$ $(7 \times 7 = 49)$,the $\text{LCM}$ of $7$ and $49$ is $49$.
Step $2$: Find the $\text{LCM}(10, 25)$.
Prime factorization of $10 = 2 \times 5$.
Prime factorization of $25 = 5 \times 5$.
$\text{LCM}(10, 25) = 2 \times 5 \times 5 = 50$.
Therefore,$Q.1$ matches with $B$ and $Q.2$ matches with $A$ (or $D$ as they are identical).

(A-B) $1$. To find the $HCF$ of $25$ and $10$:
Prime factorization of $25 = 5^2$.
Prime factorization of $10 = 2 \times 5$.
The common factor is $5$. Therefore, $HCF$ $(25, 10)$ = $5$.
$2$. To find the $LCM$ of $17$ and $11$:
Since $17$ and $11$ are both prime numbers, their $LCM$ is their product.
$LCM$ $(17, 11)$ = $17 \times 11 = 187$.
Thus, the correct matching is $Q.1-A$ and $Q.2-B$.

(A) To find the Highest Common Factor $(HCF)$ of $30$ and $72$:
Prime factorization of $30 = 2 \times 3 \times 5$.
Prime factorization of $72 = 2^3 \times 3^2$.
The common factors are $2^1$ and $3^1$.
$HCF(30, 72) = 2 \times 3 = 6$.
To find the $HCF$ of $19$ and $23$:
Since both $19$ and $23$ are prime numbers,their only common factor is $1$.
$HCF(19, 23) = 1$.
Thus,the $HCF$ values are $6$ and $1$ respectively. The correct option is $A$.

(A) Step $1$: Find the $HCF$ of $8$ and $6$.
Prime factorization of $8 = 2^3$.
Prime factorization of $6 = 2 \times 3$.
The common factor is $2$,so $HCF$ $(8, 6) = 2$.
Step $2$: Find the $LCM$ of $26$ and $91$.
Prime factorization of $26 = 2 \times 13$.
Prime factorization of $91 = 7 \times 13$.
$LCM$ $(26, 91) = 2 \times 7 \times 13 = 182$.
Therefore,$Q.1$ matches with $C$ and $Q.2$ matches with $B$.

(A) To find the time when Sonia and Ravi 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.
Time taken by Sonia = $18$ minutes.
Time taken by Ravi = $12$ 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.

(A) We know that for any two positive integers $a$ and $b$,the relationship between their $HCF$ and $LCM$ is given by:
$HCF(a, b) \times LCM(a, b) = a \times b$
Given $a = 306$,$b = 657$,and $HCF(306, 657) = 9$.
Substituting these values into the formula:
$9 \times LCM(306, 657) = 306 \times 657$
$LCM(306, 657) = \frac{306 \times 657}{9}$
$LCM(306, 657) = 34 \times 657$
$LCM(306, 657) = 22338$
Therefore,the $LCM$ is $22338$.

(A) To find the $HCF$ and $LCM$ of $8$,$9$,and $25$ using the prime factorization method:
Step $1$: Find the prime factors of each number.
$8 = 2 \times 2 \times 2 = 2^3$
$9 = 3 \times 3 = 3^2$
$25 = 5 \times 5 = 5^2$
Step $2$: 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 $HCF$ is $1$.
Step $3$: The $LCM$ is the product of the greatest power of each prime factor involved in the numbers.
$LCM = 2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 72 \times 25 = 1800$.
Thus,$HCF$ = $1$ and $LCM$ = $1800$.

(A) The prime factorization of the given numbers is as follows:
$17 = 17 \times 1$
$23 = 23 \times 1$
$29 = 29 \times 1$
Since $17, 23,$ and $29$ are all prime numbers,they have no common factor other than $1$.
Therefore,the Highest Common Factor $(HCF)$ is $1$.
The Least Common Multiple $(LCM)$ is the product of the highest powers of all prime factors involved:
$LCM = 17 \times 23 \times 29 = 11339$.
Thus,$HCF$ = $1$ and $LCM$ = $11339$.

(A) To find the $HCF$ and $LCM$ of $12$,$15$,and $21$ using prime factorization:
Step $1$: Write the prime factorization of each number.
$12 = 2^2 \times 3^1$
$15 = 3^1 \times 5^1$
$21 = 3^1 \times 7^1$
Step $2$: Find the $HCF$ (Highest Common Factor).
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 $3$: Find the $LCM$ (Least Common Multiple).
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 $7^1$.
$LCM = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 = 420$.
Thus,$HCF = 3$ and $LCM = 420$.

(A) Step $1$: Prime factorization of $336$ and $54$.
$336 = 2^4 \times 3^1 \times 7^1$
$54 = 2^1 \times 3^3$
Step $2$: Find $HCF$.
$HCF = 2^{\min(4,1)} \times 3^{\min(1,3)} = 2^1 \times 3^1 = 6$.
Step $3$: Find $LCM$.
$LCM = 2^{\max(4,1)} \times 3^{\max(1,3)} \times 7^1 = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024$.
Step $4$: Verify $HCF \times LCM = \text{Product of integers}$.
$HCF \times LCM = 6 \times 3024 = 18144$.
$\text{Product of integers} = 336 \times 54 = 18144$.
Since $18144 = 18144$, the relationship is verified.

(A) Step $1$: Find the prime factorization of $510$ and $92$.
$510 = 2 \times 3 \times 5 \times 17$
$92 = 2^2 \times 23$
Step $2$: Find $HCF$ and $LCM$.
$HCF = 2^1 = 2$
$LCM = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$
Step $3$: Verify the relationship $HCF \times LCM = \text{product of the two numbers}$.
$HCF \times LCM = 2 \times 23460 = 46920$
$\text{Product of numbers} = 510 \times 92 = 46920$
Since $46920 = 46920$,the relationship is verified.

(A) Step $1$: Prime factorization of $26$ and $91$.
$26 = 2 \times 13$
$91 = 7 \times 13$
Step $2$: Find $HCF$ and $LCM$.
$HCF = \text{Product of the smallest power of each common prime factor} = 13$.
$LCM = \text{Product of the greatest power of each prime factor involved} = 2 \times 7 \times 13 = 182$.
Step $3$: Verify the relationship $HCF \times LCM = \text{Product of the two numbers}$.
$HCF \times LCM = 13 \times 182 = 2366$.
$\text{Product of the two numbers} = 26 \times 91 = 2366$.
Since $2366 = 2366$, the relationship is verified.

(B) To express $140$ as a product of its prime factors,we perform prime factorization:
$1$. Divide $140$ by the smallest prime number,$2$: $140 \div 2 = 70$.
$2$. Divide $70$ by $2$: $70 \div 2 = 35$.
$3$. Divide $35$ by the next smallest prime number,$5$: $35 \div 5 = 7$.
$4$. Divide $7$ by the prime number $7$: $7 \div 7 = 1$.
Thus,$140 = 2 \times 2 \times 5 \times 7$.
In exponential form,this is $2^2 \times 5 \times 7$.

(A) To find the $HCF$ of $867$ and $255$ using Euclid's division algorithm,we follow these steps:
Step $1$: Since $867 > 255$,we apply Euclid's division lemma to $867$ and $255$:
$867 = 255 \times 3 + 102$
Step $2$: Since the remainder $102 \neq 0$,we apply the lemma to $255$ and $102$:
$255 = 102 \times 2 + 51$
Step $3$: Since the remainder $51 \neq 0$,we apply the lemma to $102$ and $51$:
$102 = 51 \times 2 + 0$
Since the remainder is now $0$,the divisor at this stage is the $HCF$.
Therefore,the $HCF$ of $867$ and $255$ is $51$.

(A) To find the $HCF$ of $135$ and $225$ using Euclid's division algorithm,we follow these steps:
Step $1$: Since $225 > 135$,we apply Euclid's division lemma to $225$ and $135$:
$225 = 135 \times 1 + 90$
Step $2$: Since the remainder $90 \neq 0$,we apply the lemma to $135$ and $90$:
$135 = 90 \times 1 + 45$
Step $3$: Since the remainder $45 \neq 0$,we apply the lemma to $90$ and $45$:
$90 = 45 \times 2 + 0$
Since the remainder is $0$,the divisor at this stage is the $HCF$.
Therefore,the $HCF$ of $135$ and $225$ is $45$.

(A) To find the $HCF$ and $LCM$ of $6$,$72$,and $120$ using prime factorization:
Step $1$: Find the prime factors of each number.
$6 = 2^1 \times 3^1$
$72 = 2^3 \times 3^2$
$120 = 2^3 \times 3^1 \times 5^1$
Step $2$: The $HCF$ is the product of the smallest power of each common prime factor.
$HCF = 2^1 \times 3^1 = 6$
Step $3$: The $LCM$ is the product of the greatest power of each prime factor involved.
$LCM = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360$
Thus,the $HCF$ is $6$ and the $LCM$ is $360$.

(A) Step $1$: Find the prime factorization of $96$ and $404$.
$96 = 2^5 \times 3^1$
$404 = 2^2 \times 101^1$
Step $2$: Find the $HCF$ by taking the smallest power of each common prime factor.
$HCF(96, 404) = 2^2 = 4$.
Step $3$: Use the relationship $LCM(a, b) \times HCF(a, b) = a \times b$.
$LCM(96, 404) = \frac{96 \times 404}{HCF(96, 404)}$
$LCM(96, 404) = \frac{96 \times 404}{4} = 96 \times 101 = 9696$.

(A) Step $1$: Find the prime factorization of the given numbers.
$6 = 2 \times 3$
$20 = 2^2 \times 5$
Step $2$: To find the $HCF$,take the product of the smallest power of each common prime factor.
Common prime factor is $2$. The smallest power is $2^1$.
Therefore,$HCF$ = $2$.
Step $3$: To find the $LCM$,take the product of the highest power of each prime factor involved.
Prime factors involved are $2, 3,$ and $5$.
Highest powers are $2^2, 3^1,$ and $5^1$.
$LCM$ = $2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$.
Thus,$HCF$ = $2$ and $LCM$ = $60$.