HCF and LCM by Prime Factorisation Questions in English

(A) To find the $HCF$ of $6$ and $20$,we first find the prime factorization of each number:
$6 = 2 \times 3$
$20 = 2^2 \times 5$
The $HCF$ is the product of the smallest power of each common prime factor.
The only common prime factor is $2$,and its smallest power is $2^1 = 2$.
Therefore,the $HCF$ of $6$ and $20$ is $2$.

(C) The Least Common Multiple $(LCM)$ of two numbers is the smallest positive integer that is divisible by both numbers.
For any two prime numbers $p$ and $q$,since they are distinct,they have no common factors other than $1$.
Therefore,the $LCM$ of two distinct prime numbers is simply their product.
$LCM(p, q) = p \times q = pq$.

(C) To find the $LCM$ of $7, 11,$ and $17$,we first observe that all three numbers are prime numbers.
Since $7, 11,$ and $17$ are prime numbers,they have no common factors other than $1$.
The $LCM$ of a set of prime numbers is simply the product of those numbers.
Therefore,$LCM(7, 11, 17) = 7 \times 11 \times 17 = 1309$.

(D) To find the $HCF(510, 92)$,we perform prime factorization of both numbers:
$510 = 2 \times 3 \times 5 \times 17$
$92 = 2 \times 2 \times 23 = 2^2 \times 23$
The common prime factor with the lowest exponent is $2^1 = 2$.
Therefore,$HCF(510, 92) = 2$.

(A) We know that for any two positive integers $a$ and $b$,the relationship between their $HCF$ and $LCM$ is given by:
$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$
Given:
$\text{HCF} = 9$
$\text{Product of numbers} (a \times b) = 288$
Substituting the values in the formula:
$9 \times \text{LCM} = 288$
$\text{LCM} = \frac{288}{9}$
$\text{LCM} = 32$
Therefore,the $LCM$ of the two numbers is $32$.

(A) To find the $LCM$ of $6$ and $20$,we first find their prime factorizations:
$6 = 2 \times 3$
$20 = 2^2 \times 5$
The $LCM$ is the product of the highest powers of all prime factors involved:
$LCM = 2^2 \times 3^1 \times 5^1$
$LCM = 4 \times 3 \times 5 = 60$
Therefore,the $LCM$ of $6$ and $20$ is $60$.

(B) For any two positive integers $a$ and $b$,the fundamental theorem of arithmetic states that the product of the two numbers is equal to the product of their Highest Common Factor $(HCF)$ and Least Common Multiple $(LCM)$.
Mathematically,this is expressed as: $HCF(a, b) \times LCM(a, b) = a \times b$.

(B) To find the $LCM$ of $6, 72,$ and $120$,we first find the prime factorization of each number:
$6 = 2^1 \times 3^1$
$72 = 8 \times 9 = 2^3 \times 3^2$
$120 = 8 \times 15 = 2^3 \times 3^1 \times 5^1$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM$ $= 2^3 \times 3^2 \times 5^1$
$LCM$ $= 8 \times 9 \times 5$
$LCM$ $= 72 \times 5 = 360$
Therefore,the $LCM$ of $6, 72,$ and $120$ is $360$.

(D) We know that for any two positive integers $a$ and $b$,the product of the numbers is equal to the product of their $HCF$ and $LCM$.
$a \times b = HCF(a, b) \times LCM(a, b)$
Given $HCF(a, b) = 8$ and $LCM(a, b) = 64$,we have:
$a \times b = 8 \times 64 = 512$
Since $8$ is the $HCF$,both $a$ and $b$ must be multiples of $8$. Let $a = 8x$ and $b = 8y$,where $x$ and $y$ are coprime and $x > y$ (since $a > b$).
Substituting these into the product equation:
$(8x) \times (8y) = 512$
$64xy = 512$
$xy = 8$
Possible pairs $(x, y)$ such that $x$ and $y$ are coprime and $x > y$ are $(8, 1)$.
If $x = 8$ and $y = 1$,then $a = 8 \times 8 = 64$ and $b = 8 \times 1 = 8$.
Since $a > b$,$a = 64$ is the correct value.

(D) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
$LCM(2, 4) = 2^2 = 4$.
Therefore,the $LCM$ of the smallest prime number and the smallest composite number is $4$.

(A) To find the Greatest Common Divisor $(GCD)$ of $13, 23,$ and $31$,we first identify the prime factorization of each number:
$13 = 13 \times 1$
$23 = 23 \times 1$
$31 = 31 \times 1$
Since $13, 23,$ and $31$ are all prime numbers and they have no common factors other than $1$,their $GCD$ is $1$.

(B) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $HCF$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
The $HCF$ is the product of the smallest power of each common prime factor,which is $2^1 = 2$.
Therefore,the $HCF$ of the smallest prime number and the smallest composite number is $2$.

(B) We know the fundamental property of two positive integers $a$ and $b$:
$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$
Given that $\text{HCF}(a, b) = 7$ and $\text{LCM}(a, b) = 385$.
Therefore,$a \times b = 7 \times 385$.
Calculating the product: $7 \times 385 = 2695$.
Thus,the product of the two integers is $2695$.

(A) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
$LCM(2, 4) = 2^2 = 4$.
Therefore,the $LCM$ of the smallest prime number and the smallest composite number is $4$.

(B) For any two positive integers $a$ and $b$,the product of their Highest Common Factor $(HCF)$ and Least Common Multiple $(LCM)$ is always equal to the product of the two numbers themselves.
Mathematically,this is expressed as: $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$.

(B) To find the $LCM$ of $6$,$72$,and $120$,we first find the prime factorization of each number:
$6 = 2^1 \times 3^1$
$72 = 8 \times 9 = 2^3 \times 3^2$
$120 = 8 \times 15 = 2^3 \times 3^1 \times 5^1$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM = 2^3 \times 3^2 \times 5^1$
$LCM = 8 \times 9 \times 5$
$LCM = 72 \times 5 = 360$
Therefore,the $LCM$ of $6$,$72$,and $120$ is $360$.

(D) We know that for any two positive integers $a$ and $b$,the product of the numbers is equal to the product of their $HCF$ and $LCM$.
$a \times b = HCF(a, b) \times LCM(a, b)$
Given $HCF(a, b) = 8$ and $LCM(a, b) = 64$,we have:
$a \times b = 8 \times 64 = 512$
Since $HCF(a, b) = 8$,both $a$ and $b$ must be multiples of $8$. Let $a = 8x$ and $b = 8y$,where $x$ and $y$ are coprime and $x > y$ (since $a > b$).
Substituting these into the product equation:
$(8x) \times (8y) = 512$
$64xy = 512$
$xy = 8$
The possible pairs of $(x, y)$ such that $x > y$ and $gcd(x, y) = 1$ are $(8, 1)$.
If $x = 8$ and $y = 1$,then $a = 8 \times 8 = 64$ and $b = 8 \times 1 = 8$.
Checking the $LCM$: $LCM(64, 8) = 64$,which matches the given condition.
Thus,$a = 64$.

(B) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM(2, 4)$:
Prime factorization of $2 = 2^1$.
Prime factorization of $4 = 2^2$.
$LCM$ is the product of the highest power of each prime factor involved,which is $2^2 = 4$.
Therefore,the $LCM$ is $4$.

(A) To find the $HCF$ (Highest Common Factor) of $13$,$23$,and $31$:
$1$. Prime factorization of $13 = 13 \times 1$.
$2$. Prime factorization of $23 = 23 \times 1$.
$3$. Prime factorization of $31 = 31 \times 1$.
Since $13$,$23$,and $31$ are all prime numbers,they have no common factors other than $1$.
Therefore,the $HCF$ of $13$,$23$,and $31$ is $1$.

(A) To find the greatest positive integer that divides $70$ and $125$ leaving remainders $5$ and $8$ respectively,we need to find the Highest Common Factor $(HCF)$ of $(70 - 5)$ and $(125 - 8)$.
Step $1$: Subtract the remainders from the given numbers.
$70 - 5 = 65$
$125 - 8 = 117$
Step $2$: Find the $HCF$ of $65$ and $117$.
Prime factorization of $65 = 5 \times 13$.
Prime factorization of $117 = 3^2 \times 13$.
Step $3$: The common factor with the lowest power is $13$.
Therefore,the greatest positive integer is $13$.

(B) First,find the $\text{HCF}$ of $65$ and $117$ using prime factorization:
$65 = 5 \times 13$
$117 = 9 \times 13 = 3^2 \times 13$
Therefore,$\text{HCF}(65, 117) = 13$.
Given the equation: $\text{HCF}(65, 117) = 65m - 117$.
Substitute the value of $\text{HCF}$:
$13 = 65m - 117$
$13 + 117 = 65m$
$130 = 65m$
$m = \frac{130}{65} = 2$.
Thus,the value of $m$ is $2$.

(B) We know the fundamental property of two positive integers $a$ and $b$ is that the product of the numbers is equal to the product of their $\text{HCF}$ and $\text{LCM}$.
Mathematically,$a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)$.
Given that $\text{HCF}(a, b) = 7$ and $\text{LCM}(a, b) = 385$.
Therefore,$a \times b = 7 \times 385$.
$a \times b = 2695$.
Thus,the product of the two integers is $2695$.

(C) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
The $LCM$ is the product of the highest power of each prime factor involved,which is $2^2 = 4$.
Therefore,the $LCM$ of $2$ and $4$ is $4$.

(B) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $HCF$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
The $HCF$ is the product of the smallest power of each common prime factor,which is $2^1 = 2$.
Therefore,the $HCF$ of the smallest prime number and the smallest composite number is $2$.

(C) The least common multiple $(LCM)$ of two numbers is the smallest positive integer that is divisible by both numbers.
Since $p$ and $q$ are distinct prime integers,they have no common factors other than $1$.
Therefore,the $LCM$ of two distinct prime numbers is simply their product.
Thus,$\text{LCM}(p, q) = p \times q = pq$.

(A) To find the $LCM(26, 91)$,we first find the prime factorization of both numbers:
$26 = 2 \times 13$
$91 = 7 \times 13$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM(26, 91) = 2 \times 7 \times 13 = 182$
Therefore,the correct option is $A$.

(D) By definition,a prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.
Since $m$ and $n$ are distinct prime integers,they do not share any common factors other than $1$.
Therefore,the Highest Common Factor $(HCF)$ of two distinct prime numbers is always $1$.

(B) The $LCM$ of two numbers must always be divisible by their $HCF$.
Given $HCF(a, b) = 25$.
We check the divisibility of the given options by $25$:
$1$. $50 / 25 = 2$ (Divisible)
$2$. $105 / 25 = 4.2$ (Not divisible)
$3$. $100 / 25 = 4$ (Divisible)
Since $105$ is not divisible by $25$,it cannot be the $LCM$ of two numbers whose $HCF$ is $25$.

(A) For any two positive integers $a$ and $b$,the relationship between their $LCM$ and $GCD$ is given by: $a \times b = LCM(a, b) \times GCD(a, b)$.
Given that $LCM(a, b) = a \times b$.
Substituting this into the formula: $a \times b = (a \times b) \times GCD(a, b)$.
Dividing both sides by $(a \times b)$,we get: $GCD(a, b) = 1$.
Therefore,the $GCD$ of the two numbers is $1$.

(A) To find the $HCF$ of $12, 15$,and $21$,we first find the prime factorization of each number:
$12 = 2^2 \times 3^1$
$15 = 3^1 \times 5^1$
$21 = 3^1 \times 7^1$
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,$\text{HCF}(12, 15, 21) = 3$.

(A) To find the $LCM$ of $6$ and $20$,we first find their prime factorizations:
$6 = 2 \times 3$
$20 = 2^2 \times 5$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM = 2^2 \times 3^1 \times 5^1$
$LCM = 4 \times 3 \times 5 = 60$
Therefore,the $LCM$ of $6$ and $20$ is $60$.

(A) We know the relationship between two numbers, their $HCF$, and their $LCM$ is:
$\text{Product of two numbers} = HCF \times LCM$
Given:
$HCF = 9$
$\text{Product of two numbers} = 288$
Substituting the values in the formula:
$288 = 9 \times LCM$
$LCM = \frac{288}{9}$
$LCM = 32$
Therefore, the $LCM$ of the two numbers is $32$.

(B) To find the $HCF(510, 92)$,we perform prime factorization of both numbers:
$510 = 2 \times 3 \times 5 \times 17$
$92 = 2 \times 2 \times 23 = 2^2 \times 23$
The $HCF$ is the product of the smallest power of each common prime factor.
The only common prime factor is $2$,and its smallest power is $2^1 = 2$.
Therefore,$HCF(510, 92) = 2$.

(D) To find the $LCM$ of prime numbers,we multiply them together because they have no common factors other than $1$.
Since $7$,$11$,and $17$ are all prime numbers,their $LCM$ is their product.
$LCM = 7 \times 11 \times 17$
$LCM = 77 \times 17$
$LCM = 1309$
Therefore,the correct option is $D$.

(C) The Least Common Multiple $(LCM)$ of two numbers is the smallest positive integer that is divisible by both numbers.
Since $p$ and $q$ are distinct prime integers,they have no common factors other than $1$.
Therefore,the $LCM$ of two distinct prime numbers is simply their product.
Thus,$\text{LCM}(p, q) = p \times q = pq$.

(A) To find the $HCF$ of $6$ and $20$,we first find the prime factorization of each number:
$6 = 2 \times 3$
$20 = 2^2 \times 5$
The $HCF$ is the product of the smallest power of each common prime factor.
The only common prime factor is $2$,and its smallest power is $2^1 = 2$.
Therefore,the $HCF$ of $6$ and $20$ is $2$.

(A) The $LCM$ of two numbers must always be divisible by their $HCF$.
Given $HCF(a, b) = 12$.
We check each option to see if it is divisible by $12$:
$(A) 90 / 12 = 7.5$ (Not divisible)
$(B) 24 / 12 = 2$ (Divisible)
$(C) 48 / 12 = 4$ (Divisible)
$(D) 60 / 12 = 5$ (Divisible)
Since $90$ is not divisible by $12$,it cannot be the $LCM$ of two numbers whose $HCF$ is $12$.

(B) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
$LCM(2, 4) = 2^2 = 4$.
Therefore,the $LCM$ of the smallest prime number and the smallest composite number is $4$.

(B) The fundamental property of $HCF$ and $LCM$ of two numbers $a$ and $b$ is that the $LCM$ must be perfectly divisible by the $HCF$.
In other words, $\frac{LCM(a, b)}{HCF(a, b)} = \text{integer}$.
Given $HCF(a, b) = 16$.
We check each option:
$1$. For $32$: $\frac{32}{16} = 2$ (Possible).
$2$. For $72$: $\frac{72}{16} = 4.5$ (Not an integer, hence impossible).
$3$. For $64$: $\frac{64}{16} = 4$ (Possible).
Therefore, $72$ cannot be the $LCM$.

(B) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM(2, 4)$:
Prime factorization of $2 = 2^1$.
Prime factorization of $4 = 2^2$.
$LCM$ is the product of the highest power of each prime factor involved.
$LCM(2, 4) = 2^2 = 4$.
Therefore,the $LCM$ of the smallest prime number and the smallest composite number is $4$.

(A) To find the Least Common Multiple $(LCM)$ of $23, 35,$ and $46$:
Step $1$: Find the prime factorization of each number.
$23 = 23^1$ (since $23$ is a prime number).
$35 = 5^1 \times 7^1$.
$46 = 2^1 \times 23^1$.
Step $2$: Identify the highest power of each prime factor present in the numbers.
The prime factors involved are $2, 5, 7,$ and $23$.
The highest powers are $2^1, 5^1, 7^1,$ and $23^1$.
Step $3$: Multiply these highest powers to find the $LCM$.
$LCM = 2^1 \times 5^1 \times 7^1 \times 23^1$
$LCM = 10 \times 7 \times 23$
$LCM = 70 \times 23 = 1610$.

(A) To find the largest positive integer that divides $70$ and $125$ leaving remainders $5$ and $8$ respectively,we subtract the remainders from the numbers first.
$70 - 5 = 65$
$125 - 8 = 117$
Now,we need to find the Greatest Common Divisor $(GCD)$ of $65$ and $117$.
Prime factorization of $65 = 5 \times 13$
Prime factorization of $117 = 3^2 \times 13$
The common factor is $13$.
Therefore,the largest positive integer is $13$.

(B) First,find the $HCF$ of $65$ and $117$ using prime factorization:
$65 = 5 \times 13$
$117 = 9 \times 13 = 3^2 \times 13$
Therefore,$HCF(65, 117) = 13$.
Given the equation: $65m - 117 = HCF(65, 117)$.
Substitute the $HCF$ value: $65m - 117 = 13$.
Add $117$ to both sides: $65m = 13 + 117$.
$65m = 130$.
Divide by $65$: $m = 130 / 65$.
$m = 2$.

(A) We know that for any two positive integers $a$ and $b$,the product of the numbers is equal to the product of their $\text{HCF}$ and $\text{LCM}$.
Mathematically,$a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)$.
Given that $\text{HCF}(a, b) = 7$ and $\text{LCM}(a, b) = 385$.
Therefore,$a \times b = 7 \times 385$.
Calculating the product: $7 \times 385 = 2695$.
Thus,$a \times b = 2695$.

(B) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
The $LCM$ is the product of the highest power of each prime factor involved,which is $2^2 = 4$.
Therefore,the $LCM$ of $2$ and $4$ is $4$.

(A) The smallest two-digit prime number is $11$.
The smallest two-digit composite number is $10$.
To find the $GCD$ of $11$ and $10$:
Prime factorization of $11 = 11^1$.
Prime factorization of $10 = 2^1 \times 5^1$.
Since there are no common prime factors,the $GCD(11, 10) = 1$.

(C) By definition,a prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.
Since $a$ and $b$ are distinct prime numbers,they have no common factors other than $1$.
The Least Common Multiple $(LCM)$ of two numbers is the smallest positive integer that is divisible by both numbers.
For any two numbers $a$ and $b$,the relationship between their product,$HCF$,and $LCM$ is given by: $a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)$.
Since $a$ and $b$ are distinct prime numbers,their Highest Common Factor $(HCF)$ is $1$.
Therefore,$a \times b = 1 \times \text{LCM}(a, b)$.
This implies that $\text{LCM}(a, b) = ab$.

(B) We know that for any two positive integers $p$ and $q$,the product of the numbers is equal to the product of their $HCF$ and $LCM$.
$p \times q = HCF(p, q) \times LCM(p, q)$
Given that $HCF(p, q) = p$.
Substituting this into the formula:
$p \times q = p \times LCM(p, q)$
Dividing both sides by $p$ (assuming $p \neq 0$):
$LCM(p, q) = q$
Therefore,the correct option is $B$.

(C) For any two prime numbers $p_1$ and $p_2$,the Highest Common Factor $(HCF)$ is always $1$ because they have no common factors other than $1$.
The Least Common Multiple $(LCM)$ of two prime numbers is their product,i.e.,$p_1 \times p_2$.
Given prime numbers are $7$ and $11$.
$HCF(7, 11) = 1$.
$LCM(7, 11) = 7 \times 11 = 77$.
Difference = $LCM - HCF = 77 - 1 = 76$.

(A) Step $1$: Find 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 $2$: The $LCM$ is the product of the highest powers of all prime factors involved:
$LCM(12, 15, 21) = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 = 420$
Step $3$: Solve the given equation for $x$:
$420 = 500 + 2x$
$2x = 420 - 500$
$2x = -80$
$x = -40$