Fundamental Theorem Of Arithmetic PYQs
Overview
This page provides comprehensive Class 10 Maths Fundamental Theorem Of Arithmetic PYQs | Real Numbers. Fundamental Theorem Of Arithmetic previous year questions for Class 10 Maths Real Numbers. Practice CBSE board PYQs with step-by-step solutions on SJMaths.
Practice Previous Year Questions Topic-wise
Q1
2026
00:00
Assertion (A): HCF(36m², 18m) = 18m, where m is a prime number.
Reason (R): HCF of two numbers is always less than or equal to the smaller number.
Reason (R): HCF of two numbers is always less than or equal to the smaller number.
Find HCF of $36m^2$ and $18m$.
$36m^2 = 2^2 \times 3^2 \times m^2$
$18m = 2 \times 3^2 \times m$
$HCF = 2 \times 3^2 \times m = 18m$. So, Assertion is true.
Reason: HCF of $a$ and $b$ divides both $a$ and $b$, so it must be $\le \min(a, b)$. This is generally true.
Is R the correct explanation for A? Yes, $18m$ is the smaller number and divides the larger number, so it is the HCF.
Final Answer: (A) Both A and R are true and R is the correct explanation of A.
Q3
2025
00:00
Which of the following CANNOT be the unit digit of $8^n$ (n is a natural number)?
For a number to end with the digit 0, its prime factorization must contain both 2 and 5.
$8^n = (2^3)^n = 2^{3n}$.
The prime factorization of $8^n$ contains only the prime number 2.
Since it does not contain the prime number 5, $8^n$ cannot end with the digit 0 for any natural number $n$.
Final Answer: (C) 0
Q4
2025
00:00
Assertion (A): For primes x, y with $x < y$: HCF(x,y) = x and LCM(x,y) = y.
Reason (R): HCF(x,y) is always $\le$ LCM(x,y) for any two natural numbers.
Reason (R): HCF(x,y) is always $\le$ LCM(x,y) for any two natural numbers.
If x and y are prime numbers, their only common factor is 1. Thus, HCF(x,y) = 1.
Their LCM is their product: LCM(x,y) = xy.
The Assertion states HCF = x and LCM = y, which is False.
Reason: For any natural numbers a and b, $HCF(a,b) \le a \le LCM(a,b)$. So Reason is True.
Final Answer: (D) Assertion (A) is false, but Reason (R) is true.
Q5
2025
00:00
Assertion (A): For any two natural numbers, HCF $\times$ LCM = product of the numbers.
Reason (R): LCM(x,y) is always a multiple of HCF(x,y).
Reason (R): LCM(x,y) is always a multiple of HCF(x,y).
Assertion: For any two natural numbers $a$ and $b$, it is a proven property that $HCF(a,b) \times LCM(a,b) = a \times b$. So, Assertion is true.
Reason: Since HCF is a factor of both numbers, and LCM is a multiple of both numbers, the LCM is always a multiple of the HCF. So, Reason is true.
However, Reason is not the correct explanation for the Assertion. The Assertion is a specific formula.
Final Answer: (B) Both A and R are true, but R is NOT the correct explanation of A.
Q6
2024
00:00
LCM of two numbers is 14 times their HCF. Sum of LCM and HCF is 600. If one number is 280, find the other.
Let HCF = $h$ and LCM = $l$. Given: $l = 14h$.
Also given: $l + h = 600$.
Substitute $l$: $14h + h = 600 \implies 15h = 600 \implies h = 40$.
So, HCF = 40. And LCM = $14 \times 40 = 560$.
We know: $LCM \times HCF = \text{Product of two numbers}$.
Let the other number be $x$.
$560 \times 40 = 280 \times x$
$x = \frac{560 \times 40}{280} = 2 \times 40 = 80$.
Final Answer: 80
Q7
2024
00:00
If $p = a^2b^3$ and $q = a^3b$ (a, b are prime numbers), then HCF(p, q) is:
$p = a^2 \times b^3$
$q = a^3 \times b$
HCF is the product of the smallest powers of each common prime factor.
Smallest power of $a$ is $a^2$. Smallest power of $b$ is $b^1 = b$.
$HCF(p, q) = a^2b$.
Final Answer: (C) a²b
Q8
2023
00:00
The LCM of the smallest prime number and the smallest composite number is __.
Smallest prime number = 2
Smallest composite number = 4
$LCM(2, 4) = 4$
Final Answer: 4
Q9
2023
00:00
If HCF(a, b) = 12 and a $\times$ b = 1800, then LCM(a, b) = __.
We know that $HCF(a,b) \times LCM(a,b) = a \times b$.
$12 \times LCM = 1800$
$LCM = \frac{1800}{12} = 150$
Final Answer: 150
Q11
2021
00:00
LCM of two numbers is 182 and HCF is 13. If one number is 26, find the other.
We know that $LCM \times HCF = \text{Product of numbers}$.
$182 \times 13 = 26 \times x$
$x = \frac{182 \times 13}{26} = \frac{182}{2} = 91$
Final Answer: 91
Q13
2019
00:00
Express 98 as a product of its prime factors.
Divide 98 by 2: $98 = 2 \times 49$
Divide 49 by 7: $49 = 7 \times 7 = 7^2$
So, $98 = 2 \times 7^2$
Final Answer: 2 $\times$ 7²
Q14
2018
00:00
If HCF(26, 169) = 13, then LCM(26, 169) = __.
$LCM \times HCF = \text{Product of numbers}$
$LCM \times 13 = 26 \times 169$
$LCM = \frac{26 \times 169}{13} = 2 \times 169 = 338$
Final Answer: 338
Q15
2017
00:00
Is $7 \times 11 \times 13 \times 15 + 15$ a composite number? State True/False with justification.
Let the number be $N = 7 \times 11 \times 13 \times 15 + 15$.
Take 15 common: $N = 15(7 \times 11 \times 13 + 1)$
$N = 15 \times (1001 + 1) = 15 \times 1002$.
Since $N$ can be expressed as a product of integers other than 1 and itself, it has more than two factors.
Therefore, it is a composite number.
Final Answer: True
Q19
2022
00:00
Find the HCF and LCM of 404 and 96 using the Fundamental Theorem of Arithmetic.
Prime factorization of 404:
$404 = 2 \times 202 = 2^2 \times 101$
Prime factorization of 96:
$96 = 2 \times 48 = 2^2 \times 24 = 2^3 \times 12 = 2^4 \times 6 = 2^5 \times 3$
$HCF(404, 96)$ = product of the smallest power of each common prime factor involved in the numbers.
$HCF = 2^2 = 4$
$LCM(404, 96)$ = product of the greatest power of each prime factor involved in the numbers.
$LCM = 2^5 \times 3 \times 101 = 32 \times 3 \times 101 = 96 \times 101 = 9696$
Final Answer: HCF = 4, LCM = 9696
Q21
2020
00:00
Find HCF of 96 and 404 by prime factorisation. Hence, find their LCM.
Prime factorization: $96 = 2^5 \times 3$, $404 = 2^2 \times 101$
$HCF(96, 404) = 2^2 = 4$
Using the formula $LCM \times HCF = a \times b$
$LCM \times 4 = 96 \times 404$
$LCM = \frac{96 \times 404}{4} = 96 \times 101 = 9696$
Final Answer: HCF = 4, LCM = 9696
Q22
2019
00:00
Given HCF(306, 657) = 9, find LCM(306, 657).
We know: $LCM(a,b) \times HCF(a,b) = a \times b$
$LCM(306, 657) \times 9 = 306 \times 657$
$LCM = \frac{306 \times 657}{9} = 34 \times 657 = 22338$
Final Answer: 22338
Q23
2019
00:00
Find HCF and LCM of 306 and 54. Verify HCF $\times$ LCM = Product of both numbers.
Prime factorization:
$306 = 2 \times 153 = 2 \times 3 \times 51 = 2 \times 3^2 \times 17$
$54 = 2 \times 27 = 2 \times 3^3$
$HCF = 2 \times 3^2 = 18$
$LCM = 2 \times 3^3 \times 17 = 2 \times 27 \times 17 = 918$
Verification:
$HCF \times LCM = 18 \times 918 = 16524$
Product of numbers = $306 \times 54 = 16524$
LHS = RHS. Verified.
Final Answer: HCF = 18, LCM = 918. Verified.
Q28
2022
00:00
Find the LCM and HCF of 12, 15 and 21 by prime factorisation method.
Prime factorization:
$12 = 2^2 \times 3$
$15 = 3 \times 5$
$21 = 3 \times 7$
HCF is the product of common prime factors with smallest power.
$HCF = 3$
LCM is the product of all prime factors with highest power.
$LCM = 2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420$
Final Answer: HCF = 3, LCM = 420
Q30
2018
00:00
Explain why $(7\times11\times13 + 13)$ and $(7\times6\times5\times4\times3\times2\times1 + 5)$ are composite numbers.
Let $N_1 = 7\times11\times13 + 13$
$N_1 = 13 \times (7\times11 + 1) = 13 \times (77 + 1) = 13 \times 78$.
Since $N_1$ can be expressed as a product of two integers greater than 1, it has more than two factors. Hence, it is composite.
Let $N_2 = 7\times6\times5\times4\times3\times2\times1 + 5$
$N_2 = 5 \times (7\times6\times4\times3\times2\times1 + 1) = 5 \times (1008 + 1) = 5 \times 1009$.
Similarly, $N_2$ has more than two factors and is composite.
Final Answer: Both numbers can be factorized into primes other than 1 and themselves, so they are composite.
Assertion (A): HCF(36m², 18m) = 18m, where m is a prime number.
Reason (R): HCF of two numbers is always less than or equal to the smaller number.
Reason (R): HCF of two numbers is always less than or equal to the smaller number.
Find HCF of $36m^2$ and $18m$.
$36m^2 = 2^2 \times 3^2 \times m^2$
$18m = 2 \times 3^2 \times m$
$HCF = 2 \times 3^2 \times m = 18m$. So, Assertion is true.
Reason: HCF of $a$ and $b$ divides both $a$ and $b$, so it must be $\le \min(a, b)$. This is generally true.
Is R the correct explanation for A? Yes, $18m$ is the smaller number and divides the larger number, so it is the HCF.
Final Answer: (A) Both A and R are true and R is the correct explanation of A.
Q3
2025
00:00
Which of the following CANNOT be the unit digit of $8^n$ (n is a natural number)?
For a number to end with the digit 0, its prime factorization must contain both 2 and 5.
$8^n = (2^3)^n = 2^{3n}$.
The prime factorization of $8^n$ contains only the prime number 2.
Since it does not contain the prime number 5, $8^n$ cannot end with the digit 0 for any natural number $n$.
Final Answer: (C) 0
Q4
2025
00:00
Assertion (A): For primes x, y with $x < y$: HCF(x,y) = x and LCM(x,y) = y.
Reason (R): HCF(x,y) is always $\le$ LCM(x,y) for any two natural numbers.
Reason (R): HCF(x,y) is always $\le$ LCM(x,y) for any two natural numbers.
If x and y are prime numbers, their only common factor is 1. Thus, HCF(x,y) = 1.
Their LCM is their product: LCM(x,y) = xy.
The Assertion states HCF = x and LCM = y, which is False.
Reason: For any natural numbers a and b, $HCF(a,b) \le a \le LCM(a,b)$. So Reason is True.
Final Answer: (D) Assertion (A) is false, but Reason (R) is true.
Q5
2025
00:00
Assertion (A): For any two natural numbers, HCF $\times$ LCM = product of the numbers.
Reason (R): LCM(x,y) is always a multiple of HCF(x,y).
Reason (R): LCM(x,y) is always a multiple of HCF(x,y).
Assertion: For any two natural numbers $a$ and $b$, it is a proven property that $HCF(a,b) \times LCM(a,b) = a \times b$. So, Assertion is true.
Reason: Since HCF is a factor of both numbers, and LCM is a multiple of both numbers, the LCM is always a multiple of the HCF. So, Reason is true.
However, Reason is not the correct explanation for the Assertion. The Assertion is a specific formula.
Final Answer: (B) Both A and R are true, but R is NOT the correct explanation of A.
Q6
2024
00:00
LCM of two numbers is 14 times their HCF. Sum of LCM and HCF is 600. If one number is 280, find the other.
Let HCF = $h$ and LCM = $l$. Given: $l = 14h$.
Also given: $l + h = 600$.
Substitute $l$: $14h + h = 600 \implies 15h = 600 \implies h = 40$.
So, HCF = 40. And LCM = $14 \times 40 = 560$.
We know: $LCM \times HCF = \text{Product of two numbers}$.
Let the other number be $x$.
$560 \times 40 = 280 \times x$
$x = \frac{560 \times 40}{280} = 2 \times 40 = 80$.
Final Answer: 80
Q7
2024
00:00
If $p = a^2b^3$ and $q = a^3b$ (a, b are prime numbers), then HCF(p, q) is:
$p = a^2 \times b^3$
$q = a^3 \times b$
HCF is the product of the smallest powers of each common prime factor.
Smallest power of $a$ is $a^2$. Smallest power of $b$ is $b^1 = b$.
$HCF(p, q) = a^2b$.
Final Answer: (C) a²b
Q8
2023
00:00
The LCM of the smallest prime number and the smallest composite number is __.
Smallest prime number = 2
Smallest composite number = 4
$LCM(2, 4) = 4$
Final Answer: 4
Q9
2023
00:00
If HCF(a, b) = 12 and a $\times$ b = 1800, then LCM(a, b) = __.
We know that $HCF(a,b) \times LCM(a,b) = a \times b$.
$12 \times LCM = 1800$
$LCM = \frac{1800}{12} = 150$
Final Answer: 150
Q11
2021
00:00
LCM of two numbers is 182 and HCF is 13. If one number is 26, find the other.
We know that $LCM \times HCF = \text{Product of numbers}$.
$182 \times 13 = 26 \times x$
$x = \frac{182 \times 13}{26} = \frac{182}{2} = 91$
Final Answer: 91
Q13
2019
00:00
Express 98 as a product of its prime factors.
Divide 98 by 2: $98 = 2 \times 49$
Divide 49 by 7: $49 = 7 \times 7 = 7^2$
So, $98 = 2 \times 7^2$
Final Answer: 2 $\times$ 7²
Q14
2018
00:00
If HCF(26, 169) = 13, then LCM(26, 169) = __.
$LCM \times HCF = \text{Product of numbers}$
$LCM \times 13 = 26 \times 169$
$LCM = \frac{26 \times 169}{13} = 2 \times 169 = 338$
Final Answer: 338
Q15
2017
00:00
Is $7 \times 11 \times 13 \times 15 + 15$ a composite number? State True/False with justification.
Let the number be $N = 7 \times 11 \times 13 \times 15 + 15$.
Take 15 common: $N = 15(7 \times 11 \times 13 + 1)$
$N = 15 \times (1001 + 1) = 15 \times 1002$.
Since $N$ can be expressed as a product of integers other than 1 and itself, it has more than two factors.
Therefore, it is a composite number.
Final Answer: True
Q19
2022
00:00
Find the HCF and LCM of 404 and 96 using the Fundamental Theorem of Arithmetic.
Prime factorization of 404:
$404 = 2 \times 202 = 2^2 \times 101$
Prime factorization of 96:
$96 = 2 \times 48 = 2^2 \times 24 = 2^3 \times 12 = 2^4 \times 6 = 2^5 \times 3$
$HCF(404, 96)$ = product of the smallest power of each common prime factor involved in the numbers.
$HCF = 2^2 = 4$
$LCM(404, 96)$ = product of the greatest power of each prime factor involved in the numbers.
$LCM = 2^5 \times 3 \times 101 = 32 \times 3 \times 101 = 96 \times 101 = 9696$
Final Answer: HCF = 4, LCM = 9696
Q21
2020
00:00
Find HCF of 96 and 404 by prime factorisation. Hence, find their LCM.
Prime factorization: $96 = 2^5 \times 3$, $404 = 2^2 \times 101$
$HCF(96, 404) = 2^2 = 4$
Using the formula $LCM \times HCF = a \times b$
$LCM \times 4 = 96 \times 404$
$LCM = \frac{96 \times 404}{4} = 96 \times 101 = 9696$
Final Answer: HCF = 4, LCM = 9696
Q22
2019
00:00
Given HCF(306, 657) = 9, find LCM(306, 657).
We know: $LCM(a,b) \times HCF(a,b) = a \times b$
$LCM(306, 657) \times 9 = 306 \times 657$
$LCM = \frac{306 \times 657}{9} = 34 \times 657 = 22338$
Final Answer: 22338
Q23
2019
00:00
Find HCF and LCM of 306 and 54. Verify HCF $\times$ LCM = Product of both numbers.
Prime factorization:
$306 = 2 \times 153 = 2 \times 3 \times 51 = 2 \times 3^2 \times 17$
$54 = 2 \times 27 = 2 \times 3^3$
$HCF = 2 \times 3^2 = 18$
$LCM = 2 \times 3^3 \times 17 = 2 \times 27 \times 17 = 918$
Verification:
$HCF \times LCM = 18 \times 918 = 16524$
Product of numbers = $306 \times 54 = 16524$
LHS = RHS. Verified.
Final Answer: HCF = 18, LCM = 918. Verified.
Q28
2022
00:00
Find the LCM and HCF of 12, 15 and 21 by prime factorisation method.
Prime factorization:
$12 = 2^2 \times 3$
$15 = 3 \times 5$
$21 = 3 \times 7$
HCF is the product of common prime factors with smallest power.
$HCF = 3$
LCM is the product of all prime factors with highest power.
$LCM = 2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420$
Final Answer: HCF = 3, LCM = 420
Q30
2018
00:00
Explain why $(7\times11\times13 + 13)$ and $(7\times6\times5\times4\times3\times2\times1 + 5)$ are composite numbers.
Let $N_1 = 7\times11\times13 + 13$
$N_1 = 13 \times (7\times11 + 1) = 13 \times (77 + 1) = 13 \times 78$.
Since $N_1$ can be expressed as a product of two integers greater than 1, it has more than two factors. Hence, it is composite.
Let $N_2 = 7\times6\times5\times4\times3\times2\times1 + 5$
$N_2 = 5 \times (7\times6\times4\times3\times2\times1 + 1) = 5 \times (1008 + 1) = 5 \times 1009$.
Similarly, $N_2$ has more than two factors and is composite.
Final Answer: Both numbers can be factorized into primes other than 1 and themselves, so they are composite.
Assertion (A): HCF(36m², 18m) = 18m, where m is a prime number.
Reason (R): HCF of two numbers is always less than or equal to the smaller number.
Reason (R): HCF of two numbers is always less than or equal to the smaller number.
Find HCF of $36m^2$ and $18m$.
$36m^2 = 2^2 \times 3^2 \times m^2$
$18m = 2 \times 3^2 \times m$
$HCF = 2 \times 3^2 \times m = 18m$. So, Assertion is true.
Reason: HCF of $a$ and $b$ divides both $a$ and $b$, so it must be $\le \min(a, b)$. This is generally true.
Is R the correct explanation for A? Yes, $18m$ is the smaller number and divides the larger number, so it is the HCF.
Final Answer: (A) Both A and R are true and R is the correct explanation of A.
Q3
2025
00:00
Which of the following CANNOT be the unit digit of $8^n$ (n is a natural number)?
For a number to end with the digit 0, its prime factorization must contain both 2 and 5.
$8^n = (2^3)^n = 2^{3n}$.
The prime factorization of $8^n$ contains only the prime number 2.
Since it does not contain the prime number 5, $8^n$ cannot end with the digit 0 for any natural number $n$.
Final Answer: (C) 0
Q4
2025
00:00
Assertion (A): For primes x, y with $x < y$: HCF(x,y) = x and LCM(x,y) = y.
Reason (R): HCF(x,y) is always $\le$ LCM(x,y) for any two natural numbers.
Reason (R): HCF(x,y) is always $\le$ LCM(x,y) for any two natural numbers.
If x and y are prime numbers, their only common factor is 1. Thus, HCF(x,y) = 1.
Their LCM is their product: LCM(x,y) = xy.
The Assertion states HCF = x and LCM = y, which is False.
Reason: For any natural numbers a and b, $HCF(a,b) \le a \le LCM(a,b)$. So Reason is True.
Final Answer: (D) Assertion (A) is false, but Reason (R) is true.
Q5
2025
00:00
Assertion (A): For any two natural numbers, HCF $\times$ LCM = product of the numbers.
Reason (R): LCM(x,y) is always a multiple of HCF(x,y).
Reason (R): LCM(x,y) is always a multiple of HCF(x,y).
Assertion: For any two natural numbers $a$ and $b$, it is a proven property that $HCF(a,b) \times LCM(a,b) = a \times b$. So, Assertion is true.
Reason: Since HCF is a factor of both numbers, and LCM is a multiple of both numbers, the LCM is always a multiple of the HCF. So, Reason is true.
However, Reason is not the correct explanation for the Assertion. The Assertion is a specific formula.
Final Answer: (B) Both A and R are true, but R is NOT the correct explanation of A.
Q6
2024
00:00
LCM of two numbers is 14 times their HCF. Sum of LCM and HCF is 600. If one number is 280, find the other.
Let HCF = $h$ and LCM = $l$. Given: $l = 14h$.
Also given: $l + h = 600$.
Substitute $l$: $14h + h = 600 \implies 15h = 600 \implies h = 40$.
So, HCF = 40. And LCM = $14 \times 40 = 560$.
We know: $LCM \times HCF = \text{Product of two numbers}$.
Let the other number be $x$.
$560 \times 40 = 280 \times x$
$x = \frac{560 \times 40}{280} = 2 \times 40 = 80$.
Final Answer: 80
Q7
2024
00:00
If $p = a^2b^3$ and $q = a^3b$ (a, b are prime numbers), then HCF(p, q) is:
$p = a^2 \times b^3$
$q = a^3 \times b$
HCF is the product of the smallest powers of each common prime factor.
Smallest power of $a$ is $a^2$. Smallest power of $b$ is $b^1 = b$.
$HCF(p, q) = a^2b$.
Final Answer: (C) a²b
Q8
2023
00:00
The LCM of the smallest prime number and the smallest composite number is __.
Smallest prime number = 2
Smallest composite number = 4
$LCM(2, 4) = 4$
Final Answer: 4
Q9
2023
00:00
If HCF(a, b) = 12 and a $\times$ b = 1800, then LCM(a, b) = __.
We know that $HCF(a,b) \times LCM(a,b) = a \times b$.
$12 \times LCM = 1800$
$LCM = \frac{1800}{12} = 150$
Final Answer: 150
Q11
2021
00:00
LCM of two numbers is 182 and HCF is 13. If one number is 26, find the other.
We know that $LCM \times HCF = \text{Product of numbers}$.
$182 \times 13 = 26 \times x$
$x = \frac{182 \times 13}{26} = \frac{182}{2} = 91$
Final Answer: 91
Q13
2019
00:00
Express 98 as a product of its prime factors.
Divide 98 by 2: $98 = 2 \times 49$
Divide 49 by 7: $49 = 7 \times 7 = 7^2$
So, $98 = 2 \times 7^2$
Final Answer: 2 $\times$ 7²
Q14
2018
00:00
If HCF(26, 169) = 13, then LCM(26, 169) = __.
$LCM \times HCF = \text{Product of numbers}$
$LCM \times 13 = 26 \times 169$
$LCM = \frac{26 \times 169}{13} = 2 \times 169 = 338$
Final Answer: 338
Q15
2017
00:00
Is $7 \times 11 \times 13 \times 15 + 15$ a composite number? State True/False with justification.
Let the number be $N = 7 \times 11 \times 13 \times 15 + 15$.
Take 15 common: $N = 15(7 \times 11 \times 13 + 1)$
$N = 15 \times (1001 + 1) = 15 \times 1002$.
Since $N$ can be expressed as a product of integers other than 1 and itself, it has more than two factors.
Therefore, it is a composite number.
Final Answer: True
Q19
2022
00:00
Find the HCF and LCM of 404 and 96 using the Fundamental Theorem of Arithmetic.
Prime factorization of 404:
$404 = 2 \times 202 = 2^2 \times 101$
Prime factorization of 96:
$96 = 2 \times 48 = 2^2 \times 24 = 2^3 \times 12 = 2^4 \times 6 = 2^5 \times 3$
$HCF(404, 96)$ = product of the smallest power of each common prime factor involved in the numbers.
$HCF = 2^2 = 4$
$LCM(404, 96)$ = product of the greatest power of each prime factor involved in the numbers.
$LCM = 2^5 \times 3 \times 101 = 32 \times 3 \times 101 = 96 \times 101 = 9696$
Final Answer: HCF = 4, LCM = 9696
Q21
2020
00:00
Find HCF of 96 and 404 by prime factorisation. Hence, find their LCM.
Prime factorization: $96 = 2^5 \times 3$, $404 = 2^2 \times 101$
$HCF(96, 404) = 2^2 = 4$
Using the formula $LCM \times HCF = a \times b$
$LCM \times 4 = 96 \times 404$
$LCM = \frac{96 \times 404}{4} = 96 \times 101 = 9696$
Final Answer: HCF = 4, LCM = 9696
Q22
2019
00:00
Given HCF(306, 657) = 9, find LCM(306, 657).
We know: $LCM(a,b) \times HCF(a,b) = a \times b$
$LCM(306, 657) \times 9 = 306 \times 657$
$LCM = \frac{306 \times 657}{9} = 34 \times 657 = 22338$
Final Answer: 22338
Q23
2019
00:00
Find HCF and LCM of 306 and 54. Verify HCF $\times$ LCM = Product of both numbers.
Prime factorization:
$306 = 2 \times 153 = 2 \times 3 \times 51 = 2 \times 3^2 \times 17$
$54 = 2 \times 27 = 2 \times 3^3$
$HCF = 2 \times 3^2 = 18$
$LCM = 2 \times 3^3 \times 17 = 2 \times 27 \times 17 = 918$
Verification:
$HCF \times LCM = 18 \times 918 = 16524$
Product of numbers = $306 \times 54 = 16524$
LHS = RHS. Verified.
Final Answer: HCF = 18, LCM = 918. Verified.
Q28
2022
00:00
Find the LCM and HCF of 12, 15 and 21 by prime factorisation method.
Prime factorization:
$12 = 2^2 \times 3$
$15 = 3 \times 5$
$21 = 3 \times 7$
HCF is the product of common prime factors with smallest power.
$HCF = 3$
LCM is the product of all prime factors with highest power.
$LCM = 2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420$
Final Answer: HCF = 3, LCM = 420
Q30
2018
00:00
Explain why $(7\times11\times13 + 13)$ and $(7\times6\times5\times4\times3\times2\times1 + 5)$ are composite numbers.
Let $N_1 = 7\times11\times13 + 13$
$N_1 = 13 \times (7\times11 + 1) = 13 \times (77 + 1) = 13 \times 78$.
Since $N_1$ can be expressed as a product of two integers greater than 1, it has more than two factors. Hence, it is composite.
Let $N_2 = 7\times6\times5\times4\times3\times2\times1 + 5$
$N_2 = 5 \times (7\times6\times4\times3\times2\times1 + 1) = 5 \times (1008 + 1) = 5 \times 1009$.
Similarly, $N_2$ has more than two factors and is composite.
Final Answer: Both numbers can be factorized into primes other than 1 and themselves, so they are composite.