HCF and LCM using prime factorisation

Step 1: HCF takes the smaller powers (2^3) and (3^2). Step 2: LCM takes (2^5), (3^4), (5), and (11), so the ratio becomes (2^2\times3^2\times5\times11). Step 3: Subtract powers when dividing prime factor forms.

Step 1: The smallest number divisible by all given numbers is their LCM. Step 2: (24=2^3\times3), (36=2^2\times3^2), and (54=2\times3^3), so LCM (=2^3\times3^3=216). Step 3: For the smallest exactly divisible number, find the LCM.

Step 1: The greatest common divisor is the HCF. Step 2: (168=2^3\times3\times7), (252=2^2\times3^2\times7), and (420=2^2\times3\times5\times7), so HCF (=2^2\times3\times7=84). Step 3: Use only primes common to all numbers.

Step 1: For two numbers, HCF (\times) LCM (=) product of the two numbers. Step 2: This direct relation is for two numbers. Step 3: Do not apply this formula blindly to three numbers in exams.

Step 1: In HCF, the smaller power of each common prime is used. Step 2: For prime (2), the smaller power must be (2), so (a=2) is possible; for prime (3), (b=2) is possible. Step 3: In unknown power questions, focus on the minimum-power condition.

Step 1: The HCF must divide the LCM. Step 2: (420) is not exactly divisible by (12), so no such pair of whole numbers is possible. Step 3: Check this necessary condition before trying pairs.

Step 1: For the maximum equal number, find the HCF. Step 2: (72=2^3\times3^2) and (96=2^5\times3), so HCF (=2^3\times3=24). Step 3: For maximum equal grouping, use HCF.

Step 1: The next common ringing time is the LCM of the intervals. Step 2: (18=2\times3^2), (24=2^3\times3), and (30=2\times3\times5), so LCM (=2^3\times3^2\times5=360). Step 3: For repeated time events, use LCM.

Step 1: The product of two numbers equals HCF (\times) LCM. Step 2: The powers of (2) are (5) and (2), so the total power is (7). Step 3: When multiplying prime powers with the same base, add the powers.

Step 1: Coprime numbers have HCF (1). Step 2: Since product (=) HCF (\times) LCM, the LCM is (391). Step 3: For coprime numbers, LCM equals the product.

Step 1: A number divisible by both must be a multiple of their LCM. Step 2: The LCM contains (2^3), (3^4), (5), and (7). Step 3: For divisibility, every required prime power must be present.

Step 1: Write both numbers in prime factor form. Step 2: (180=2^2\times3^2\times5) and (252=2^2\times3^2\times7), so the common smaller powers give (2^2\times3^2). Step 3: Do not include non-common factors like (5) or (7) in HCF.

Step 1: For LCM, choose the larger power of each prime. Step 2: The larger power of (2) is (2^5), and that of (3) is (3^2), so LCM (=2^5\times3^2). Step 3: Do not assume the larger given number is always the LCM.

Step 1: HCF includes only common prime factors. Step 2: The common primes are (2) and (3), with smaller powers (2^3) and (3^2). Step 3: A prime appearing in only one number is not written in HCF.

Step 1: Product of two numbers equals HCF (\times) LCM. Step 2: (15\times630=9450), so the product is (9450). Step 3: This formula is directly reliable for two numbers.

Step 1: LCM uses the higher power of each prime. Step 2: The powers of (5) are (3) and (1), so (5^3) appears in the LCM. Step 3: Compare powers separately for each prime base.

Step 1: The maximum number of identical packets is found by HCF. Step 2: (126=2\times3^2\times7) and (210=2\times3\times5\times7), so HCF (=2\times3\times7=42). Step 3: For maximum equal distribution, find the HCF.

Step 1: Remainder (0) means the number is divisible by all the given numbers. Step 2: (45=3^2\times5), (60=2^2\times3\times5), and (75=3\times5^2), so LCM (=2^2\times3^2\times5^2=900). Step 3: For the smallest such number, use LCM.

Step 1: Product of two numbers equals HCF (\times) LCM. Step 2: The other number is (\frac{(2^2\times3)(2^4\times3^3\times5)}{2^4\times3\times5}=2^2\times3^3). Step 3: While dividing prime forms, subtract powers of the same base.

Step 1: HCF contains only common prime factors. Step 2: If there is no common prime factor, the only common divisor is (1). Step 3: Treat such numbers as coprime to solve quickly.