HCF and LCM using prime factorisation

Step 1: (a=2^2\times5=20). Step 2: Since (ab=420), (b=\frac{420}{20}=21), and (20) and (21) are coprime. Step 3: The coprime condition helps verify the final answer.

Step 1: Product of two numbers equals HCF times LCM. Step 2: The powers of (3) are (1) and (3), so the total power is (4). Step 3: When multiplying powers with the same base, add the exponents.

Step 1: Prime factorise: (48=2^4\times3), (72=2^3\times3^2), and (108=2^2\times3^3). Step 2: The common smallest powers are (2^2) and (3), so HCF is (12). Step 3: For three numbers, take the smallest power across all.

Step 1: (54=2\times3^3), (81=3^4), and (135=3^3\times5). Step 2: Using highest powers, LCM (=2\times3^4\times5=810). Step 3: A prime appearing in only one number still appears in the LCM.

Step 1: Product of two numbers equals HCF times LCM. Step 2: The other number is (\frac{28\times840}{140}=168). Step 3: Simplify the division to reduce mistakes.

Step 1: Product equals HCF (\times) LCM. Step 2: LCM (=\frac{24000}{20}=1200). Step 3: Check the answer using the given relation.

Step 1: (75=3\times5^2) and (125=5^3). Step 2: HCF (=5^2=25) and LCM (=3\times5^3=375), so (L-H=350). Step 3: Find both values separately first.

Step 1: The ratio means dividing LCM by HCF. Step 2: Subtract powers of the same bases: (2^{4-2}\times3^{2-1}\times5=2^2\times3\times5). Step 3: Remember the exponent rule for division.

Step 1: When (r) and (s) are coprime, LCM is (6rs). Step 2: (7) and (8) are coprime, so LCM (=6\times7\times8=336). Step 3: Factor out the HCF and check the remaining numbers.

Step 1: LCM contains the highest power of every prime present. Step 2: (11) appears only in the second number as (11^1), so its power in (L) is (1). Step 3: A prime appearing in only one number is still included in the LCM.

Step 1: Remainder (0) means the number is exactly divisible by all three numbers. Step 2: (36=2^2\times3^2), (54=2\times3^3), and (90=2\times3^2\times5), so LCM (=2^2\times3^3\times5=540). Step 3: The smallest possible value is always the LCM.

Step 1: HCF includes only the common prime factors. Step 2: The common primes are (2), (3), and (5), with smaller powers (2^4), (3), and (5). Step 3: For HCF, always choose the smaller power.

Step 1: Prime factorise: (84=2^2\times3\times7), (126=2\times3^2\times7), and (198=2\times3^2\times11). Step 2: The highest powers are (2^2), (3^2), (7), and (11), so the LCM is (2772). Step 3: Include every distinct prime factor.

Step 1: Product of two numbers equals HCF times LCM. Step 2: LCM (=\frac{30240}{42}=720). Step 3: Simplifying the division first reduces mistakes.

Step 1: HCF is (2^3\times3^2\times7), and LCM is (2^5\times3^4\times5\times7^2). Step 2: On division, subtract powers to get (2^2\times3^2\times5\times7). Step 3: Prime factor form is the fastest method for ratio questions.

Step 1: The maximum number of identical groups is found by HCF. Step 2: (156=2^2\times3\times13) and (234=2\times3^2\times13), so HCF (=2\times3\times13=78). Step 3: For maximum equal grouping, use HCF.

Step 1: The next common flashing time is the LCM of all intervals. Step 2: (12=2^2\times3), (18=2\times3^2), (27=3^3), and (36=2^2\times3^2), so LCM (=2^2\times3^3=108). Step 3: Use LCM for repeated-time situations.

Step 1: Product of two numbers (=) HCF (\times) LCM. Step 2: The other number is (\frac{35\times1470}{210}=245). Step 3: Divide first to keep the calculation simple.

Step 1: (70=14\times5) and (182=14\times13). Step 2: Since (5) and (13) are coprime, HCF is (14) and LCM is (14\times5\times13=910). Step 3: Factor out the HCF while checking options.

Step 1: HCF uses the smaller power of each common prime. Step 2: For (2), the smaller power must be (4), so (a=4) is possible; for (3), (b=2) is possible. Step 3: Check unknown powers separately.