HCF and LCM using prime factorisation

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.

Step 1: LCM takes the highest power of every prime. Step 2: The highest power of (3) must be (5), so (a=5); the highest power of (5) must be (2), so (b=2). Step 3: Focus on the maximum powers in LCM.

Step 1: Subtracting (7) makes the number divisible by all three numbers. Step 2: (40=2^3\times5), (54=2\times3^3), and (72=2^3\times3^2), so LCM (=2^3\times3^3\times5=1080). Hence the number is (1080+7=1087). Step 3: Add the common remainder at the end.

Step 1: Compare the powers of (2). Step 2: (96=2^5\times3), (144=2^4\times3^2), and (240=2^4\times3\times5), so the smallest power is (4). Step 3: HCF uses the smallest power.

Step 1: Coprime numbers have HCF (1). Step 2: Product (=) HCF (\times) LCM, so the LCM is (437). Step 3: For coprime numbers, LCM equals the product.

Step 1: The common prime factors are (3) and (11), so HCF (=33). Step 2: LCM (=3\times5\times7\times11=1155), so the sum is (33+1155=1188). Step 3: When sum is asked, find both values separately.

Step 1: The smallest such number is the LCM of the two given numbers. Step 2: The powers of (3) are (2) and (4), so the higher power (4) is used. Step 3: For divisibility, choose the higher power.

Step 1: After factoring out HCF (27), the remaining numbers are coprime. Step 2: LCM (=27rs=1215), so (rs=45). Step 3: In such questions, divide by the given HCF to simplify.

Step 1: (112=2^4\times7) and (168=2^3\times3\times7). Step 2: The common smaller powers are (2^3) and (7), so HCF (=56). Step 3: Compare prime factors before choosing the statement.

Step 1: LCM uses the higher power of a prime. Step 2: The powers of (5) are (2) and (1), so the higher power is (2). Step 3: Compare powers only for the same base.

Step 1: Check prime factors: (45=3^2\times5), (60=2^2\times3\times5), and (84=2^2\times3\times7). Step 2: The distinct primes in the LCM are (2), (3), (5), and (7), so the count is (4). Step 3: Do not count powers as separate primes.

Step 1: (32=2^5) and (768=2^8\times3). Step 2: Product equals HCF times LCM, so the power of (2) should be (5+8=13). Step 3: Add exponents of the same base carefully.

Step 1: (32=2^5) and (768=2^8\times3). Step 2: Product (=) HCF (\times) LCM, so the power of (2) is (5+8=13). Step 3: Exponents with the same base add during multiplication.

Step 1: For maximum equal length, find the HCF. Step 2: (72=2^3\times3^2) and (120=2^3\times3\times5), so HCF (=2^3\times3=24). Step 3: For maximum equal cutting, use HCF.

Step 1: The smallest number divisible by all is the LCM. Step 2: (25=5^2), (40=2^3\times5), and (64=2^6), so LCM (=2^6\times5^2=1600). Step 3: Do not miss (2^6) because of (64).

Step 1: For two numbers, HCF (\times) LCM equals the product of the two numbers. Step 2: Therefore (LH) equals the product of the two given numbers. Step 3: Use this relation directly only for two numbers.

Step 1: HCF takes the smaller power of a common prime. Step 2: The powers of (3) are (3) and (1), so the smaller power is (1). Step 3: A power of (1) matters even when it is not usually written.