HCF and LCM using prime factorisation

Step 1: For the same remainder, take the HCF of the differences of the given numbers. Step 2: The differences are (165), (255), and (420); their HCF is (15). Step 3: In same-remainder divisor questions, work with differences.

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

Step 1: LCM uses the highest power of each prime. Step 2: The highest power of (3) must be (6), so (a=6); the highest power of (5) must be (4), so (b=4). Step 3: In unknown-power LCM questions, identify the maximum power.

Step 1: The maximum number of identical groups is found by HCF. Step 2: (198=2\times3^2\times11) and (330=2\times3\times5\times11), so HCF (=2\times3\times11=66). Step 3: Use HCF for maximum equal distribution.

Step 1: The next common ringing time is the LCM of the intervals. Step 2: (45=3^2\times5), (60=2^2\times3\times5), and (84=2^2\times3\times7), so LCM (=2^2\times3^2\times5\times7=1260). Step 3: Use LCM for repeated-time questions.

Step 1: Product of two numbers equals HCF times LCM. Step 2: The powers of (3) are (2) and (5), so the total power is (7). Step 3: Exponents of the same base add during multiplication.

Step 1: LCM (=\frac{154\times231}{77}=462). Step 2: (\frac{462}{77}=6), so the ratio is (6:1). Step 3: Always write ratios in simplest form.

Step 1: The maximum number of equal parts is found by HCF. Step 2: (96=2^5\times3), (160=2^5\times5), and (224=2^5\times7), so HCF (=2^5=32). Step 3: In maximum equal division questions, identify HCF.

Step 1: The smallest number exactly divisible by all is the LCM. Step 2: (64=2^6), (72=2^3\times3^2), and (125=5^3), so LCM (=2^6\times3^2\times5^3=72000). Step 3: Keeping the highest powers correctly gives the right answer.

Step 1: (B=A\times11), so (A) exactly divides (B). Step 2: When one number divides the other exactly, the smaller number is the HCF. Step 3: Identifying a multiple relation saves time.

Step 1: After factoring out HCF (40), (r) and (s) are coprime. Step 2: LCM (=40rs=1680), so (rs=42). Step 3: Factor out the HCF to make the question shorter.

Step 1: Compare the powers of (3). Step 2: (135=3^3\times5), (180=2^2\times3^2\times5), and (225=3^2\times5^2), so the smallest power is (2). Step 3: HCF uses the smallest power.

Step 1: Prime factorise: (80=2^4\times5), (144=2^4\times3^2), and (225=3^2\times5^2). Step 2: The distinct primes in the LCM are (2), (3), and (5), so the count is (3). Step 3: Do not count powers as separate primes.

Step 1: LCM takes the higher power, while HCF takes the lower power. Step 2: The powers of (5) are (1) and (3), so (L) has power (3) and (H) has power (1). Step 3: Do not interchange the two rules.

Step 1: (121=11^2), (143=11\times13), and (169=13^2). Step 2: No prime factor is common to all three numbers, so the HCF is (1). Step 3: For three numbers, a common factor must appear in all of them.

Step 1: (32=2^5) and (1536=2^9\times3). Step 2: Product equals HCF times LCM, so the power of (2) is (5+9=14). Step 3: Add exponents when multiplying powers with the same base.

Step 1: LCM (=\frac{180\times252}{36}=1260). Step 2: The difference is (1260-36=1224). Step 3: When difference is asked, find both values clearly first.

Step 1: Coprime numbers have HCF (1). Step 2: Product (=) HCF (\times) LCM, so the LCM is (899). Step 3: For coprime numbers, the LCM equals the product.

Step 1: HCF includes only primes common to all three numbers. Step 2: (2) and (3) are common, but (5) is absent in the third number; the smallest powers are (2^2) and (3). Step 3: Do not include a prime that is not present in every number.

Step 1: Let the numbers be (15m) and (15n), where (m) and (n) are coprime. Step 2: (15mn=420), so (mn=28=2^2\times7); the unordered coprime pairs are ((1,28)) and ((4,7)), so the count is (2). Step 3: (m) and (n) must not share a prime factor.