HCF and LCM using prime factorisation

Step 1: HCF uses the smaller power of each common prime. Step 2: The smaller power of (2) must be (6), so (a=6) is possible; the smaller power of (3) must be (4), so (b=4) is possible. Step 3: Check each base condition separately.

Step 1: LCM takes the highest power of every prime. Step 2: The highest power of (3) must be (7), so (a=7); the highest power of (5) must be (3), so (b=3). Step 3: Identify the maximum powers in LCM.

Step 1: The maximum number of identical packets is found by HCF. Step 2: (312=2^3\times3\times13) and (468=2^2\times3^2\times13), so HCF (=2^2\times3\times13=156). Step 3: Use HCF for maximum equal distribution.

Step 1: The next common signal time is the LCM of all intervals. Step 2: (28=2^2\times7), (36=2^2\times3^2), (63=3^2\times7), and (84=2^2\times3\times7), so the LCM is (252). 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 (4), so the total power is (6). Step 3: When multiplying powers with the same base, add the exponents.

Step 1: LCM (=\frac{330\times462}{66}=2310). Step 2: The ratio (2310:66=35:1). Step 3: Do not forget to reduce the ratio to simplest form.

Step 1: The maximum number of equal parts is found by HCF. Step 2: (192=2^6\times3), (288=2^5\times3^2), and (480=2^5\times3\times5), so HCF (=2^5\times3=96). Step 3: In maximum equal division, take the smallest common powers.

Step 1: The smallest number exactly divisible by all is the LCM. Step 2: (121=11^2), (144=2^4\times3^2), and (250=2\times5^3), so LCM (=2^4\times3^2\times5^3\times11^2=2178000). Step 3: Multiply the highest powers carefully.

Step 1: (q=p\times19), so (p) exactly divides (q). Step 2: When one number exactly divides the other, the smaller number is the HCF. Step 3: Identifying a multiple relation saves time.

Step 1: After taking out HCF (84), (r) and (s) are coprime. Step 2: LCM (=84rs=5460), so (rs=65). Step 3: First divide the LCM by the HCF.

Step 1: Compare the powers of (3). Step 2: (216=2^3\times3^3), (324=2^2\times3^4), and (540=2^2\times3^3\times5), so the smallest power is (3). Step 3: HCF uses the smallest power.

Step 1: Prime factorise: (196=2^2\times7^2), (225=3^2\times5^2), and (308=2^2\times7\times11). Step 2: The distinct primes in the LCM are (2), (3), (5), (7), and (11), so there are (5). Step 3: Count distinct prime bases, not powers.

Step 1: LCM takes the higher power, and HCF takes the lower power. Step 2: The powers of (5) are (4) and (1), so (L) has (4) and (H) has (1). Step 3: Do not interchange the two rules.

Step 1: (221=13\times17), (323=17\times19), and (437=19\times23). Step 2: No prime factor is common to all three numbers, so the HCF is (1). Step 3: Only a factor present in all three numbers is taken.

Step 1: LCM (=\frac{264\times396}{132}=792). Step 2: The difference is (792-132=660). Step 3: When difference is asked, first find the LCM correctly.

Step 1: The smallest number divisible by both is their LCM. Step 2: The powers of (3) are (2) and (5), so the higher power (5) is used. Step 3: For divisibility, choose the required highest power.

Step 1: Coprime numbers have HCF (1). Step 2: Product (=) HCF (\times) LCM, so the LCM is (1517). Step 3: For coprime numbers, LCM equals the product.

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

Step 1: (128=2^7), (192=2^6\times3), and (343=7^3). Step 2: (7) appears from (343) as (7^3), so its power in the LCM is (3). Step 3: A prime appearing in only one number is still included in the LCM.

Step 1: Let the numbers be (18m) and (18n), where (m) and (n) are coprime. Step 2: (18mn=6930), so (mn=385=5\times7\times11). Three distinct prime factors give (4) unordered coprime pairs. Step 3: Do not count the reversed order again.