HCF and LCM using prime factorisation

Step 1: Compare the powers of (3). Step 2: (162=2\times3^4), (270=2\times3^3\times5), and (378=2\times3^3\times7), so the smallest power is (3). Step 3: HCF uses the smallest power.

Step 1: Prime factorise: (112=2^4\times7), (180=2^2\times3^2\times5), and (225=3^2\times5^2). Step 2: The distinct primes in the LCM are (2), (3), (5), and (7), so there are (4). 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 (1) and (4), so (L) has (4) and (H) has (1). Step 3: Do not interchange the two rules.

Step 1: (169=13^2), (221=13\times17), and (299=13\times23). Step 2: The common prime in all three is (13), so the HCF should be (13). Step 3: Carefully identify the factor common to all three numbers.

Step 1: (169=13^2), (221=13\times17), and (299=13\times23). Step 2: (13) is common to all three and no larger common factor exists, so HCF is (13). Step 3: Do not miss a prime that appears in all three numbers.

Step 1: (64=2^6) and (5120=2^{10}\times5). Step 2: Product equals HCF times LCM, so the power of (2) is (6+10=16). Step 3: Add exponents when multiplying powers with the same base.

Step 1: LCM (=\frac{224\times336}{112}=672). Step 2: The difference is (672-112=560). Step 3: When difference is asked, find the LCM correctly first.

Step 1: The smallest number divisible by both is their LCM. Step 2: The powers of (3) are (3) 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 (1147). 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 (2^2) and (3). Step 3: Do not include a prime that is not present in every number.

Step 1: (68=2^2\times17), (85=5\times17), and (289=17^2). Step 2: The highest power of (17) in the LCM is (2). Step 3: For LCM, choose the highest power.

Step 1: Let the numbers be (21m) and (21n), where (m) and (n) are coprime. Step 2: (21mn=2310), so (mn=110=2\times5\times11). Three distinct prime factors give (4) unordered coprime pairs. Step 3: Do not count the reversed order again.

Step 1: The HCF must exactly divide the LCM. Step 2: (1950\div75=26), which is a whole number, so such whole numbers can exist. Step 3: For existence checks, first inspect this quotient.

Step 1: (132=2^2\times3\times11) and (220=2^2\times5\times11). Step 2: (5) appears only in the second number as (5^1), so its power in the LCM is (1). Step 3: A prime appearing in only one number is included in the LCM.

Step 1: The smallest number divisible by both is their LCM. Step 2: The highest powers are (2^7), (3^4), (5), and (11). Step 3: Include all required prime powers together.

Step 1: The smaller power of (2) in the HCF must be (4). Step 2: The second number has (2^6), so (a=4) makes the smaller power (4). Step 3: Apply the minimum-power condition in HCF.

Step 1: The highest power of (3) in the LCM must be (6). Step 2: The second number has (3^3), so (b=6) gives the required highest power (6). Step 3: For LCM, check the maximum-power condition.

Step 1: (a=2^4\times3=48). Step 2: (ab=2496), so (b=\frac{2496}{48}=52), but (48) and (52) are not coprime. Step 3: The coprime condition is essential for checking the answer.

Step 1: (98=2\times7^2), (147=3\times7^2), and (245=5\times7^2). Step 2: The highest powers are (2), (3), (5), and (7^2), so LCM (=2\times3\times5\times49=1470). Step 3: Take the common highest power (7^2) only once.

Step 1: Subtracting (15) makes the number divisible by all three numbers. Step 2: (40=2^3\times5), (56=2^3\times7), and (88=2^3\times11), so LCM (=2^3\times5\times7\times11=3080). Hence the number is (3080+15=3095). Step 3: Add the common remainder at the end.