HCF and LCM using prime factorisation

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.

Step 1: In the LCM, take the highest power of (2). Step 2: The powers of (2) are (5), (7), and (4), so the highest power is (7). Step 3: In LCM, do not add powers; take the highest power.

Step 1: Product of two numbers equals HCF times LCM. Step 2: The other number is (\frac{33\times2145}{165}=429). Step 3: Use (165=33\times5) to simplify the division.

Step 1: Remainder (0) means the number is exactly divisible by all three numbers. Step 2: (168=2^3\times3\times7), (210=2\times3\times5\times7), and (280=2^3\times5\times7), so LCM (=2^3\times3\times5\times7=840). Step 3: For the smallest divisible number, find the LCM.

Step 1: HCF uses the smaller power of a common prime. Step 2: The powers of (5) are (3) and (1), so the smaller power is (1). Step 3: Recognise power (1) correctly in the answer.

Step 1: LCM takes the higher power, while HCF takes the lower power. Step 2: The powers of (3) are (2) and (6), so (L) has (6) and (H) has (2). Step 3: Keep the higher-power and lower-power rules separate.

Step 1: (286=26\times11) and (260=26\times10). Step 2: Since (11) and (10) are coprime, HCF is (26) and LCM is (26\times11\times10=2860). Step 3: In options, factor out the HCF and check if the remaining numbers are coprime.

Step 1: For two numbers, HCF (\times) LCM equals the product of the two numbers. Step 2: Therefore, here the product equals (288\times432). Step 3: Apply this relation directly for two numbers.

Step 1: (143=11\times13), (187=11\times17), and (253=11\times23). Step 2: The common prime in all three is (11), so the HCF is (11). Step 3: Identify the prime common to all three numbers.

Step 1: LCM uses the highest power of (3). Step 2: The powers of (3) are (5), (2), and (4), so the highest power is (5). Step 3: Choose the highest power instead of adding powers.

Step 1: After factoring out HCF (81), (r) and (s) are coprime. Step 2: LCM (=81rs=4617), so (rs=57). Step 3: In such questions, first divide the LCM by the HCF.

Step 1: (H=2^4\times3^3\times5) and (L=2^7\times3^6\times5^2). Step 2: (\frac{L}{H}=2^{7-4}\times3^{6-3}\times5^{2-1}=2^3\times3^3\times5). Step 3: In division, subtract powers of the same base.

Step 1: (330=2\times3\times5\times11), (462=2\times3\times7\times11), and (770=2\times5\times7\times11). Step 2: (H=2\times11=22) and (L=2\times3\times5\times7\times11=2310), so (L\div H=105). Step 3: For three numbers, first check common primes and then all distinct primes.

Step 1: The LCM contains all distinct primes appearing in both numbers. Step 2: The distinct primes are (2), (3), (5), and (17), so there are (4). Step 3: Count only distinct prime bases, not their powers.