HCF and LCM using prime factorisation

Step 1: The HCF must exactly divide the LCM. Step 2: (990) is not exactly divisible by (44), so such whole numbers are not possible. Step 3: Check this necessary condition before searching for pairs.

Step 1: LCM takes the higher power of every prime. Step 2: The powers of (11) are (1) and (2), so (L) contains (11^2). Step 3: Compare powers only for the same base.

Step 1: The smallest number divisible by both is their LCM. Step 2: The highest powers are (2^6), (3^2), (5), and (7), so the value is (64\times9\times5\times7=20160). Step 3: Include all required prime powers together.

Step 1: The smaller power of (2) in the HCF must be (5). Step 2: The second number has (2^7), so (a=5) makes the smaller power (5). 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=2^3\times7=56). Step 2: Since (ab=1736), (b=\frac{1736}{56}=31), and (56) and (31) are coprime. Step 3: Use the coprime condition to verify the final answer.

Step 1: (102=2\times3\times17), (170=2\times5\times17), and (255=3\times5\times17). Step 2: The required primes are (2), (3), (5), and (17), so the LCM is (510). Step 3: Take a common prime once and include every distinct prime.

Step 1: Subtracting (17) makes the number divisible by all three numbers. Step 2: (64=2^6), (80=2^4\times5), and (96=2^5\times3), so the LCM is (960). Hence the number is (960+17=977). 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 (6), (8), and (5), so the highest power is (8). 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{55\times3575}{275}=715). Step 3: Use (275=55\times5) to simplify the division.

Step 1: Remainder (0) means the number is exactly divisible by all three numbers. Step 2: (225=3^2\times5^2), (270=2\times3^3\times5), and (315=3^2\times5\times7), so the LCM is (9450). 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 (2) are (8) and (5), so the smaller power is (5). Step 3: Do not add the powers; take only the smaller one.

Step 1: LCM takes the higher power, while HCF takes the lower power. Step 2: The powers of (3) are (7) and (3), so (L) has (7) and (H) has (3). Step 3: Keep the higher-power and lower-power rules separate.

Step 1: (78=26\times3) and (286=26\times11). Step 2: Since (3) and (11) are coprime, HCF is (26) and LCM is (26\times3\times11=858). Step 3: In options, factor out the HCF and check whether the remaining numbers are coprime.

Step 1: (221=13\times17), (323=17\times19), and (391=17\times23). Step 2: The common prime in all three is (17), so the HCF is (17). Step 3: Identify the prime common to all three numbers.

Step 1: After factoring out HCF (90), (r) and (s) are coprime. Step 2: LCM (=90rs=6930), so (rs=77). Step 3: In such questions, first divide the LCM by the HCF.

Step 1: (H=2^5\times3^3\times5) and (L=2^8\times3^7\times5^2). Step 2: (\frac{L}{H}=2^{8-5}\times3^{7-3}\times5^{2-1}=2^3\times3^4\times5). Step 3: In division, subtract powers of the same base.

Step 1: (420=2^2\times3\times5\times7), (660=2^2\times3\times5\times11), and (924=2^2\times3\times7\times11). Step 2: (H=2^2\times3=12) and (L=2^2\times3\times5\times7\times11=4620), so (L\div H=385). Step 3: For three numbers, check common primes first and then all distinct primes.

Step 1: The LCM contains all distinct primes appearing in the numbers. Step 2: The distinct primes are (2), (3), (5), (7), and (13), so there are (5). Step 3: Count only distinct prime bases, not their powers.

Step 1: The HCF must exactly divide the LCM. Step 2: (4620\div84=55), which is a whole number, so such whole numbers can exist. Step 3: For existence checks, first test this necessary condition.