HCF and LCM using prime factorisation

Step 1: The HCF must exactly divide the LCM. Step 2: (1890\div54=35) is not exact, so such whole numbers are not possible. Step 3: Check this necessary condition before searching for pairs.

Step 1: (120=2^3\times3\times5) and (168=2^3\times3\times7). Step 2: In the LCM, (7) appears from the second number as (7^1). Step 3: A prime present in one number is included in the LCM.

Step 1: The smallest number divisible by both is their LCM. Step 2: The highest powers are (2^6), (3^3), (5), and (7). Step 3: For divisibility, all required prime powers must be present.

Step 1: The smaller power of (2) in the HCF must be (3). Step 2: The second number has (2^5), so (a=3) makes the smaller power (3). Step 3: Apply the minimum-power condition directly in HCF questions.

Step 1: The highest power of (3) in the LCM must be (5). Step 2: The second number has (3^2), so (b=5) gives the required highest power (5). Step 3: For LCM, check the maximum-power condition.

Step 1: (a=2^3\times5=40). Step 2: Since (ab=1720), (b=\frac{1720}{40}=43), and (40) and (43) are coprime. Step 3: Use the coprime condition to verify the final answer.

Step 1: In each case, divisor minus remainder is (5), so adding (5) to the number makes it divisible by all three divisors. Step 2: The LCM of (36), (48), and (60) is (720), so the number is (720-5=715). Step 3: When remainders are equally less than divisors, subtract that difference from the LCM.

Step 1: In the LCM, take the highest power of (2). Step 2: The powers of (2) are (4), (6), and (5), so the highest power is (6). 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{27\times1701}{189}=243). Step 3: Notice (189=27\times7) to simplify quickly.

Step 1: Remainder (0) means the number is exactly divisible by all three numbers. Step 2: (144=2^4\times3^2), (180=2^2\times3^2\times5), and (216=2^3\times3^3), so LCM (=2^4\times3^3\times5=2160). 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 (2) and (1), so the smaller power is (1). Step 3: A power of (1) is still an important part of the answer.

Step 1: LCM takes the higher power of (3), and HCF takes the lower power. Step 2: The powers of (3) are (2) and (5), so (L) has (5) and (H) has (2). Step 3: Remember the higher-power and lower-power rules separately.

Step 1: (154=22\times7) and (264=22\times12). Step 2: Since (7) and (12) are coprime, HCF is (22) and LCM is (22\times7\times12=1848). Step 3: In options, factor out the HCF and check whether the remaining numbers are coprime.

Step 1: For two numbers, HCF (\times) LCM equals the product of the two numbers. Step 2: So here the product equals (216\times360). Step 3: Apply this formula directly only for two numbers.

Step 1: (99=3^2\times11), (165=3\times5\times11), and (231=3\times7\times11). Step 2: The common smaller powers are (3) and (11), so HCF (=33). Step 3: Choose only primes present in all three numbers.

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

Step 1: After factoring out HCF (63), (r) and (s) are coprime. Step 2: LCM (=63rs=2079), so (rs=33). Step 3: In such questions, first divide the LCM by the HCF.

Step 1: (H=2^4\times3^2\times5) and (L=2^6\times3^5\times5^2). Step 2: (\frac{L}{H}=2^{6-4}\times3^{5-2}\times5^{2-1}=2^2\times3^3\times5). Step 3: In division, subtract exponents of the same base.

Step 1: (252=2^2\times3^2\times7), (315=3^2\times5\times7), and (420=2^2\times3\times5\times7). Step 2: (H=3\times7=21) and (L=2^2\times3^2\times5\times7=1260), so (L\div H=60). Step 3: For three numbers, find HCF and LCM by their separate rules.

Step 1: The LCM includes all distinct primes appearing in the two numbers. Step 2: The distinct primes are (2), (3), (5), and (13), so there are (4). Step 3: Do not count powers; count only distinct prime bases.