HCF and LCM using prime factorisation

Step 1: (108=2^2\times3^3), (162=2\times3^4), and (270=2\times3^3\times5). Step 2: The highest powers are (2^2), (3^4), and (5), so LCM (=4\times81\times5=1620). Step 3: Use the highest powers to get the final value.

Step 1: After subtracting (11), the number must be divisible by all three numbers. Step 2: (32=2^5), (48=2^4\times3), and (72=2^3\times3^2), so LCM (=2^5\times3^2=288). Hence the number is (288+11=299). Step 3: Add the common remainder at the end.

Step 1: The highest power of (2) in the LCM must be (6). Step 2: The second number has power (4), so (a=6) gives the required highest power (6). Step 3: For LCM, check the maximum-power condition.

Step 1: Product (=) HCF (\times) LCM. Step 2: LCM (=\frac{70560}{28}=2520). Step 3: You can divide first by (7) and then by (4) to simplify.

Step 1: For two numbers, LCM (=\frac{\text{first number}\times\text{second number}}{\text{HCF}}). Step 2: (\frac{154\times231}{77}=462). Step 3: First calculate (154\div77=2) for a quicker solution.

Step 1: HCF takes the smaller power of each common prime. Step 2: The smaller powers of (2), (3), and (11) are (3), (2), and (1), so HCF (=2^3\times3^2\times11). Step 3: Compare the powers for each base separately.

Step 1: (91=7\times13), (143=11\times13), and (187=11\times17). Step 2: There is no prime factor common to all three, so the HCF should be (1). Step 3: For three numbers, a common factor must be present in every number.

Step 1: (91=7\times13), (143=11\times13), and (187=11\times17). Step 2: No prime factor is common to all three numbers, so the HCF is (1). Step 3: A factor common to only two numbers is not enough for the HCF of all three.

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

Step 1: The HCF must exactly divide the LCM. Step 2: (420\div15=28), which is a whole number, so such two whole numbers can exist. Step 3: For existence checks, test divisibility first.

Step 1: Product of two numbers equals HCF times LCM. Step 2: The other number is (\frac{21\times1386}{198}=147). Step 3: Simplifying the division first makes the calculation faster.

Step 1: Compare the powers of (2). Step 2: (120=2^3\times3\times5), (180=2^2\times3^2\times5), and (300=2^2\times3\times5^2), so the smallest power is (2). Step 3: HCF uses the smallest power.

Step 1: Look at the powers of (5). (48) has no factor (5), (75=3\times5^2), and (125=5^3). Step 2: The highest power of (5) in the LCM is (3). Step 3: A missing prime has power zero, but still compare the existing highest power.

Step 1: HCF uses only common prime factors. Step 2: The common primes are (2), (3), and (5), with smaller powers (2^4), (3^2), and (5). Step 3: In exponent questions, choose the smaller power for each base separately.

Step 1: LCM uses the highest power of every prime factor present. Step 2: The highest powers are (2^5), (3^4), (5), (7), and (11). Step 3: Include a prime even if it appears in only one number.

Step 1: For two numbers, product equals HCF times LCM. Step 2: The other number is (\frac{36\times1620}{180}=324). Step 3: Write the relation first and then simplify the division.

Step 1: HCF is (2^2\times3^3\times7), and LCM is (2^5\times3^5\times5\times7^2). Step 2: On division, subtract powers to get (2^3\times3^2\times5\times7). Step 3: Exponent subtraction is very useful in ratio questions.

Step 1: For two whole numbers, the HCF must exactly divide the LCM. Step 2: (1320\div24=55), which is a whole number, so such numbers can exist. Step 3: For existence checks, test this divisibility condition first.

Step 1: Let the numbers be (18m) and (18n), where (m) and (n) are coprime. Step 2: (18mn=1260), so (mn=70=2\times5\times7); this gives (4) unordered coprime factor pairs. Step 3: For a square-free product, split prime factors into two groups to count unordered pairs.

Step 1: Subtracting (13) makes the number divisible by (42), (56), and (70). Step 2: Their LCM is (280), so the smallest number is (280+13=293). Step 3: In common-remainder questions, add the remainder at the end.