HCF and LCM using prime factorisation

Step 1: The smallest number divisible by all is the LCM. Step 2: (36=2^2\times3^2), (100=2^2\times5^2), and (150=2\times3\times5^2), so LCM (=2^2\times3^2\times5^2=900). Step 3: Always verify the final multiplication before choosing an option.

Step 1: Such a smallest number is the LCM of the three numbers. Step 2: (36=2^2\times3^2), (100=2^2\times5^2), and (150=2\times3\times5^2), so LCM (=2^2\times3^2\times5^2=900). Step 3: Take the highest power of each prime.

Step 1: In (\frac{L}{H}), divide the LCM by the HCF. Step 2: Subtract powers of the same bases: (2^{6-2}\times3^{3-1}\times5^{2-1}\times7=2^4\times3^2\times5\times7). Step 3: Use exponent subtraction in division.

Step 1: Product of two numbers equals HCF times LCM. Step 2: LCM (=\frac{286\times429}{143}=858). Step 3: Divide (286) by (143) first to get (2), then multiply (2\times429).

Step 1: (q=p\times7), so (p) exactly divides (q). Step 2: When one number exactly divides the other, the smaller number is the HCF. Step 3: First check whether one number is a multiple of the other.

Step 1: The maximum number of equal parts is found using HCF. Step 2: (96=2^5\times3), (144=2^4\times3^2), and (192=2^6\times3), so HCF (=2^4\times3=48). Step 3: For maximum equal division, use HCF.

Step 1: HCF includes only primes common to all three numbers. Step 2: (2) and (3) are common, but (5) is not in the third number; the smallest powers are (2^2) and (3^2). Step 3: Do not include a prime that is not present in every number.

Step 1: (44=2^2\times11), (77=7\times11), and (121=11^2). Step 2: The highest power of (11) in the LCM is (2). Step 3: For LCM, choose the highest power.

Step 1: When (u) and (v) are coprime, the LCM of (30u) and (30v) is (30uv). Step 2: Since (uv=26), LCM (=30\times26=780). Step 3: Factoring out the HCF simplifies the question.

Step 1: LCM (=\frac{216\times360}{72}=1080). Step 2: Then (\frac{1080}{72}=15), so the value of the ratio is (15). Step 3: First find the LCM, then simplify the ratio.

Step 1: A common multiple must contain all prime powers required by both numbers. Step 2: The highest powers are (2^6), (3^3), (5), and (7). Step 3: Do not miss any required prime factor while checking a multiple.

Step 1: HCF contains only primes present in both numbers. Step 2: (2), (3), and (7) are common, but (5) appears only in the second number. Step 3: First identify common primes, then choose their smaller powers.

Step 1: LCM takes the higher power of (5), and HCF takes the lower power. Step 2: The powers of (5) are (1) and (3), so (L) has power (3) and (H) has power (1). Step 3: Do not interchange the two rules.

Step 1: Product of two numbers equals HCF times LCM. Step 2: (11) appears only in the LCM as (11^1), so its power in the product is (1). Step 3: In multiplication, add powers of the same prime.

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.