HCF and LCM using prime factorisation

Step 1: For two whole numbers, the HCF must exactly divide the LCM. Step 2: (1080\div24=45), so such numbers are possible; the impossible statement is not correct. Step 3: Check all statements before selecting.

Step 1: The HCF must divide the LCM. Step 2: (1080\div24=45), which is a whole number, so such two whole numbers can exist. Step 3: For existence checks, first test divisibility.

Step 1: The HCF must be an exact divisor of the LCM. Step 2: (420) is not exactly divisible by (20), so such two whole numbers are not possible. Step 3: Check this condition before trying to form a pair.

Step 1: LCM uses the higher power. Step 2: The powers of (5) are (1) and (3), so (L) contains (5^3). Step 3: Compare exponents only when the base is the same.

Step 1: Remainder (0) means the number is divisible by all three numbers. Step 2: (48=2^4\times3), (80=2^4\times5), and (112=2^4\times7), so LCM (=2^4\times3\times5\times7=1680). Step 3: For the smallest divisible number, use LCM.

Step 1: In (\frac{L}{H}), divide the LCM by the HCF. Step 2: Subtract powers of the same bases: (2^{5-3}\times3^{4-2}\times5=2^2\times3^2\times5). Step 3: Remember subtraction of exponents during division.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: LCM (=\frac{252\times378}{126}=756). Step 3: Simplify (252) by (126) first to calculate faster.

Step 1: (p) divides (q) because (q=p\times5). Step 2: When one number exactly divides the other, the smaller number is the HCF. Step 3: First identify divisibility between the two numbers.

Step 1: The maximum number of equal parts is found by HCF. Step 2: (96=2^5\times3), (128=2^7), and (160=2^5\times5), so HCF (=2^5=32). 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 present in the third number; the smallest powers are (2^2) and (3). Step 3: Do not include a prime that is not in every number.

Step 1: (35=5\times7), (49=7^2), and (63=3^2\times7). Step 2: The highest power of (7) in the LCM is (2). Step 3: LCM uses the highest power of each prime.

Step 1: When (u) and (v) are coprime, the LCM of (18u) and (18v) is (18uv). Step 2: Since (uv=40), LCM (=18\times40=720). Step 3: Factoring out the HCF is useful in such questions.

Step 1: LCM (=\frac{180\times252}{36}=1260). Step 2: Then (\frac{1260}{36}=35), so the value is (35). Step 3: First find the LCM, then simplify the ratio.

Step 1: A common multiple must contain all required prime powers of both numbers. Step 2: The required highest powers are (2^4), (3^3), and (5). 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), (5), and (7) are common, but (3) appears only in the second number. Step 3: Identify common primes before choosing powers.

Step 1: LCM takes the higher power of (2), and HCF takes the lower power. Step 2: The powers are (6) and (4), so (L) has power (6) and (H) has power (4). Step 3: Do not interchange the power rules of LCM and HCF.

Step 1: Product of two numbers equals HCF times LCM. Step 2: (7) appears only in the LCM as (7^1), so its power in the product is (1). Step 3: In multiplication, add exponents of the same prime.

Step 1: (75=3\times5^2), (125=5^3), and (200=2^3\times5^2). Step 2: The highest powers are (2^3), (3), and (5^3), so LCM (=8\times3\times125=3000). Step 3: Do not forget that (5^3=125).

Step 1: Prime factorise: (75=3\times5^2), (125=5^3), and (200=2^3\times5^2). Step 2: LCM (=2^3\times3\times5^3=3000). Step 3: Use the highest powers to get the final value.

Step 1: Subtracting (4) makes the number divisible by (21), (28), and (36). Step 2: (21=3\times7), (28=2^2\times7), and (36=2^2\times3^2), so LCM (=2^2\times3^2\times7=252). Hence the smallest number is (252+4=256). Step 3: Add the common remainder at the end.