HCF and LCM using prime factorisation

Step 1: After subtracting (4), the number must be divisible by all three numbers. Step 2: The LCM of (21), (28), and (36) is (252), so the number is (252+4=256). Step 3: In common-remainder questions, find the LCM first.

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

Step 1: Product (=) HCF (\times) LCM. Step 2: LCM (=\frac{9504}{12}=792). Step 3: You can simplify (12) as (3\times4) while dividing.

Step 1: Product of two numbers equals HCF times LCM. Step 2: LCM (=\frac{140\times196}{28}=980). Step 3: Divide (196) by (28) first to get (7), then multiply (140\times7).

Step 1: HCF takes the smaller power of each common prime. Step 2: The smaller powers of (2), (3), and (7) are (3), (1), and (1), so HCF (=2^3\times3\times7). Step 3: Match each smaller power with the correct base.

Step 1: HCF contains only the common prime factors. Step 2: The common primes are (2) and (3), with smaller powers (2^5) and (3^2). Step 3: For HCF, always take the smaller powers.

Step 1: Prime factorise: (63=3^2\times7), (98=2\times7^2), and (154=2\times7\times11). Step 2: The highest powers are (2), (3^2), (7^2), and (11), so the LCM is (9702). Step 3: Include a prime even if it appears in only one number.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: LCM (=\frac{66528}{72}=924). Step 3: For large numbers, simplify the division in small steps.

Step 1: Product of two numbers equals HCF times LCM. Step 2: The other number is (\frac{48\times1440}{240}=288). Step 3: First simplify (1440) by (240) to make the calculation easier.

Step 1: The smallest number exactly divisible by all given numbers is their LCM. Step 2: (54=2\times3^3), (72=2^3\times3^2), and (90=2\times3^2\times5), so LCM (=2^3\times3^3\times5=1080). Step 3: Choose the highest powers carefully.

Step 1: The greatest common divisor is the HCF. Step 2: (225=3^2\times5^2), (375=3\times5^3), and (525=3\times5^2\times7), so HCF (=3\times5^2=75). Step 3: Use only the smallest powers common to all numbers.

Step 1: HCF is (2^2\times3^2\times13), and LCM is (2^4\times3^5\times5\times13). Step 2: On division, subtract the powers of the same bases, giving (2^2\times3^3\times5). Step 3: In prime-power division, subtract exponents.

Step 1: The next common signal time is the LCM of the intervals. Step 2: (16=2^4), (24=2^3\times3), and (40=2^3\times5), so LCM (=2^4\times3\times5=240). Step 3: For repeated-time questions, use LCM.

Step 1: HCF uses the smaller power of each common prime. Step 2: The smaller power of (2) must be (5), so (a=5) is possible; the smaller power of (3) must be (3), so (b=3) is possible. Step 3: Check unknown powers separately.

Step 1: LCM contains the highest power of each prime. Step 2: The highest power of (3) must be (4), so (a=4); the highest power of (7) must be (3), so (b=3). Step 3: For LCM, focus on the maximum-power condition.

Step 1: Subtracting (9) makes the number divisible by (28), (42), and (56). Step 2: (28=2^2\times7), (42=2\times3\times7), and (56=2^3\times7), so LCM (=2^3\times3\times7=168). Hence the number is (168+9=177). Step 3: Add the common remainder at the end.

Step 1: The maximum number of identical boxes is found using HCF. Step 2: (168=2^3\times3\times7) and (252=2^2\times3^2\times7), so HCF (=2^2\times3\times7=84). Step 3: For maximum equal distribution, use HCF.

Step 1: Coprime numbers have HCF (1). Step 2: Product (=) HCF (\times) LCM, so product (=1\times667=667). Step 3: For coprime numbers, the LCM equals the product.

Step 1: HCF (=2^3\times11=88). Step 2: LCM (=2^4\times3\times11=528), so the sum is (88+528=616). Step 3: When sum is asked, find both values separately.

Step 1: The smallest such number is the LCM of the two given numbers. Step 2: The powers of (3) are (2) and (5), so the higher power (5) will be used. Step 3: For divisibility, choose the higher power.