HCF and LCM using prime factorisation

Step 1: Let the numbers be (42m) and (42n), where (m) and (n) are coprime. Step 2: (42mn=2772), so (mn=66=2\times3\times11). Splitting three distinct prime factors into two groups gives (4) unordered pairs. Step 3: While counting pairs, make sure (m) and (n) remain coprime.

Step 1: In each case, the difference between divisor and remainder is (5), so adding (5) to the number makes it divisible by (48), (72), and (120). Step 2: Their LCM is (720), so the smallest number is (720-5=715). Step 3: In such questions, spotting the common difference makes the calculation much easier.

Step 1: The smallest divisible number (N) is the LCM, and (M) is the HCF. Step 2: (N=2^5\times3^4\times5\times7) and (M=2^3\times3^2), so (\frac{N}{M}=2^2\times3^2\times5\times7). Step 3: While dividing, subtract powers of the same base.

Step 1: HCF contains only common prime factors. Step 2: The common primes are (2), (3), and (5), with smaller powers (2^3), (3^3), and (5). Step 3: Choose the smaller power for each prime base separately.

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

Step 1: Product of two numbers equals HCF times LCM. Step 2: The other number is (\frac{54\times2970}{270}=594). Step 3: Simplify the division first to reduce calculation work.

Step 1: HCF is (2^3\times3^2\times5), and LCM is (2^6\times3^5\times5^2\times13). Step 2: On division, subtract powers to get (2^3\times3^3\times5\times13). Step 3: In ratios, subtract exponents of the same base.

Step 1: Let the numbers be (48m) and (48n), where (m) and (n) are coprime. Step 2: (48mn=2112), so (mn=44=2^2\times11); the unordered coprime pairs are ((1,44)) and ((4,11)), so the count is (2). Step 3: Do not split the same prime factor into both parts.

Step 1: In each case, divisor minus remainder is (7), so adding (7) to the number makes it divisible by all three divisors. Step 2: The LCM of (36), (54), and (90) is (540), so the number is (540-7=533). Step 3: In such questions, identify the common difference and subtract it from the LCM.

Step 1: For the same remainder, take the HCF of the differences. Step 2: The differences are (276), (432), and (708); their HCF is (12). Step 3: Use the differences, not the original numbers directly.

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 (2), so (b=2) is possible. Step 3: Check the condition for each base separately.

Step 1: LCM takes the highest power of every prime. Step 2: The highest power of (3) must be (5), so (a=5); the highest power of (5) must be (4), so (b=4). Step 3: For LCM, identify the maximum power.

Step 1: The maximum number of identical groups is found by HCF. Step 2: (276=2^2\times3\times23) and (414=2\times3^2\times23), so HCF (=2\times3\times23=138). Step 3: Use HCF for maximum equal distribution.

Step 1: The next common ringing time is the LCM of the intervals. Step 2: (18=2\times3^2), (27=3^3), (45=3^2\times5), and (60=2^2\times3\times5), so LCM (=2^2\times3^3\times5=540). Step 3: For repeated-time questions, use LCM.

Step 1: Product of two numbers equals HCF times LCM. Step 2: The powers of (2) are (4) and (7), so the total power is (11). Step 3: When multiplying powers with the same base, add the exponents.

Step 1: LCM (=\frac{264\times396}{132}=792). Step 2: The ratio (792:132=6:1). Step 3: Always reduce the ratio to its simplest form.

Step 1: The maximum number of equal parts is found by HCF. Step 2: (128=2^7), (192=2^6\times3), and (320=2^6\times5), so HCF (=2^6=64). Step 3: For maximum equal division, take the smallest common power.

Step 1: The smallest number exactly divisible by all is the LCM. Step 2: (81=3^4), (96=2^5\times3), and (125=5^3), so LCM (=2^5\times3^4\times5^3=324000). Step 3: Multiply the highest powers carefully.

Step 1: (q=p\times17), so (p) exactly divides (q). Step 2: When one number exactly divides the other, the smaller number is the HCF. Step 3: Identifying a multiple relation saves time in such questions.

Step 1: After taking out HCF (72), (r) and (s) are coprime. Step 2: LCM (=72rs=4680), so (rs=65). Step 3: First divide the LCM by the HCF.