HCF and LCM using prime factorisation

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.

Step 1: Compare the powers of (3). Step 2: (162=2\times3^4), (270=2\times3^3\times5), and (378=2\times3^3\times7), so the smallest power is (3). Step 3: HCF uses the smallest power.

Step 1: Prime factorise: (112=2^4\times7), (180=2^2\times3^2\times5), and (225=3^2\times5^2). Step 2: The distinct primes in the LCM are (2), (3), (5), and (7), so there are (4). Step 3: Count distinct prime bases, not powers.

Step 1: LCM takes the higher power, and HCF takes the lower power. Step 2: The powers of (5) are (1) and (4), so (L) has (4) and (H) has (1). Step 3: Do not interchange the two rules.

Step 1: (169=13^2), (221=13\times17), and (299=13\times23). Step 2: The common prime in all three is (13), so the HCF should be (13). Step 3: Carefully identify the factor common to all three numbers.

Step 1: (169=13^2), (221=13\times17), and (299=13\times23). Step 2: (13) is common to all three and no larger common factor exists, so HCF is (13). Step 3: Do not miss a prime that appears in all three numbers.

Step 1: (64=2^6) and (5120=2^{10}\times5). Step 2: Product equals HCF times LCM, so the power of (2) is (6+10=16). Step 3: Add exponents when multiplying powers with the same base.

Step 1: LCM (=\frac{224\times336}{112}=672). Step 2: The difference is (672-112=560). Step 3: When difference is asked, find the LCM correctly first.

Step 1: The smallest number divisible by both is their LCM. Step 2: The powers of (3) are (3) and (5), so the higher power (5) is used. Step 3: For divisibility, choose the required highest power.