HCF and LCM using prime factorisation

Step 1: A correct pair must give both HCF (24) and LCM (720). Step 2: (72=2^3\times 3^2) and (240=2^4\times 3\times 5). Their HCF is (2^3\times 3=24) and LCM is (2^4\times 3^2\times 5=720). Step 3: For option checking, prime factorise first.

Step 1: Subtract the remainder (5) from each number to get (132), (180), and (252). Step 2: Find their HCF. (132=2^2\times 3\times 11), (180=2^2\times 3^2\times 5), (252=2^2\times 3^2\times 7), so the common part is (2^2\times 3=12). Step 3: In same-remainder problems, subtract the remainder first.

Step 1: The time when they ring together again is the LCM of (18), (24), and (30). Step 2: (18=2\times 3^2), (24=2^3\times 3), (30=2\times 3\times 5). The LCM is (2^3\times 3^2\times 5=360). Step 3: Use LCM for repeated-time meeting problems.

Step 1: Since all sweets must be divided equally, find the HCF. Step 2: (96=2^5\times 3), (144=2^4\times 3^2), and (240=2^4\times 3\times 5). The common smallest part is (2^4\times 3=48). Step 3: For greatest equal grouping, use HCF.

Step 1: The HCF is (2^3\times 3^2\times 5) and the LCM is (2^5\times 3^4\times 5^2). Step 2: On division, subtract exponents, so (\frac{\text{LCM}}{\text{HCF}}=2^2\times 3^2\times 5). Step 3: In such ratios, subtract smaller exponents from larger exponents.

Step 1: For two numbers, (xy=\text{HCF}\times \text{LCM}). Step 2: Therefore, dividing (xy) by HCF gives the LCM. Step 3: This relation is very useful in two-number problems.

Step 1: If the HCF of two numbers is (1), the numbers are co-prime. Step 2: For two numbers, product (=) HCF (\times) LCM, so (7429=1\times) LCM. Hence the LCM is (7429). Step 3: The LCM of co-prime numbers equals their product.

Step 1: The other number (=\frac{\text{HCF}\times \text{LCM}}{\text{first number}}). Step 2: Using exponents, (\frac{(2^2\times 3^2)(2^5\times 3^2\times 5\times 7)}{2^5\times 3^2\times 5}=2^2\times 3^2\times 7). Step 3: In such problems, simplify by subtracting exponents.

Step 1: First find the LCM of (36), (48), and (60). Step 2: (36=2^2\times 3^2), (48=2^4\times 3), (60=2^2\times 3\times 5), so the LCM is (2^4\times 3^2\times 5=720). The smallest multiple greater than (1000) is (1440). Step 3: Find the LCM first, then choose its multiple according to the limit.

Step 1: HCF takes the smaller exponents of common prime factors. Step 2: The HCF is (2^4\times 3^2). Counting repetition, the total number of prime factors is (4+2=6). Step 3: First form the HCF, then add its exponents.

Step 1: LCM includes every prime factor appearing in either number. Step 2: The primes are (2), (3), (5), and (11). So there are (4) distinct prime factors. Step 3: While counting distinct factors, do not count powers separately.

Step 1: In the HCF, the exponent of (2) must be (\min(a,4)=3), so (a=3) fits. Step 2: The exponent of (3) must be (\min(2,b)=2), so (b\geq 2); among the options, ((3,2)) fits. Step 3: For unknown exponents, apply the smaller-exponent rule.

Step 1: LCM takes the greater exponents. Step 2: For (2), (\max(a,2)=5), so (a=5). For (5), (\max(2,b)=3), so (b=3). Step 3: To match an LCM, compare the required largest exponents.

Step 1: The greatest number that divides all exactly is the HCF. Step 2: (72=2^3\times 3^2), (108=2^2\times 3^3), and (180=2^2\times 3^2\times 5). The common smallest part is (2^2\times 3^2=36). Step 3: When the greatest exact divisor is asked, find the HCF.

Step 1: The smallest such number is the LCM of the three numbers. Step 2: (45=3^2\times 5), (54=2\times 3^3), and (72=2^3\times 3^2). The LCM is (2^3\times 3^3\times 5=1080). Step 3: Remainder (0) means exact divisibility by all numbers.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: Therefore, the product is (15\times 630=9450). The sum or the exact numbers are not fixed without more information. Step 3: In relation-based questions, choose only what is definitely proved.

Step 1: For HCF, take only the common prime factors. Step 2: The common factors are (2) and (3), and the smaller powers are (2^2) and (3^2), so the HCF is (36). Step 3: In exams, remember that HCF uses the minimum powers of common primes.

Step 1: For two numbers, product of numbers (=) HCF (\times) LCM. Step 2: The other number is (\frac{18\times540}{90}=108). Step 3: In such questions, write the formula first and then calculate.

Step 1: For LCM, take the highest power of every prime factor present. Step 2: The highest powers are (2^4), (3^3), (5^2), and (7). Step 3: Do not miss a prime factor that appears in only one number.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: So LCM (=\frac{12960}{36}=360). Step 3: While dividing, check place value carefully to avoid calculation errors.