HCF and LCM using prime factorisation

Step 1: (2160=2^4\times 3^3\times 5) and (3780=2^2\times 3^3\times 5\times 7). Step 2: For HCF, take the smaller exponents of common prime factors, so (2^2\times 3^3\times 5=540). Step 3: In such questions, carefully choose the smaller powers.

Step 1: For LCM, take the greater exponent of every prime factor present. Step 2: This gives (2^3\times 3^2\times 5\times 7=2520). Step 3: While finding LCM, do not leave out any prime factor.

Step 1: Since the product is directly given, the other number is (15120\div 216). Step 2: (15120\div 216=70). On checking, the HCF of (216) and (70) is (2), so the given HCF (36) is inconsistent. Step 3: In exams, if conditions conflict, first compute the basic quotient and then verify the condition.

Step 1: Divide the total product by the given number to get the other number. Step 2: (15120\div 216=70), so the other number is (70). Step 3: In such questions, find the quotient accurately before doing any extra work.

Step 1: For two numbers, product of the numbers equals HCF times LCM. Step 2: The other number (=\frac{18\times 540}{90}=108). Step 3: Apply this relation directly only for two numbers.

Step 1: For two numbers, HCF times LCM equals the product of the two numbers. Step 2: Adding exponents gives (2^{3+2}\times 3^{2+4}\times 5\times 7=2^5\times 3^6\times 5\times 7). Step 3: When the product is asked, you need not find HCF and LCM separately.

Step 1: HCF contains only the prime factors common to all three numbers. Step 2: The smallest exponent of (2) is (3) and of (3) is (1). (5) and (7) are not common to all. Hence the answer is (2^3\times 3). Step 3: For three numbers, first identify prime factors common to all.

Step 1: LCM takes the greatest exponent of every prime factor present. Step 2: The greatest exponent of (2) is (4), of (3) is (5), and of (5) is (2). So the answer is (2^4\times 3^5\times 5^2). Step 3: Any prime appearing in at least one number must appear in the LCM.

Step 1: For two numbers, their product equals HCF times LCM. Step 2: The exponent of (2) in the product is (2+5=7). Step 3: If only one prime exponent is asked, add only that prime's exponents.

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.