Fundamental Theorem of Arithmetic

Step 1: Uniqueness means the set of prime factors remains fixed. Step 2: Changing the order does not change the product, so (2\times3\times5) and (5\times3\times2) are the same prime factorisation. Step 3: In exams, do not treat order change as a different answer.

Step 1: Write (55440=5040\times11). Step 2: Since (5040=2^4\times3^2\times5\times7), the complete factorisation is (2^4\times3^2\times5\times7\times11). Step 3: Do not leave a composite factor like 5040 in the final answer.

Step 1: For HCF, take the smaller powers of only common prime factors. Step 2: The common factors are 2 and 3, with smaller powers (2^3) and (3^4). Step 3: (2^3\times3^4=8\times81=648), so the answer is 648.

Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^5), (3^6), (5^2), and (7^2). Step 3: (32\times729\times25\times49=28576800), so the answer is 28576800.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: So LCM (=181440\div144=1260). Step 3: Apply this relation directly only for two numbers.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (84\times1980=166320). Step 3: In such questions, multiply the two given values directly.

Step 1: (3528=72\times49=2^3\times3^2\times7^2). Step 2: For a perfect square, all exponents must be even, but the exponent of 2 is 3. Step 3: Multiplying by 2 makes the exponent 4, so the smallest number is 2.

Step 1: (13230=2\times3^3\times5\times7^2). Step 2: For a perfect square, all exponents must be even, so the odd powers of 2, 3, and 5 must be reduced. Step 3: Dividing by (2\times3\times5=30) leaves (3^2\times7^2), so the smallest number is 30.

Step 1: (5400=54\times100=2^3\times3^3\times5^2). Step 2: For a perfect cube, every exponent must be a multiple of 3. Step 3: The exponent of 5 is 2, so one more 5 is needed; the smallest number is 5.

Step 1: Recognise (74088=42^3). Step 2: Since (42=2\times3\times7), (42^3=2^3\times3^3\times7^3). Step 3: 42 is composite, so (42^3) is not the final prime factorisation.

Step 1: For a perfect square, all exponents must be even. Step 2: The exponents of 2, 5, and 11 are odd. Step 3: Multiplying by (2\times5\times11=110) makes all exponents even.

Step 1: For a perfect cube, each exponent must be a multiple of 3. Step 2: We need (2), (3^2), and (5) to make the powers 6, 9, and 3. Step 3: So the smallest multiplier is (2\times3^2\times5).

Step 1: In multiplication, exponents of the same prime base are added. Step 2: The power of 3 in (a) is 3 and in (b) is 5. Step 3: In (ab), the power of 3 is (3+5=8).

Step 1: In (xy), powers with the same base 5 are added. Step 2: The power of 5 in (x) is 3 and in (y) is 2. Step 3: The total power is (3+2=5).

Step 1: HCF uses only common prime factors. Step 2: The common factors are 2 and 3; the smaller powers are (2^6) and (3^5). Step 3: Therefore, the HCF is (2^6\times3^5).

Step 1: LCM uses the highest powers of all prime factors. Step 2: The highest powers are (2^9), (3^7), (5^2), and (11). Step 3: So the correct form is (2^9\times3^7\times5^2\times11).

Step 1: Product of the two numbers is (36\times1260=45360). Step 2: One number is 180, so the other number is (45360\div180=252). Step 3: To check, the HCF of 180 and 252 is 36.

Step 1: Product of the two numbers is (42\times1386=58212). Step 2: The other number is (58212\div198=294). Step 3: As a check, the HCF of 198 and 294 is 42.

Step 1: Prime factorise 5400. Step 2: (5400=54\times100=2^3\times3^3\times5^2). Step 3: Comparing with the given form gives (a=3).

Step 1: Check the fixed part of the given form. Step 2: (2^4\times7^2=16\times49=784), and (15876\div784) is not a whole number. Step 3: So the given form does not match the number; none of the listed (b) values can be correct.

Step 1: To count with repetition, add the exponents. Step 2: (2^6) gives 6, (3^4) gives 4, (5^3) gives 3, and 7 gives 1 factor. Step 3: Total (6+4+3+1=14), so the answer is 14.

Step 1: When counting distinct prime factors, do not add exponents. Step 2: The prime factors here are 2, 3, and 11. Step 3: Therefore, the number of distinct prime factors is 3.

Step 1: The first number has prime factors 2 and 3. Step 2: The second number has prime factors 5 and 7. Step 3: There is no common prime factor, so they are co-prime.

Step 1: Co-prime numbers have HCF 1. Step 2: For two numbers, product (=) HCF (\times) LCM. Step 3: Therefore, the LCM will be (1547).

Step 1: Divide 8192 repeatedly by 2. Step 2: (8192=2^{13}). Step 3: 4 and 16 are composite, so they should not be written in the final prime form.