Fundamental Theorem of Arithmetic

Step 1: Write (945=9\times105). Step 2: (9=3^2) and (105=3\times5\times7), so (945=3^3\times5\times7). Step 3: Do not keep composite factors like 9 or 105 in the final answer.

Step 1: Calculate (2^3=8) and (3^2=9). Step 2: (8\times9\times5=360). Step 3: When converting prime factorisation into a number, multiply all factors.

Step 1: Look at distinct prime factors, not powers. Step 2: In (m), the distinct prime factors are 2, 3, and 5. Step 3: In (2^4), count 2 only once in the distinct list.

Step 1: (16=2^4) and (45=3^2\times5). Step 2: They have no common prime factor, so they are co-prime. Step 3: Co-prime numbers have HCF 1.

Step 1: Co-prime numbers have HCF 1. Step 2: For two numbers, product (=) HCF (\times) LCM. Step 3: Therefore, the LCM is (221).

Step 1: In prime factorisation, every factor must be prime. Step 2: (2), (3), and (11) are prime, so (2^2\times3\times11) is correct. Step 3: 4, 6, and 12 are composite, so they should not remain in the final form.

Step 1: The common prime factors are 2 and 3. Step 2: The smaller powers are (2^3) and (3^1), so (8\times3=24). Step 3: For HCF, take smaller powers of only common factors.

Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^3), (3^2), and (5^1). Step 3: (8\times9\times5=360), so the answer is 360.

Step 1: Write (1260=126\times10). Step 2: (126=2\times3^2\times7) and (10=2\times5). Step 3: Therefore, (1260=2^2\times3^2\times5\times7).

Step 1: Calculate (2^5=32) and (3^2=9). Step 2: (32\times9=288). Step 3: In prime factorisation with powers, evaluate powers first.

Step 1: When repetition is counted, add the exponents. Step 2: (2^4) gives 4 factors, (3^2) gives 2 factors, and (5) gives 1 factor. Step 3: Total (4+2+1=7), so the answer is 7.

Step 1: Prime factorise 108. Step 2: (108=4\times27=2^2\times3^3), so the given form (2^a\times3^2) does not match 108. Step 3: The power of 3 should also be 3; therefore no value of (a) alone can make it correct.

Step 1: Write 216 as (8\times27). Step 2: (8=2^3) and (27=3^3), so (216=2^3\times3^3). Step 3: Therefore, (a=3).

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

Step 1: (18=2\times3^2) and (35=5\times7), so they are co-prime. Step 2: The LCM of co-prime numbers equals their product. Step 3: (18\times35=630), so the answer is 630.

Step 1: Divide 1024 repeatedly by 2. Step 2: (1024=2^{10}). Step 3: (4^5) and (32^2) can give the value, but 4 and 32 are not prime.

Step 1: Calculate (3^4=81) and (5^2=25). Step 2: (81\times25=2025). Step 3: In questions with powers, simplify the powers first.

Step 1: Write (675=27\times25). Step 2: (27=3^3) and (25=5^2), so (675=3^3\times5^2). Step 3: 27 and 25 are composite, so write their prime powers in the final form.

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

Step 1: For HCF, take the smaller powers of common prime factors. Step 2: The smaller powers are (2^3) and (3^2). Step 3: (2^3\times3^2=8\times9=72), so the answer is 72.