Fundamental Theorem of Arithmetic

Step 1: In HCF, take the smaller power of the common prime. Step 2: The powers of 5 are 1 and 2. Step 3: The smaller power is 1, so the answer is 1.

Step 1: (3969=63^2). Step 2: Since (63=3^2\times7), (63^2=3^4\times7^2). Step 3: (63\times63) gives the product, but it is not prime factorisation.

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

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: Dividing by 2 gives (2^2\times3^2\times7^2), so the smallest number is 2.

Step 1: In multiplication, add the exponents of the same prime base. Step 2: The power of 3 in (x) is 2 and in (y) is 3. Step 3: In (xy), the power of 3 is (2+3=5).

Step 1: Powers of the same base 2 are added in multiplication. Step 2: The power of 2 in (x) is 5 and in (y) is 3. Step 3: The total power is (5+3=8).

Step 1: Write (2310=231\times10). Step 2: (231=3\times7\times11) and (10=2\times5). Step 3: The distinct prime factors are 2, 3, 5, 7, and 11.

Step 1: The common prime factors are 2, 3, and 11. Step 2: The smaller powers are (2^2), (3^1), and (11^1). Step 3: (4\times3\times11=132), so the HCF is 132.

Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^3), (3^2), (5), and (11^2). Step 3: (8\times9\times5\times121=43560), so the answer is 43560.

Step 1: The product is (36\times900). Step 2: (36=2^2\times3^2) and (900=2^2\times3^2\times5^2). Step 3: The power of 2 in the product is (2+2=4).

Step 1: The product is (30\times2310). Step 2: 30 has no factor 11 and (2310=2\times3\times5\times7\times11). Step 3: Therefore, the power of 11 in the product is (0+1=1).

Step 1: (2401=49\times49). Step 2: Since (49=7^2), (2401=7^4). Step 3: 49 is composite, so (49^2) is not the final prime factorisation.

Step 1: In a perfect cube, each exponent must be a multiple of 3. Step 2: (2^2) and (3^5) cause the issue; reduce them to 0 and 3. Step 3: Dividing by (2^2\times3^2) gives exponents (0,3,3).

Step 1: First calculate (2^3=8) and (3^2=9). Step 2: (8\times9\times5\times7=2520). Step 3: To get the number from prime factorisation, multiply all factors.

Step 1: For two numbers, HCF (\times) LCM equals the product of the two numbers. Step 2: This also follows from prime powers because the smaller and higher exponents together give the total exponent. Step 3: Therefore, the answer is (ab).

Step 1: HCF is (2^3\times3=24). Step 2: LCM is (2^5\times3^2=288). Step 3: The ratio is (288\div24=12).

Step 1: In (xy), exponents of the same bases are added. Step 2: Powers become (2^{6}), (3^{4}), and (5^{3}). Step 3: Counting with repetition gives (6+4+3=13).

Step 1: In LCM, take the higher power. Step 2: The powers of 7 are 1 and 3. Step 3: The higher power is 3, so the answer is 3.

Step 1: In HCF, take the smaller power. Step 2: The powers of 3 are 2 and 4. Step 3: The smaller power is 2, so the answer is 2.

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.