Fundamental Theorem of Arithmetic

Step 1: In HCF, take the smaller power of the common prime. Step 2: The powers of 7 are 1 and 2. Step 3: The smaller power is 1, so the answer is 1.

Step 1: Write (20412=2916\times7). Step 2: (2916=4\times729=2^2\times3^6). Step 3: Therefore, (20412=2^2\times3^6\times7).

Step 1: For a perfect cube, exponents must be multiples of 3. Step 2: We must make (2^8), (3^4), and (5^2) into powers 9, 6, and 3. Step 3: The smallest multiplier is (2\times3^2\times5).

Step 1: (21168=2^4\times3^3\times7^2). Step 2: For a perfect square, all exponents must be even, but the exponent of 3 is 3. Step 3: Dividing by 3 gives (2^4\times3^2\times7^2), a perfect square.

Step 1: In multiplication, exponents of the same prime base are added. Step 2: The power of 7 in (x) is 3 and in (y) is 4. Step 3: In (xy), the power of 7 is (3+4=7).

Step 1: Powers of the same base 2 are added in multiplication. Step 2: The power of 2 in (x) is 7 and in (y) is 6. Step 3: The total power is (7+6=13).

Step 1: Write (45045=3465\times13). Step 2: (3465=3^2\times5\times7\times11). Step 3: The distinct prime factors are 3, 5, 7, 11, and 13.

Step 1: The common prime factors are 2, 3, and 11. Step 2: The smaller powers are (2^3), (3^2), and (11^1). Step 3: (8\times9\times11=792), so the HCF is 792.

Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^5), (3^4), (5), and (11^2). Step 3: (32\times81\times5\times121=1568160), so the answer is 1568160.

Step 1: The product is (120\times8400). Step 2: (120=2^3\times3\times5) and (8400=2^4\times3\times5^2\times7). Step 3: The power of 2 in the product is (3+4=7).

Step 1: The product is (66\times30030). Step 2: 66 has no factor 13 and (30030=2\times3\times5\times7\times11\times13). Step 3: Therefore, the power of 13 in the product is (0+1=1).

Step 1: (28561=169\times169). Step 2: Since (169=13^2), (28561=13^4). Step 3: 169 is composite, so (169^2) is not the final prime factorisation.

Step 1: In a perfect cube, every exponent must be a multiple of 3. Step 2: Reduce (2^8) to (2^6) by dividing by (2^2), and reduce (3^5) to (3^3) by dividing by (3^2). Step 3: So the smallest divisor is (2^2\times3^2).

Step 1: First calculate (2^5=32) and (3^3=27). Step 2: (32\times27\times5\times7=30240). 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: In prime powers, the smaller and higher exponents together give the total exponent. Step 3: Therefore, the answer is (ab).

Step 1: HCF is (2^5\times3^2=288). Step 2: LCM is (2^8\times3^4=20736). Step 3: The ratio is (20736\div288=72).

Step 1: In (xy), exponents of the same bases are added. Step 2: Powers become (2^{10}), (3^{7}), and (5^{7}). Step 3: Counting with repetition gives (10+7+7=24).

Step 1: In LCM, take the higher power. Step 2: The powers of 5 are 2 and 4. Step 3: The higher power is 4, so the answer is 4.

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

Step 1: The theorem says that prime factorisation of a number greater than 1 is fixed. Step 2: The order may change, but the prime bases and their powers do not change. Step 3: In exams, do not treat a change of order as a new factorisation.