Fundamental Theorem of Arithmetic

Step 1: The product is (78\times510510). Step 2: (78=2\times3\times13), and 510510 also has 13 to power 1. Step 3: Therefore, the power of 13 in the product is (1+1=2).

Step 1: (83521=289\times289). Step 2: Since (289=17^2), (83521=17^4). Step 3: 289 is composite, so (289^2) is not the final prime factorisation.

Step 1: In a perfect cube, every exponent must be a multiple of 3. Step 2: Reduce (2^{10}) to (2^9) by dividing by 2, and (3^8) to (3^6) by dividing by (3^2). Step 3: The smallest divisor is (2\times3^2).

Step 1: First calculate (2^6=64) and (3^4=81). Step 2: (64\times81\times5\times7=181440). 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 (2^6\times3^2). Step 2: LCM is (2^9\times3^5). Step 3: The ratio is (2^{9-6}\times3^{5-2}=2^3\times3^3=216).

Step 1: In (xy), exponents of the same bases are added. Step 2: Powers become (2^{12}), (3^9), and (5^9). Step 3: Counting with repetition gives (12+9+9=30).

Step 1: In LCM, take the higher power. Step 2: The powers of 5 are 3 and 6. Step 3: The higher power is 6, so the answer is 6.

Step 1: In HCF, take the smaller power. Step 2: The powers of 3 are 4 and 7. Step 3: The smaller power is 4, so the answer is 4.

Step 1: The theorem gives certainty of prime factorisation for numbers greater than 1. Step 2: Prime factors and their powers remain fixed; only their order may change. Step 3: In the final answer, do not leave composite factors.

Step 1: Write (221760=20160\times11). Step 2: Since (20160=2^6\times3^2\times5\times7), the complete prime form is (2^6\times3^2\times5\times7\times11). Step 3: 20160 is composite, so do not keep it in the final answer.

Step 1: For HCF, take the smaller powers of common prime factors only. Step 2: The common factors are 2 and 3, with smaller powers (2^5) and (3^5). Step 3: (32\times243=7776), so the answer is 7776.

Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^8), (3^7), (5^2), and (7^3). Step 3: Their product is 38482790400, so that is the answer.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: LCM (=1814400\div720=2520). Step 3: Use this relation directly only for two numbers.

Step 1: The product of two numbers equals the product of their HCF and LCM. Step 2: (210\times9240=1940400). Step 3: Apply this formula directly only when exactly two numbers are involved.

Step 1: (21600=216\times100=2^5\times3^3\times5^2). Step 2: For a perfect square, all exponents must be even, but powers of 2 and 3 are odd. Step 3: Multiplying by (2\times3=6) makes all powers even.

Step 1: (52920=2^3\times3^3\times5\times7^2). Step 2: For a perfect square, odd powers of 2, 3, and 5 must be reduced. Step 3: Dividing by (2\times3\times5=30) leaves (2^2\times3^2\times7^2).

Step 1: (30240=2^5\times3^3\times5\times7). Step 2: For a perfect cube, exponents must be multiples of 3. Step 3: We need (2), (5^2), and (7^2), so the smallest multiplier is (2\times25\times49=2450).

Step 1: Write (500094=2\times250047). Step 2: (250047=3^3\times7^3\times13), so the full form is (2\times3^3\times7^3\times13). Step 3: Composite bases should not remain in final prime form.

Step 1: For a perfect square, every exponent must be even. Step 2: Powers of 2 and 5 are odd, while the others are even. Step 3: Multiplying by (2\times5=10) makes all powers even.