Fundamental Theorem of Arithmetic

Step 1: The product is (42\times30030). Step 2: 42 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: (14641=121\times121). Step 2: Since (121=11^2), (14641=11^4). Step 3: 121 is composite, so (121^2) is not the final prime factorisation.

Step 1: In a perfect cube, every exponent must be a multiple of 3. Step 2: Reduce (2^5) to (2^3) by dividing by (2^2), and reduce (3^4) to (3^3) by dividing by 3. Step 3: So the smallest divisor is (2^2\times3).

Step 1: First calculate (2^4=16) and (3^3=27). Step 2: (16\times27\times5\times7=15120). 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 (2^4\times3=48). Step 2: LCM is (2^7\times3^3=3456). Step 3: The ratio is (3456\div48=72).

Step 1: In (xy), exponents of the same bases are added. Step 2: Powers become (2^8), (3^6), and (5^5). Step 3: Counting with repetition gives (8+6+5=19).

Step 1: In LCM, take the higher power. Step 2: The powers of 7 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: This theorem says that the prime factorisation of a number greater than 1 is fixed. Step 2: The order may change, but the prime factors do not change. Step 3: In exams, do not treat changed order as a new factorisation.

Step 1: Write (83160=7560\times11). Step 2: Since (7560=2^3\times3^3\times5\times7), the full factorisation is (2^3\times3^3\times5\times7\times11). Step 3: Do not leave a composite factor like 7560 in the final answer.

Step 1: For HCF, take the smaller powers of common prime factors. Step 2: The common factors are 2 and 3, with smaller powers (2^4) and (3^5). Step 3: (2^4\times3^5=16\times243=3888), so the answer is 3888.

Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^6), (3^7), (5^3), and (7^2). Step 3: (64\times2187\times125\times49=857304000), so the answer is 857304000.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: LCM (=362880\div144=2520). Step 3: Use this relation directly only for two numbers.

Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (63\times4620=291060). Step 3: In such questions, directly multiply the two given values.

Step 1: (10800=108\times100=2^4\times3^3\times5^2). Step 2: For a perfect square, all exponents must be even, but the exponent of 3 is 3. Step 3: Multiplying by 3 makes the exponent 4, so the smallest number is 3.

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), a perfect square.

Step 1: (7560=2^3\times3^3\times5\times7). Step 2: For a perfect cube, each exponent must be a multiple of 3. Step 3: The powers of 5 and 7 are 1, so multiply by (5^2\times7^2=1225).

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 form.

Step 1: For a perfect square, all exponents must be even. Step 2: The exponents of 2 and 3 are odd, while the others are even. Step 3: Multiplying by (2\times3=6) makes all exponents even.