Fundamental Theorem of Arithmetic

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.

Step 1: For a perfect cube, every exponent must be a multiple of 3. Step 2: We must make (2^7), (3^5), and (5^4) into powers 9, 6, and 6. Step 3: So the smallest multiplier is (2^2\times3\times5^2).

Step 1: In multiplication, exponents of the same prime base are added. Step 2: The power of 3 in (a) is 4 and in (b) is 6. Step 3: In (ab), the power of 3 will be (4+6=10).

Step 1: In (xy), powers with the same base 5 are added. Step 2: The power of 5 in (x) is 5 and in (y) is 2. Step 3: The total power is (5+2=7).

Step 1: HCF uses only common prime factors. Step 2: The common factors are 2 and 3, with smaller powers (2^7) and (3^6). Step 3: Therefore, the HCF is (2^7\times3^6).

Step 1: LCM uses the highest powers of all prime factors. Step 2: The highest powers are (2^{10}), (3^8), (5^2), and (7^3). Step 3: So the correct form is (2^{10}\times3^8\times5^2\times7^3).

Step 1: Product of the two numbers is (48\times3360=161280). Step 2: One number is 240, so the other is (161280\div240=672). Step 3: As a check, the HCF of 240 and 672 is 48.

Step 1: Product of the two numbers is (72\times2520=181440). Step 2: The other number is (181440\div360=504). Step 3: To check, the HCF of 360 and 504 is 72.

Step 1: Prime factorise 7560. Step 2: (7560=756\times10=2^3\times3^3\times5\times7). Step 3: Comparing with the given form gives (a=3).

Step 1: Write (15120=16\times945). Step 2: (945=3^3\times5\times7), so (15120=2^4\times3^3\times5\times7). Step 3: Comparing gives (b=3).

Step 1: To count with repetition, add the exponents. Step 2: (2^7) gives 7, (3^5) gives 5, (5^4) gives 4, and 11 gives 1 factor. Step 3: Total (7+5+4+1=17), so the answer is 17.

Step 1: When counting distinct prime factors, do not add exponents. Step 2: The prime factors here are 2, 3, 5, and 11. Step 3: Therefore, the number of distinct prime factors is 4.

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

Step 1: Co-prime numbers have HCF 1. Step 2: For two numbers, product (=) HCF (\times) LCM. Step 3: Therefore, the LCM will be 2431.

Step 1: Divide 16384 repeatedly by 2. Step 2: (16384=2^{14}). Step 3: 4 and 16 are composite, so they should not be written in the final prime form.