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In this Class 10 Mathematics topic from the chapter Real Numbers, students learn that every integer greater than 1 can be expressed as a product of prime numbers, and that this prime factorisation is unique apart from the order of the factors. They practise finding prime factors and use the theorem to understand and determine the HCF and LCM of numbers. The topic builds clear reasoning about the structure of whole numbers and supports later work with divisibility and number relationships.
TOPIC PRACTICE
Quiz this set
Up to 20 questions from this page. Select your focus, then start.
If (N=2^{10}\times3^8\times5^5\times7), by which smallest number must (N) be multiplied to make it a perfect cube?
Correct answer: A
Step 1: For a perfect cube, every exponent must be a multiple of 3. Step 2: We must make 10 to 12, 8 to 9, 5 to 6, and 1 to 3. Step 3: So the smallest multiplier is (2^2\times3\times5\times7^2).
If (a=2^7\times3^6\times5^2) and (b=2^5\times3^8\times7^3), what will be the power of 3 in (ab)?
Correct answer: C
Step 1: In multiplication, exponents of the same prime base are added. Step 2: The power of 3 in (a) is 6 and in (b) is 8. Step 3: In (ab), the power of 3 is (6+8=14).
If (A=2^{12}\times3^8\times5^4) and (B=2^9\times3^{10}\times7^5), which is the HCF of (A) and (B)?
Correct answer: A
Step 1: HCF uses the smaller powers of common prime factors only. Step 2: The common factors are 2 and 3, with smaller powers (2^9) and (3^8). Step 3: So the correct form is (2^9\times3^8).
If (A=2^{12}\times3^8\times5^4) and (B=2^9\times3^{10}\times7^5), which is the LCM of (A) and (B)?
Correct answer: A
Step 1: LCM uses the highest powers of all prime factors. Step 2: The highest powers are (2^{12}), (3^{10}), (5^4), and (7^5). Step 3: So the correct form is (2^{12}\times3^{10}\times5^4\times7^5).
The HCF of two numbers is 120 and their LCM is 9240. If one number is 840, what is the other number?
Correct answer: A
Step 1: Product of the two numbers is (120\times9240=1108800). Step 2: One number is 840, so the other is (1108800\div840=1320). Step 3: As a check, the HCF of 840 and 1320 is 120.
The HCF of two numbers is 144 and their LCM is 10080. If one number is 720, what is the other number?
Correct answer: A
Step 1: Product of the two numbers is (144\times10080=1451520). Step 2: The other number is (1451520\div720=2016). Step 3: To check, the HCF of 720 and 2016 is 144.
If (q=2^6\times3^b\times5^2\times7) and (q=1814400), what is the value of (b)?
Correct answer: B
Step 1: Write (1814400=18144\times100). Step 2: (18144=2^5\times3^4\times7) and (100=2^2\times5^2), so the actual form is (2^7\times3^4\times5^2\times7). Step 3: The power of 3 is (b=4).
If a number has prime factorisation (2^9\times3^7\times5^6\times17^2), how many prime factors are there if repetition is counted?
Correct answer: C
Step 1: To count with repetition, add the exponents. Step 2: (2^9) gives 9, (3^7) gives 7, (5^6) gives 6, and (17^2) gives 2 factors. Step 3: Total (9+7+6+2=24), so the answer is 24.
If a number has prime factorisation (2^{11}\times3^8\times5^3\times7^2\times13), how many distinct prime factors does it have?
Correct answer: A
Step 1: When counting distinct prime factors, exponents are not added. Step 2: The prime bases are 2, 3, 5, 7, and 13. Step 3: Therefore, the number of distinct prime factors is 5.
If the two numbers are (2^8\times3^4\times13) and (5^5\times7^3\times11), which statement about them is correct?
Correct answer: A
Step 1: The prime factors of the first number are 2, 3, and 13. Step 2: The prime factors of the second number are 5, 7, and 11. Step 3: There is no common prime factor, so they are co-prime.
Step 1: Recognise (705600=840^2). Step 2: Since (840=2^3\times3\times5\times7), (840^2=2^6\times3^2\times5^2\times7^2). Step 3: In a perfect square, all prime exponents are even.
If (n=2^7\times3^6\times5^3\times19), by which number must (n) be divisible?
Correct answer: A
Step 1: (69255=3^6\times5\times19). Step 2: All these prime factors are present in (n) with sufficient powers. Step 3: Therefore, (n) must be divisible by 69255.
If (n=2^9\times3^5\times5^3), by which of the following will (n) not be divisible?
Correct answer: D
Step 1: In (n), the powers are (2^9), (3^5), and (5^3). Step 2: (16875=3^3\times5^4), which needs power 4 of 5. Step 3: Since (n) has only (5^3), (n) is not divisible by 16875.
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