What is the prime factorisation of 1024?
Correct answer: A
Step 1: Divide 1024 repeatedly by 2. Step 2: (1024=2^{10}). Step 3: (4^5) and (32^2) can give the value, but 4 and 32 are not prime.
Mathematics
Fundamental Theorem of Arithmetic
अंकगणित का मौलिक प्रमेय
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
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Medium · Level 2
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- (2^{10})
- (2^8)
- (4^5)
- (32^2)
Medium · Level 2
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- 1625
- 2025
- 2250
- 4050
Medium · Level 2
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- (3^3\times5^2)
- (3^2\times5^3)
- (3^3\times5)
- (27\times25)
Medium · Level 2
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- 3
- 5
- 6
- 10
Medium · Level 2
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- 72
- 216
- 288
- 648
Medium · Level 2
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- 1296
- 2592
- 648
- 2304
Medium · Level 2
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- 3, 5, 7 and 11
- 2, 3, 5 and 11
- 3, 5 and 13
- 5, 7 and 11
Medium · Level 2
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- 441
- 8820
- 20
- 399
Medium · Level 2
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- (3^4\times5^2)
- (3^2\times5^4)
- (3^3\times5^2)
- (45\times45)
Medium · Level 2
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- 1 is neither prime nor composite
- 1 is prime because it is the smallest number
- 1 is composite because it divides every number
- 1 has no factor
Medium · Level 2
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- 10
- 20
- 50
- 100
Medium · Level 2
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- 2100
- 4200
- 8400
- 1400
Medium · Level 2
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- (2^3\times3\times7\times11)
- (2^2\times3^2\times7\times11)
- (2^3\times3^2\times7)
- (168\times11)
Medium · Level 2
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- Every composite number has a unique prime factorisation except for order
- Every number has only one prime factor
- Every number has LCM 1
- Every number is co-prime
Medium · Level 2
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- 2400
- 4800
- 9600
- 1200
Medium · Level 2
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- 1
- Their sum
- Their product
- The larger number
Medium · Level 2
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- 168
- 252
- 336
- 672
Medium · Level 2
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- 1
- 2
- 3
- 4
Medium · Level 2
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- 1
- 2
- 3
- 4
Medium · Level 2
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- 2
- 3
- 5
- 6
Medium · Level 2
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- 10
- 3
- 7
- 9
Medium · Level 2
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- (3^3\times7^2)
- (3^2\times7^3)
- (3\times7^2\times9)
- (27\times49)
Medium · Level 2
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- 3
- 4
- 5
- 6
Medium · Level 2
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- 60
- 120
- 180
- 240
Medium · Level 2
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- Prime factorisation is unique except for order
- Every number always has prime factorisation in two factors
- Every number is factorised only into even numbers
- Composite factors may remain in final prime factorisation
If (n=3^4\times5^2), what is the value of (n)?
Correct answer: B
Step 1: Calculate (3^4=81) and (5^2=25). Step 2: (81\times25=2025). Step 3: In questions with powers, simplify the powers first.
Which option gives the correct prime factorisation of 675?
Correct answer: A
Step 1: Write (675=27\times25). Step 2: (27=3^3) and (25=5^2), so (675=3^3\times5^2). Step 3: 27 and 25 are composite, so write their prime powers in the final form.
If a number has prime factorisation (2^2\times3^3\times5), how many prime factors does it have including repetition?
Correct answer: C
Step 1: To count with repetition, add the exponents. Step 2: (2^2) gives 2, (3^3) gives 3, and (5) gives 1 factor. Step 3: Total (2+3+1=6), so the answer is 6.
If (A=2^5\times3^2) and (B=2^3\times3^4), what is the HCF of (A) and (B)?
Correct answer: A
Step 1: For HCF, take the smaller powers of common prime factors. Step 2: The smaller powers are (2^3) and (3^2). Step 3: (2^3\times3^2=8\times9=72), so the answer is 72.
If (A=2^5\times3^2) and (B=2^3\times3^4), what is the LCM of (A) and (B)?
Correct answer: B
Step 1: For LCM, take the highest powers. Step 2: The highest powers are (2^5) and (3^4). Step 3: (32\times81=2592), so the answer is 2592.
What are the distinct prime factors of 1155?
Correct answer: A
Step 1: Write (1155=105\times11). Step 2: (105=3\times5\times7), so (1155=3\times5\times7\times11). Step 3: The distinct prime factors are 3, 5, 7, and 11.
If the HCF of two numbers is 21 and their LCM is 420, what is their product?
Correct answer: B
Step 1: For two numbers, product (=) HCF (\times) LCM. Step 2: (21\times420=8820). Step 3: Use the HCF-LCM relation in the correct situation.
What is the prime factorisation of 2025?
Correct answer: A
Step 1: Write (2025=45\times45). Step 2: (45=3^2\times5), so (2025=3^4\times5^2). Step 3: In perfect squares, the powers of the base factors get doubled.
Which option gives the correct statement about 1?
Correct answer: A
Step 1: A prime number must have exactly two positive factors. Step 2: 1 has only one positive factor, so it is not prime. Step 3: A composite number needs more than two factors, so 1 is not composite either.
If (p=2^2\times3\times5^2) and (q=2^3\times5\times7), what is the HCF of (p) and (q)?
Correct answer: B
Step 1: The common prime factors are 2 and 5. Step 2: The smaller powers are (2^2) and (5^1). Step 3: (2^2\times5=4\times5=20), so the HCF is 20.
If (p=2^2\times3\times5^2) and (q=2^3\times5\times7), what is the LCM of (p) and (q)?
Correct answer: B
Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^3), (3), (5^2), and (7). Step 3: (8\times3\times25\times7=4200), so the answer is 4200.
Which is the prime factorisation of 1848?
Correct answer: A
Step 1: Write (1848=168\times11). Step 2: (168=2^3\times3\times7), so (1848=2^3\times3\times7\times11). Step 3: 168 is composite, so do not keep it in the final form.
Which conclusion correctly follows from the Fundamental Theorem of Arithmetic?
Correct answer: A
Step 1: The main idea of the theorem is prime factorisation. Step 2: Every composite number greater than 1 can be written in a fixed way as prime factors. Step 3: The order may change, but the prime factors do not change.
If (r=2^6\times3\times5^2), what is the value of (r)?
Correct answer: B
Step 1: Calculate (2^6=64) and (5^2=25). Step 2: (64\times3\times25=4800). Step 3: Simplifying powers first makes multiplication easier.
If two numbers have no common prime factor in their prime factorisations, what will their HCF be?
Correct answer: A
Step 1: Having no common prime factor means the numbers are co-prime. Step 2: Co-prime numbers have HCF 1. Step 3: Prime factorisation helps identify co-primality quickly.
Which number has prime factorisation (2^4\times3\times7)?
Correct answer: C
Step 1: Calculate (2^4=16). Step 2: (16\times3\times7=336). Step 3: To get the number from prime factorisation, multiply all factors.
If the prime factorisation of a number is (2^3\times3^a\times5) and the number is 1080, what is the value of (a)?
Correct answer: C
Step 1: Write (1080=108\times10). Step 2: (108=2^2\times3^3) and (10=2\times5), so (1080=2^3\times3^3\times5). Step 3: Comparing with the given form gives (a=3).
If (a=2^2\times3^2\times5) and (b=2\times3\times5\times7), what will be the power of 3 in the prime factorisation of (ab)?
Correct answer: C
Step 1: In multiplication, powers with the same base are added. Step 2: The power of 3 in (a) is 2 and in (b) is 1. Step 3: In (ab), the power of 3 will be (2+1=3).
If (a=2^3\times5^2) and (b=2^2\times5^3), what will be the power of 5 in the prime factorisation of (ab)?
Correct answer: C
Step 1: While multiplying, add the exponents of the same prime base. Step 2: The power of 5 in (a) is 2 and in (b) is 3. Step 3: In (ab), the power of 5 will be (2+3=5).
If a number has only (2^3) and (5^2) in its prime factorisation, by which number must it be divisible?
Correct answer: A
Step 1: (2^3) and (5^2) contain both factors 2 and 5. Step 2: The product of 2 and 5 is 10, so the number must be divisible by 10. Step 3: Divisibility can be quickly identified from prime factors.
Which is the prime factorisation of 1323?
Correct answer: A
Step 1: Write (1323=27\times49). Step 2: (27=3^3) and (49=7^2), so (1323=3^3\times7^2). Step 3: 27 and 49 are composite, so write prime powers in the final form.
If (x=2^3\times3^2) and (y=2^2\times3^3), what will be the power of 2 in (xy)?
Correct answer: C
Step 1: In (xy), the powers of the same base 2 are added. Step 2: The power of 2 in (x) is 3 and in (y) is 2. Step 3: So the power of 2 in (xy) is (3+2=5).
If (x=2^4\times3\times5) and (y=2^2\times3^3\times5^2), what is the HCF of (x) and (y)?
Correct answer: A
Step 1: For HCF, take the smaller powers of common prime factors. Step 2: The smaller powers are (2^2), (3^1), and (5^1). Step 3: (4\times3\times5=60), so the HCF is 60.
According to the Fundamental Theorem of Arithmetic, which statement about prime factorisation is correct?
Correct answer: A
Step 1: The theorem tells us that prime factorisation is fixed. Step 2: The order of factors may change but the prime factors remain the same. Step 3: In exams, do not treat changed order as a new factorisation.
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