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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
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As a product of prime numbers
As a sum of only even numbers
As a difference of only odd numbers
Always as a single digit
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(2\times3^3)
(2^2\times3^2)
(3^2\times5)
(6\times9)
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(2^4\times5)
(2^3\times5^2)
(4\times20)
(2\times5\times8)
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The product does not change
The product always changes
The number always becomes prime
The number always becomes 1
Easy · Level 5View options
3
4
5
6
Easy · Level 5View options
84
126
168
42
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(3^2\times11)
(3\times11^2)
(9\times11)
(3\times33)
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47
51
57
63
Easy · Level 5View options
67
71
91
73
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(3\times5\times7)
(3^2\times5)
(5^2\times7)
(15\times7)
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2
3
4
5
Easy · Level 5View options
(13^2)
(13\times12)
(7\times24)
(169\times1)
Easy · Level 5View options
12
24
48
144
Easy · Level 5View options
48
72
144
288
Easy · Level 5View options
(2^5\times3^2)
(2^4\times3^3)
(2^3\times3^2)
(32\times9)
Easy · Level 5View options
45
90
135
150
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Because 1 is not a prime number
Because 1 is an even number
Because 1 has no value
Because 1 changes every number
Easy · Level 5View options
(3\times7^2)
(3^2\times7)
(7\times21)
(3\times49^2)
Easy · Level 5View options
2 and 31
2 and 13
4 and 31
2 and 17
Easy · Level 5View options
1
2
Their sum
Their product
Easy · Level 5View options
25 and 36
18 and 24
21 and 35
28 and 42
Easy · Level 5View options
84
126
252
504
Easy · Level 5View options
252
504
756
84
Easy · Level 5View options
(2^3\times3^2\times5)
(2^2\times3^3\times5)
(2^3\times3\times5^2)
(36\times10)
Easy · Level 5View options
30
45
60
90
Question 1EasyLevel 5
According to the Fundamental Theorem of Arithmetic, every positive integer greater than 1 can be written in which form?
Correct answer: A
Step 1: This theorem is connected with prime factorisation. Step 2: Every positive integer greater than 1 can be written as a product of prime numbers. Step 3: In exams, remember it through prime factorisation.
What happens when the order of factors is changed in prime factorisation?
Correct answer: A
Step 1: Changing the order in multiplication does not change the product. Step 2: (2\times5\times3) and (3\times2\times5) both give 30. Step 3: In prime factorisation, the set of factors matters, not the order.
Step 1: Write (99=9\times11). Step 2: (9=3^2) and 11 is prime, so (99=3^2\times11). Step 3: 9 and 33 are composite, so do not keep them in the final answer.
Step 1: A prime number has exactly two positive factors. Step 2: 47 is divisible only by 1 and 47, while 51, 57, and 63 are composite. Step 3: Check small numbers by 2, 3, 5, and 7.
Step 1: A composite number has more than two positive factors. Step 2: (91=7\times13), so it is composite. Step 3: If a number can be written as a product of smaller primes, it is composite.
Step 1: Write (105=15\times7). Step 2: (15=3\times5), so (105=3\times5\times7). Step 3: 15 is composite, so do not keep it in the final prime factorisation.
If (180=2^2\times3^2\times5), how many distinct prime factors does 180 have?
Correct answer: B
Step 1: Look at distinct prime factors, not their powers. Step 2: In (180=2^2\times3^2\times5), the distinct primes are 2, 3, and 5. Step 3: Even with a higher power, count the same prime once.
If two numbers have prime factorisations (2^4\times3) and (2^2\times3^2), what is their HCF?
Correct answer: A
Step 1: For HCF, take smaller powers of common prime factors. Step 2: The smaller power of (2) is 2 and of (3) is 1, so (2^2\times3=12). Step 3: Use smaller powers for HCF.
If two numbers have prime factorisations (2^4\times3) and (2^2\times3^2), what is their LCM?
Correct answer: C
Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest power of (2) is 4 and of (3) is 2, so (2^4\times3^2=144). Step 3: Use highest powers for LCM.
Step 1: Write (288=32\times9). Step 2: (32=2^5) and (9=3^2), so (288=2^5\times3^2). Step 3: 32 and 9 are composite, so write prime powers in the final form.
Why is 1 not written as a separate factor in prime factorisation?
Correct answer: A
Step 1: In prime factorisation, only prime factors are written. Step 2: 1 is not prime because it does not have exactly two positive factors. Step 3: Therefore, (1\times n) is not treated as prime factorisation.
Step 1: Co-prime numbers have no common factor except 1. Step 2: Therefore, their HCF is 1. Step 3: To identify co-prime numbers, check common prime factors.
Step 1: (25=5^2) and (36=2^2\times3^2). Step 2: They have no common prime factor, so they are co-prime. Step 3: If a common prime factor appears, the pair is not co-prime.
If (a=2^3\times3\times7) and (b=2^2\times3^2\times7), what is the HCF of (a) and (b)?
Correct answer: A
Step 1: The common prime factors are 2, 3, and 7. Step 2: The smaller powers are (2^2), (3^1), and (7^1), so (4\times3\times7=84). Step 3: HCF uses smaller powers.
Step 1: Write (360=36\times10). Step 2: (36=2^2\times3^2) and (10=2\times5), so (360=2^3\times3^2\times5). Step 3: Combine repeated prime factors using powers.
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