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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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The order may change, but the prime factors and their powers remain the same
Every number is equal to only one prime number
Every composite number has only two factors
Composite numbers may remain in the final factorisation
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(2^5\times3^2\times5\times7\times11)
(2^4\times3^3\times5\times7\times11)
(2^5\times3^2\times5^2\times7)
(10080\times11)
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1296
2592
5184
10368
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714420000
357210000
1428840000
238140000
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2520
2160
2880
3240
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582120
291060
462000
474600
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3
2
5
6
Expert · Level 1View options
30
42
60
70
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2450
1225
490
245
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(3^3\times7^3\times13)
(3^2\times7^4\times13)
(3^3\times7^2\times13^2)
(21^3\times13)
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110
55
22
10
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(2\times3^2\times5^2\times7)
(2^2\times3\times5\times7)
(2\times3\times5^2\times7^2)
(2^2\times3^2\times5\times7)
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10
11
12
13
Expert · Level 1View options
8
9
10
11
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(2^8\times3^7)
(2^{11}\times3^9)
(2^8\times3^7\times5\times7)
(2^3\times3^2)
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(2^{11}\times3^9\times5^3\times7^4)
(2^8\times3^7)
(2^{11}\times3^7\times5^3)
(2^8\times3^9\times7^4)
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924
840
1008
1155
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1344
960
1152
1440
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2
3
4
5
Expert · Level 1View options
2
3
4
5
Expert · Level 1View options
19
20
21
22
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5
7
10
23
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They are co-prime
Their HCF is 11
Their common prime factor is 7
Their LCM is 1
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1
247
323
4199
Expert · Level 1View options
(2^{15})
(2^{14})
(4^8)
(16^4)
Question 1ExpertLevel 1
What is the most accurate meaning of unique prime factorisation in the Fundamental Theorem of Arithmetic?
Correct answer: A
Step 1: The theorem says that prime factorisation of a number greater than 1 is fixed. Step 2: The order may change, but the prime bases and their powers do not change. Step 3: In exams, do not treat a change of order as a new factorisation.
Step 1: Write (110880=10080\times11). Step 2: Since (10080=2^5\times3^2\times5\times7), the full form is (2^5\times3^2\times5\times7\times11). Step 3: Do not leave a composite factor such as 10080 in the final answer.
If (a=2^7\times3^4\times5^3) and (b=2^4\times3^6\times7^2), 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 common factors are 2 and 3, with smaller powers (2^4) and (3^4). Step 3: (2^4\times3^4=16\times81=1296), so the answer is 1296.
If (a=2^7\times3^4\times5^3) and (b=2^4\times3^6\times7^2), what is the LCM of (a) and (b)?
Correct answer: A
Step 1: For LCM, take the highest powers of all prime factors. Step 2: The highest powers are (2^7), (3^6), (5^3), and (7^2). Step 3: (128\times729\times125\times49=714420000), so the answer is 714420000.
If the HCF of two numbers is 126 and their LCM is 4620, what is their product?
Correct answer: A
Step 1: For two numbers, the product equals HCF multiplied by LCM. Step 2: (126\times4620=582120). Step 3: In such questions, first confirm that exactly two numbers are involved.
What is the smallest positive number by which 43200 must be multiplied to get a perfect square?
Correct answer: A
Step 1: (43200=432\times100=2^6\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 it 4, so the smallest number is 3.
By which smallest number should 26460 be divided to get a perfect square?
Correct answer: C
Step 1: (26460=2^2\times3^3\times5\times7^2). Step 2: For a perfect square, the odd powers of 3 and 5 must be reduced. Step 3: Dividing by (3\times5=15) leaves (2^2\times3^2\times7^2), so the smallest number is 15.
What is the smallest number by which 15120 must be multiplied to get a perfect cube?
Correct answer: A
Step 1: (15120=2^4\times3^3\times5\times7). Step 2: For a perfect cube, exponents must be multiples of 3. Step 3: Multiplying by (2^2\times5^2\times7^2=2450) makes the powers 6, 3, 3, and 3.
Step 1: A convenient form is (250047=21^3\times13). Step 2: Since (21=3\times7), (21^3\times13=3^3\times7^3\times13). Step 3: Since 21 is composite, write 3 and 7 in the final prime form.
If (N=2^9\times3^6\times5^5\times11^3), by which smallest number must (N) be multiplied to make it a perfect square?
Correct answer: A
Step 1: For a perfect square, every exponent must be even. Step 2: The powers of 2, 5, and 11 are 9, 5, and 3, which are odd. Step 3: Multiplying by (2\times5\times11=110) makes all powers even.
If (N=2^8\times3^7\times5^4\times7^2), 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 8 to 9, 7 to 9, 4 to 6, and 2 to 3. Step 3: So the smallest multiplier is (2\times3^2\times5^2\times7).
If (a=2^6\times3^5\times7^2) and (b=2^4\times3^7\times5^3), 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 5 and in (b) is 7. Step 3: In (ab), the power of 3 is (5+7=12).
If (A=2^{11}\times3^7\times5^3) and (B=2^8\times3^9\times7^4), 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 (2^8) and (3^7). Step 3: Therefore, the HCF is (2^8\times3^7).
If (A=2^{11}\times3^7\times5^3) and (B=2^8\times3^9\times7^4), 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^{11}), (3^9), (5^3), and (7^4). Step 3: So the correct form is (2^{11}\times3^9\times5^3\times7^4).
The HCF of two numbers is 84 and their LCM is 4620. If one number is 420, what is the other number?
Correct answer: A
Step 1: Product of the two numbers is (84\times4620=388080). Step 2: One number is 420, so the other is (388080\div420=924). Step 3: As a check, the HCF of 420 and 924 is 84.
The HCF of two numbers is 96 and their LCM is 6720. If one number is 480, what is the other number?
Correct answer: A
Step 1: Product of the two numbers is (96\times6720=645120). Step 2: The other number is (645120\div480=1344). Step 3: To check, the HCF of 480 and 1344 is 96.
If (q=2^5\times3^b\times5^2\times7) and (q=604800), what is the value of (b)?
Correct answer: B
Step 1: Write (604800=6048\times100). Step 2: (6048=2^5\times3^3\times7) and (100=2^2\times5^2), so the actual form is (2^7\times3^3\times5^2\times7). Step 3: The given power of 2 does not match, but the power of 3 is (b=3).
If a number has prime factorisation (2^8\times3^6\times5^5\times13^2), how many prime factors are there if repetition is counted?
Correct answer: C
Step 1: To count with repetition, add the exponents. Step 2: (2^8) gives 8, (3^6) gives 6, (5^5) gives 5, and (13^2) gives 2 factors. Step 3: Total (8+6+5+2=21), so the answer is 21.
If a number has prime factorisation (2^{10}\times3^7\times5^2\times11^3\times13), how many distinct prime factors does it have?
Correct answer: A
Step 1: When counting distinct prime factors, do not add exponents. Step 2: The prime bases are 2, 3, 5, 11, and 13. Step 3: Therefore, the number of distinct prime factors is 5.
If the two numbers are (2^7\times3^3\times11) and (5^4\times7^2\times13), which statement about them is correct?
Correct answer: A
Step 1: The prime factors of the first number are 2, 3, and 11. Step 2: The prime factors of the second number are 5, 7, and 13. Step 3: There is no common prime factor, so they are co-prime.
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