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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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Hard · Level 5 · distinct-prime-factors,prime-factorisation,hardView options
3
5
8
15
Hard · Level 5 · coprime,prime-factorisation,hardView options
They are co-prime
Their HCF is 5
Their common prime factor is 3
Their LCM is 1
Hard · Level 5 · coprime,lcm,hardView options
1
91
119
1547
Hard · Level 5 · prime-factorisation,powers,hardView options
(2^{13})
(2^{12})
(4^7)
(16^3)
Hard · Level 5 · prime-factorisation,large-number,hardView options
(2^2\times3\times5^2\times7^2)
(2^2\times3^2\times5\times7^2)
(2^3\times3\times5^2\times7)
(147\times100)
Hard · Level 5 · divisibility,prime-factorisation,hardView options
1485
13
17
19
Hard · Level 5 · divisibility,prime-factorisation,hardView options
144
80
45
75
Hard · Level 5 · evaluate-factorisation,powers,hardView options
5760
2880
11520
6480
Hard · Level 5 · evaluate-factorisation,powers,hardView options
97200
48600
194400
32400
Hard · Level 5 · lcm,powers,hardView options
3
5
6
9
Hard · Level 5 · hcf,powers,hardView options
2
3
5
7
Hard · Level 5 · prime-factorisation,perfect-square,hardView options
(2^2\times3^4\times7^2)
(2^3\times3^3\times7^2)
(2^2\times3^3\times7^3)
(126\times126)
Hard · Level 5 · perfect-cube,prime-exponents,hardView options
(2^2\times3\times5^2)
(2\times3\times5^2)
(2^2\times3^2\times5)
(2\times3^2\times5)
Hard · Level 5 · perfect-square,division,hardView options
3
5
15
35
Hard · Level 5 · product-factorisation,powers,hardView options
5
6
7
8
Hard · Level 5 · product-factorisation,powers,hardView options
6
8
10
12
Hard · Level 5 · distinct-prime-factors,prime-factorisation,hardView options
2, 3, 5, 7, 11 and 13
2, 3, 5, 7 and 11
3, 5, 7, 11 and 13
2, 5, 7, 11 and 13
Hard · Level 5 · hcf,prime-factorisation,hardView options
156
312
468
780
Hard · Level 5 · lcm,prime-factorisation,hardView options
365040
182520
730080
121680
Hard · Level 5 · hcf-lcm-relation,powers,hardView options
3
4
5
6
Question 1HardLevel 5
If a number has prime factorisation (2^8\times3^5\times11^2), 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 factors here are 2, 3, and 11. Step 3: Therefore, the number of distinct prime factors is 3.
If the two numbers are (2^4\times3^2) and (5^3\times7), which statement about them is correct?
Correct answer: A
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: Write (14700=147\times100). Step 2: (147=3\times7^2) and (100=2^2\times5^2), so (14700=2^2\times3\times5^2\times7^2). Step 3: 147 and 100 are composite, so do not keep them in the final form.
If (n=2^4\times3^3\times5\times11), by which number must (n) be divisible?
Correct answer: A
Step 1: (1485=3^3\times5\times11). Step 2: All these factors are present in the prime factorisation of (n). Step 3: Therefore, (n) must be divisible by 1485.
If (n=2^4\times3^2\times5), by which of the following will (n) not be divisible?
Correct answer: D
Step 1: In (n), the powers are (2^4), (3^2), and (5^1). Step 2: (75=3\times5^2), which needs power 2 of 5. Step 3: Since (n) has only (5^1), (n) will not be divisible by 75.
If the prime factorisation of a number is (2^4\times3^5\times5^2), what is the number?
Correct answer: A
Step 1: Calculate (2^4=16), (3^5=243), and (5^2=25). Step 2: (16\times243\times25=97200). Step 3: Simplifying powers first makes the calculation easier.
Step 1: (15876=126^2). Step 2: Since (126=2\times3^2\times7), (126^2=2^2\times3^4\times7^2). Step 3: In a perfect square, every prime exponent is even.
If a number has prime factorisation (2^7\times3^5\times5^4), what is the smallest number by which it must be multiplied to make a perfect cube?
Correct answer: A
Step 1: For a perfect cube, exponents must be multiples 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).
By which smallest number should 6615 be divided to get a perfect square?
Correct answer: C
Step 1: (6615=3^3\times5\times7^2). Step 2: For a perfect square, all exponents must be even, but powers of 3 and 5 are odd. Step 3: Dividing by (3\times5=15) leaves (3^2\times7^2), a perfect square.
If (x=2^5\times3^3\times7) and (y=2^2\times3^4\times5), what will be the power of 3 in (xy)?
Correct answer: C
Step 1: In multiplication, exponents of the same prime base are added. Step 2: The power of 3 in (x) is 3 and in (y) is 4. Step 3: In (xy), the power of 3 is (3+4=7).
If (x=2^6\times3^2\times5^3) and (y=2^4\times3^5\times5), what will be the power of 2 in (xy)?
Correct answer: C
Step 1: Powers of the same base 2 are added in multiplication. Step 2: The power of 2 in (x) is 6 and in (y) is 4. Step 3: The total power is (6+4=10).
If (a=2^4\times3^3\times13) and (b=2^2\times3\times5\times13^2), what is the HCF of (a) and (b)?
Correct answer: A
Step 1: The common prime factors are 2, 3, and 13. Step 2: The smaller powers are (2^2), (3^1), and (13^1). Step 3: (4\times3\times13=156), so the HCF is 156.
If (a=2^4\times3^3\times13) and (b=2^2\times3\times5\times13^2), 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 (2^4), (3^3), (5), and (13^2). Step 3: (16\times27\times5\times169=365040), so the answer is 365040.
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