Which number has prime factorisation (2^3\times3^3\times11)?
Correct answer: A
Step 1: Calculate (2^3=8) and (3^3=27). Step 2: (8\times27\times11=2376). Step 3: Finding powers first is the correct method.
Mathematics
Prime Factorisation
अभाज्य गुणनखंडन
In Class 10 Mathematics, under the Real Numbers chapter, Prime Factorisation teaches students to express a composite number as a product of prime numbers. Students practise identifying prime factors and writing factorisations using multiplication and exponents. This foundational skill supports the Fundamental Theorem of Arithmetic and later work with HCF and LCM.
TOPIC PRACTICE
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Hard · Level 4
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- 2376
- 1188
- 4752
- 3564
Hard · Level 4
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- 3465
- 1732
- 6930
- 1155
Hard · Level 4
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- 5040
- 2520
- 10080
- 7560
Hard · Level 4
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- 5292
- 2646
- 10584
- 4410
Hard · Level 4
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- 10368
- 5184
- 20736
- 7776
Hard · Level 4
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- 110
- 10
- 55
- 22
Hard · Level 4
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- 3
- 9
- 21
- 7
Hard · Level 4
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- (2\times3^2\times5\times7^2)
- (2\times3\times5\times7)
- (3^2\times7^2)
- (2^2\times5\times7)
Hard · Level 4
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- (2^2\times3\times5)
- (2\times3\times5)
- (2^2\times3^2\times5)
- (2\times5)
Hard · Level 4
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- (2^6)
- (2^4\times3^2)
- (3^4\times5)
- (2^5\times7)
Hard · Level 4
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- (2^5\times3\times5^4\times11)
- (2^7\times3\times5^4)
- (3^3\times5^2)
- (2^6\times5^6)
Hard · Level 4
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- 15
- 4
- 12
- 10
Hard · Level 4
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- 4
- 16
- 7
- 5
Hard · Level 4
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- 3
- 2
- 1
- 4
Hard · Level 4
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- 3
- 2
- 1
- 4
Hard · Level 4
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- 35280
- 17640
- 70560
- 22050
Hard · Level 4
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- 20736
- 10368
- 41472
- 7776
Hard · Level 4
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- (2^3\times3^2\times5\times7\times11)
- (8\times9\times385)
- (2^3\times9\times385)
- (72\times385)
Hard · Level 4
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- (2^4\times45\times49)
- (2^4\times3^2\times5\times7^2)
- (2^3\times3^3\times5\times7)
- (2\times3\times5^2\times7^2)
Hard · Level 4
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- 5040
- 2520
- 10080
- 7560
Hard · Level 4
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- 9800
- 4900
- 19600
- 1225
Hard · Level 4
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- (2^3\times3\times5^2\times7^2)
- (2^2\times3^2\times5^2\times7)
- (600\times49)
- (8\times3675)
Hard · Level 4
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- (3^3\times5^2\times7^2)
- (3^2\times5^3\times7^2)
- (675\times49)
- (27\times1225)
Hard · Level 4
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- (2^6\times3^6)
- (2^5\times3^6)
- (64\times729)
- (6^6)
Hard · Level 4
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- Because 600 and 49 are composite forms
- Because 49 is prime
- Because 600 cannot be factorised
- Because the product will change
Which number has prime factorisation (3^2\times5\times7\times11)?
Correct answer: A
Step 1: Calculate (3^2=9). Step 2: (9\times5\times7\times11=3465). Step 3: When there are many factors, multiply in pairs.
Which number has prime factorisation (2^4\times3^2\times5\times7)?
Correct answer: A
Step 1: Calculate (2^4=16) and (3^2=9). Step 2: (16\times9\times5\times7=5040). Step 3: First do (16\times9=144), then multiply the rest.
Which number has prime factorisation (2^2\times3^3\times7^2)?
Correct answer: A
Step 1: Calculate (2^2=4), (3^3=27), and (7^2=49). Step 2: (4\times27\times49=5292). Step 3: Simplify all three powers separately.
Which number has prime factorisation (2^7\times3^4)?
Correct answer: A
Step 1: Calculate (2^7=128) and (3^4=81). Step 2: (128\times81=10368). Step 3: Solving large powers first reduces mistakes.
If the number is (2^7\times3^2\times5^3\times11), by which smallest number should it be multiplied to make a perfect square?
Correct answer: A
Step 1: In a perfect square, all exponents are even. Step 2: Powers of 2, 5, and 11 are odd. Step 3: Multiplying by (2\times5\times11=110) makes all exponents even.
If the number is (2^6\times3^5\times7^2), by which smallest number should it be divided to make a perfect square?
Correct answer: A
Step 1: In a perfect square, all exponents should be even. Step 2: Only the power of 3 is odd. Step 3: Dividing by 3 makes the power of 3 equal to 4 and the number becomes a perfect square.
If the number is (2^5\times3^4\times5^2\times7), by which smallest number should it be multiplied to make a perfect cube?
Correct answer: A
Step 1: In a perfect cube, exponents are multiples of 3. Step 2: Powers 5, 4, 2, and 1 must become 6, 6, 3, and 3. Step 3: The smallest multiplier is (2\times3^2\times5\times7^2).
If the number is (2^8\times3^7\times5^4), by which smallest number should it be divided to make a perfect cube?
Correct answer: A
Step 1: For a perfect cube, exponents should be multiples of 3. Step 2: Reducing 8 to 6, 7 to 6, and 4 to 3 is the smallest way. Step 3: Therefore, the divisor is (2^2\times3\times5).
If (n=2^5\times3^4\times5^2\times7), by which number will (n) not be divisible?
Correct answer: A
Step 1: Every prime power of a divisor must be available in the number. Step 2: (n) has power 5 of 2, but (2^6) needs 6. Step 3: Therefore, (n) is not divisible by (2^6).
If (n=2^6\times3^2\times5^5\times11), by which number must (n) be divisible?
Correct answer: A
Step 1: For divisibility, exponents in the divisor must not exceed those in the number. Step 2: (2^5), (3), (5^4), and 11 are all available in (n). Step 3: Therefore, (n) must be divisible by this number.
If a number has prime factorisation (2^6\times3^4\times5^3\times7^2), how many prime factors does it have with repetition?
Correct answer: A
Step 1: For counting with repetition, add the exponents. Step 2: (6+4+3+2=15). Step 3: Remember the difference between the number of distinct bases and the count with repetition.
If a number has prime factorisation (2^4\times3^7\times11^3\times13^2), how many distinct prime factors does it have?
Correct answer: A
Step 1: While counting distinct primes, only bases are counted. Step 2: The bases are 2, 3, 11, and 13. Step 3: Therefore, there are 4 distinct prime factors.
If (a=2^4\times3^5\times7) and (b=2^3\times3^2\times5\times7^2), what will be the power of 7 in (ab)?
Correct answer: A
Step 1: In multiplication, powers of the same prime base are added. Step 2: The power of 7 in (a) is 1 and in (b) is 2. Step 3: In (ab), the power of 7 will be (1+2=3).
If (x=2^7\times5^4\times11^2) and (y=2^2\times3^3\times5^3\times11), what will be the power of 11 in (xy)?
Correct answer: A
Step 1: Powers of the same base 11 are added in multiplication. Step 2: The power of 11 in (x) is 2 and in (y) is 1. Step 3: The total power will be (2+1=3).
Which option gives the correct value of (2^4\times3^2\times5\times7^2)?
Correct answer: A
Step 1: Calculate (2^4=16), (3^2=9), and (7^2=49). Step 2: (16\times9\times5\times49=35280). Step 3: Solve powers separately first.
Which option gives the correct value of (2^8\times3^4)?
Correct answer: A
Step 1: Calculate (2^8=256) and (3^4=81). Step 2: (256\times81=20736). Step 3: Simplify higher powers separately and multiply.
Which option gives only the final prime factorisation?
Correct answer: A
Step 1: In the final form, bases must be prime. Step 2: In the first form, the bases 2, 3, 5, 7, and 11 are prime. Step 3: 8, 9, 385, and 72 are composite, so they are not final forms.
Which option has incomplete prime factorisation?
Correct answer: A
Step 1: In an incomplete form, composite factors remain. Step 2: Both 45 and 49 are composite. Step 3: (2^4\times45\times49) must be changed into (2^4\times3^2\times5\times7^2).
If (2^4\times3^2\times5\times7) is the prime factorisation of a number, what is the number?
Correct answer: A
Step 1: Calculate (2^4=16) and (3^2=9). Step 2: (16\times9\times5\times7=5040). Step 3: Solving powers first helps find the correct option quickly.
If (2^3\times5^2\times7^2) is the prime factorisation of a number, what is the number?
Correct answer: A
Step 1: Calculate (2^3=8), (5^2=25), and (7^2=49). Step 2: (8\times25\times49=9800). Step 3: Find all three powers separately.
What is the correct prime factorisation of 29400?
Correct answer: A
Step 1: Write (29400=600\times49). Step 2: (600=2^3\times3\times5^2) and (49=7^2), so (29400=2^3\times3\times5^2\times7^2). Step 3: Break 600 and 49 completely.
Which is the prime factorisation of 33075?
Correct answer: A
Step 1: Write (33075=675\times49). Step 2: (675=3^3\times5^2) and (49=7^2), so (33075=3^3\times5^2\times7^2). Step 3: Write 675 and 49 as prime powers.
What is the correct prime factorisation of 46656?
Correct answer: A
Step 1: (46656) can be written as (64\times729). Step 2: (64=2^6) and (729=3^6), so (46656=2^6\times3^6). Step 3: Do not leave 64 and 729 in the final form.
Why will (600\times49) not be considered the final answer in prime factorisation?
Correct answer: A
Step 1: In final prime factorisation, every base should be prime. Step 2: (600=2^3\times3\times5^2) and (49=7^2). Step 3: Therefore, the final form is (2^3\times3\times5^2\times7^2).
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