Which is the prime factorisation of 11025?
Correct answer: A
Step 1: Recognise (11025=105^2). Step 2: Since (105=3\times5\times7), (105^2=3^2\times5^2\times7^2). Step 3: In perfect squares, every prime exponent is even.
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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Hard · Level 2
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- (3^2\times5^2\times7^2)
- (3^3\times5^2\times7)
- (3^2\times5^3\times7)
- (105\times105)
Hard · Level 2
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- 315
- 11
- 13
- 17
Hard · Level 2
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- 75
- 80
- 48
- 150
Hard · Level 2
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- 2880
- 1440
- 5760
- 3240
Hard · Level 2
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- 16200
- 32400
- 8100
- 64800
Hard · Level 2
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- 2
- 3
- 4
- 6
Hard · Level 2
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- 0
- 1
- 2
- 3
Hard · Level 2
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- (3^4\times7^2)
- (3^2\times7^4)
- (3^3\times7^3)
- (63\times63)
Hard · Level 2
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- (2\times3\times5^2)
- (2^2\times3\times5)
- (2\times3^2\times5)
- (2^2\times3^2\times5^2)
Hard · Level 2
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- 2
- 3
- 6
- 7
Hard · Level 2
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- 3
- 4
- 5
- 6
Hard · Level 2
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- 5
- 6
- 8
- 10
Hard · Level 2
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- 2, 3, 5, 7 and 11
- 2, 3, 5 and 7
- 3, 5, 7 and 11
- 2, 5, 7 and 13
Hard · Level 2
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- 132
- 264
- 396
- 660
Hard · Level 2
View options
- 43560
- 21780
- 87120
- 14520
Hard · Level 2
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- 2
- 3
- 4
- 5
Hard · Level 2
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- 0
- 1
- 2
- 3
Hard · Level 2
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- (7^4)
- (7^3)
- (49^2)
- (7\times343^2)
Hard · Level 2
View options
- (2^2\times3^2)
- (2\times3^2)
- (2^2\times3)
- (3^2\times5)
Hard · Level 2
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- 2520
- 1260
- 5040
- 840
Hard · Level 2
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- (ab)
- (a+b)
- (a-b)
- 1
Hard · Level 2
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- 12
- 24
- 36
- 48
Hard · Level 2
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- 10
- 11
- 12
- 13
Hard · Level 2
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- 1
- 2
- 3
- 4
Hard · Level 2
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- 1
- 2
- 3
- 4
If (n=2^3\times3^2\times5\times7), by which number must (n) be divisible?
Correct answer: A
Step 1: The prime factorisation of 315 is (3^2\times5\times7). Step 2: All these factors are present in the prime factorisation of (n). Step 3: Therefore, (n) must be divisible by 315.
If (n=2^4\times3\times5^2), by which of the following will (n) not be divisible?
Correct answer: C
Step 1: In (n), the powers are (2^4), (3^1), and (5^2). Step 2: (48=2^4\times3), so divisibility by 48 is possible. Step 3: (80=2^4\times5), (75=3\times5^2), and (150=2\times3\times5^2) are also present; hence all given options divide (n).
Which option gives the correct value of (2^6\times3^2\times5)?
Correct answer: A
Step 1: Calculate (2^6=64) and (3^2=9). Step 2: (64\times9\times5=2880). Step 3: In larger products, evaluate powers first and then multiply.
If the prime factorisation of a number is (2^3\times3^4\times5^2), what is the number?
Correct answer: A
Step 1: Calculate (2^3=8), (3^4=81), and (5^2=25). Step 2: (8\times81\times25=16200). Step 3: Simplifying powers first keeps the calculation safe.
If (a=2^2\times3^3\times5) and (b=2^4\times3\times5^2), what is the power of 2 in the LCM of (a) and (b)?
Correct answer: C
Step 1: In LCM, take the higher power of each prime. Step 2: The powers of 2 are 2 and 4. Step 3: The higher power is 4, so the answer is 4.
If (a=2^2\times3^3\times5) and (b=2^4\times3\times5^2), what is the power of 5 in the HCF of (a) and (b)?
Correct answer: B
Step 1: In HCF, take the smaller power of the common prime. Step 2: The powers of 5 are 1 and 2. Step 3: The smaller power is 1, so the answer is 1.
Which is the prime factorisation of 3969?
Correct answer: A
Step 1: (3969=63^2). Step 2: Since (63=3^2\times7), (63^2=3^4\times7^2). Step 3: (63\times63) gives the product, but it is not prime factorisation.
If a number has prime factorisation (2^5\times3^2\times5^4), what is the smallest number by which it must be multiplied to make a perfect cube?
Correct answer: B
Step 1: For a perfect cube, exponents must be multiples of 3. Step 2: We need (2), (3), and (5^2) to make the powers (6,3,6). Step 3: The smallest multiplier is (2\times3\times5^2).
By which smallest number should 3528 be divided to get a perfect square?
Correct answer: C
Step 1: (3528=72\times49=2^3\times3^2\times7^2). Step 2: For a perfect square, all exponents must be even, but the exponent of 2 is 3. Step 3: Dividing by 2 gives (2^2\times3^2\times7^2), so the smallest number is 2.
If (x=2^4\times3^2\times7) and (y=2^2\times3^3\times5), what will be the power of 3 in (xy)?
Correct answer: C
Step 1: In multiplication, add the exponents of the same prime base. Step 2: The power of 3 in (x) is 2 and in (y) is 3. Step 3: In (xy), the power of 3 is (2+3=5).
If (x=2^5\times3\times5^2) and (y=2^3\times3^4\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 5 and in (y) is 3. Step 3: The total power is (5+3=8).
Which option gives all distinct prime factors of 2310 correctly?
Correct answer: A
Step 1: Write (2310=231\times10). Step 2: (231=3\times7\times11) and (10=2\times5). Step 3: The distinct prime factors are 2, 3, 5, 7, and 11.
If (a=2^3\times3^2\times11) and (b=2^2\times3\times5\times11^2), what is the HCF of (a) and (b)?
Correct answer: A
Step 1: The common prime factors are 2, 3, and 11. Step 2: The smaller powers are (2^2), (3^1), and (11^1). Step 3: (4\times3\times11=132), so the HCF is 132.
If (a=2^3\times3^2\times11) and (b=2^2\times3\times5\times11^2), 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 (2^3), (3^2), (5), and (11^2). Step 3: (8\times9\times5\times121=43560), so the answer is 43560.
If the HCF of two numbers is 36 and their LCM is 900, what is the power of 2 in their product?
Correct answer: C
Step 1: The product is (36\times900). Step 2: (36=2^2\times3^2) and (900=2^2\times3^2\times5^2). Step 3: The power of 2 in the product is (2+2=4).
If the HCF of two numbers is 30 and their LCM is 2310, what is the power of 11 in their product?
Correct answer: B
Step 1: The product is (30\times2310). Step 2: 30 has no factor 11 and (2310=2\times3\times5\times7\times11). Step 3: Therefore, the power of 11 in the product is (0+1=1).
What is the prime factorisation of 2401?
Correct answer: A
Step 1: (2401=49\times49). Step 2: Since (49=7^2), (2401=7^4). Step 3: 49 is composite, so (49^2) is not the final prime factorisation.
If (n=2^2\times3^5\times5^3), by which smallest number should (n) be divided to make it a perfect cube?
Correct answer: A
Step 1: In a perfect cube, each exponent must be a multiple of 3. Step 2: (2^2) and (3^5) cause the issue; reduce them to 0 and 3. Step 3: Dividing by (2^2\times3^2) gives exponents (0,3,3).
Which option gives the number formed by (2^3\times3^2\times5\times7)?
Correct answer: A
Step 1: First calculate (2^3=8) and (3^2=9). Step 2: (8\times9\times5\times7=2520). Step 3: To get the number from prime factorisation, multiply all factors.
If (a=2^4\times3^2\times5) and (b=2^3\times3\times5^2), the product of their HCF and LCM will be equal to what?
Correct answer: A
Step 1: For two numbers, HCF (\times) LCM equals the product of the two numbers. Step 2: This also follows from prime powers because the smaller and higher exponents together give the total exponent. Step 3: Therefore, the answer is (ab).
If two numbers have prime factorisations (2^3\times3^2) and (2^5\times3), what is the ratio of their LCM to HCF?
Correct answer: A
Step 1: HCF is (2^3\times3=24). Step 2: LCM is (2^5\times3^2=288). Step 3: The ratio is (288\div24=12).
If (x=2^2\times3^3\times5) and (y=2^4\times3\times5^2), how many prime factors will (xy) have if repetition is counted?
Correct answer: D
Step 1: In (xy), exponents of the same bases are added. Step 2: Powers become (2^{6}), (3^{4}), and (5^{3}). Step 3: Counting with repetition gives (6+4+3=13).
If (m=2^5\times3^2\times7) and (n=2^2\times3^4\times7^3), what is the power of 7 in the LCM of (m) and (n)?
Correct answer: C
Step 1: In LCM, take the higher power. Step 2: The powers of 7 are 1 and 3. Step 3: The higher power is 3, so the answer is 3.
If (m=2^5\times3^2\times7) and (n=2^2\times3^4\times7^3), what is the power of 3 in the HCF of (m) and (n)?
Correct answer: B
Step 1: In HCF, take the smaller power. Step 2: The powers of 3 are 2 and 4. Step 3: The smaller power is 2, so the answer is 2.
QUIZ COMPLETE