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Real Numbers - The Fundamental Theorem of Arithmetic

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Fundamental Theorem of Arithmetic states that every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

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A composite number is a positive integer greater than 11 that has at least one divisor other than 11 and itself.

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Prime Factorization: The process of representing a composite number as a product of its prime factors, e.g., 12=22×312 = 2^2 \times 3.

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To find the HCFHCF (Highest Common Factor) of two numbers, we take the product of the smallest power of each common prime factor in the numbers.

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To find the LCMLCM (Least Common Multiple) of two numbers, we take the product of the greatest power of each prime factor involved in the numbers.

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For any two positive integers aa and bb, the relationship between their HCFHCF and LCMLCM is given by HCF(a,b)×LCM(a,b)=a×bHCF(a, b) \times LCM(a, b) = a \times b.

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For a number nxn^x to end with the digit 00, its prime factorization must contain both the primes 22 and 55.

📐Formulae

HCF(a,b)×LCM(a,b)=a×bHCF(a, b) \times LCM(a, b) = a \times b

HCF(a,b,c)=a⋅b⋅c⋅LCM(a,b,c)LCM(a,b)⋅LCM(b,c)⋅LCM(a,c)HCF(a, b, c) = \frac{a \cdot b \cdot c \cdot LCM(a, b, c)}{LCM(a, b) \cdot LCM(b, c) \cdot LCM(a, c)}

LCM(a,b,c)=a⋅b⋅c⋅HCF(a,b,c)HCF(a,b)⋅HCF(b,c)⋅HCF(a,c)LCM(a, b, c) = \frac{a \cdot b \cdot c \cdot HCF(a, b, c)}{HCF(a, b) \cdot HCF(b, c) \cdot HCF(a, c)}

💡Examples

Problem 1:

Find the HCFHCF and LCMLCM of 9696 and 404404 by the prime factorization method and verify that HCF×LCM=product of the two numbersHCF \times LCM = \text{product of the two numbers}.

Solution:

First, find prime factors: 96=25×396 = 2^5 \times 3 404=22×101404 = 2^2 \times 101 HCF(96,404)=22=4HCF(96, 404) = 2^2 = 4 LCM(96,404)=25×3×101=32×3×101=9696LCM(96, 404) = 2^5 \times 3 \times 101 = 32 \times 3 \times 101 = 9696 Verification: HCF×LCM=4×9696=38784HCF \times LCM = 4 \times 9696 = 38784 Product of numbers = 96×404=3878496 \times 404 = 38784 Hence, HCF×LCM=a×bHCF \times LCM = a \times b.

Explanation:

We identify the common prime factors for HCF and all occurring prime factors with highest powers for LCM. Then we multiply HCF and LCM to compare with the product of the original numbers.

Problem 2:

Check whether 6n6^n can end with the digit 00 for any natural number nn.

Solution:

If the number 6n6^n ends with the digit 00, then it must be divisible by 55. This means the prime factorization of 6n6^n must contain the prime 55. However, the prime factorization of 6n=(2×3)n=2n×3n6^n = (2 \times 3)^n = 2^n \times 3^n. The only primes in the factorization of 6n6^n are 22 and 33. By the uniqueness of the Fundamental Theorem of Arithmetic, there are no other primes in the factorization of 6n6^n. Since 55 is not a factor, 6n6^n cannot end with the digit 00.

Explanation:

A number ends in 00 if and only if its prime factorization includes both 22 and 55. We use the theorem to prove the uniqueness of the prime factors of 6n6^n.

Problem 3:

Given that HCF(306,657)=9HCF(306, 657) = 9, find LCM(306,657)LCM(306, 657).

Solution:

We know that HCF(a,b)×LCM(a,b)=a×bHCF(a, b) \times LCM(a, b) = a \times b. Substituting the values: 9×LCM(306,657)=306×6579 \times LCM(306, 657) = 306 \times 657 LCM(306,657)=306×6579LCM(306, 657) = \frac{306 \times 657}{9} LCM(306,657)=34×657=22338LCM(306, 657) = 34 \times 657 = 22338

Explanation:

Instead of performing prime factorization for large numbers, we use the property relating HCF, LCM, and the product of the two numbers.