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Real Numbers - 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 as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur. This can be visualized using a 'Factor Tree' where a composite number like 120120 splits into branches until every leaf is a prime number (2,2,2,3,52, 2, 2, 3, 5).

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A prime number is a natural number greater than 11 that has only two factors: 11 and itself. Visually, prime numbers like 33 or 77 cannot be arranged into a perfect rectangular grid (other than a 1×n1 \times n line), whereas composite numbers like 88 can be represented as a 2×42 \times 4 rectangle.

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Prime Factorization is the process of decomposing a composite number into its prime building blocks. This is often represented as a vertical 'ladder' where you repeatedly divide the number by the smallest possible prime until you reach 11. For example, 1818 becomes 2×322 \times 3^2.

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The Highest Common Factor (HCFHCF) of two numbers is the product of the smallest power of each common prime factor. In a Venn diagram representing the prime factors of two numbers, the HCFHCF is the product of all prime factors found in the intersection (the overlapping middle section) of the two circles.

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The Least Common Multiple (LCMLCM) of two numbers is the product of the greatest power of each prime factor involved in the numbers. In a Venn diagram, the LCMLCM is calculated by multiplying all the prime factors present in the union of both circles (every factor shown in either circle).

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The relationship between HCFHCF and LCMLCM for any two positive integers aa and bb is given by HCF(a,b)×LCM(a,b)=a×bHCF(a, b) \times LCM(a, b) = a \times b. This means the product of the two numbers is always equal to the product of their HCFHCF and LCMLCM.

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A number nxn^x ends with the digit 00 if and only if its prime factorization contains both 22 and 55. Visually, since 10=2×510 = 2 \times 5, the 'prime branches' of the number must include at least one pair of 22 and 55 to result in a trailing zero.

📐Formulae

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

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

HCF(a,b)=Product of the smallest power of each common prime factorHCF(a, b) = \text{Product of the smallest power of each common prime factor}

LCM(a,b)=Product of the greatest power of each prime factor involvedLCM(a, b) = \text{Product of the greatest power of each prime factor involved}

💡Examples

Problem 1:

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

Solution:

  1. Find prime factors of 9696: 96=2×2×2×2×2×3=25×396 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3.
  2. Find prime factors of 404404: 404=2×2×101=22×101404 = 2 \times 2 \times 101 = 2^2 \times 101.
  3. Calculate HCFHCF: The common factor is 22, and the smallest power is 222^2. So, HCF(96,404)=4HCF(96, 404) = 4.
  4. Calculate LCMLCM: The prime factors involved are 2,3,2, 3, and 101101. Taking the highest powers: LCM(96,404)=25×3×101=32×3×101=9696LCM(96, 404) = 2^5 \times 3 \times 101 = 32 \times 3 \times 101 = 9696.
  5. Verification: HCF×LCM=4×9696=38784HCF \times LCM = 4 \times 9696 = 38784. Product of numbers =96×404=38784= 96 \times 404 = 38784. Since 38784=3878438784 = 38784, the formula is verified.

Explanation:

We first break down both numbers into their prime components. The HCFHCF uses the lowest powers of shared factors, while the LCMLCM uses the highest powers of all factors present. Finally, we multiply the results to check against the product of the original numbers.

Problem 2:

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

Solution:

  1. For a number to end with the digit 00, it must be divisible by 1010.
  2. The prime factorization of 1010 is 2×52 \times 5.
  3. Therefore, the prime factorization of 6n6^n must contain both 22 and 55 as prime factors.
  4. Prime factorization of 6n=(2×3)n=2n×3n6^n = (2 \times 3)^n = 2^n \times 3^n.
  5. The only prime factors of 6n6^n are 22 and 33.
  6. Since 55 is not a prime factor in the factorization of 6n6^n, the number 6n6^n cannot end with the digit 00 for any natural number nn.

Explanation:

This approach uses the Uniqueness part of the Fundamental Theorem of Arithmetic. Since the prime factorization of 66 is strictly 2×32 \times 3, no matter what power we raise it to, the factor 55 will never appear, making it impossible to form a multiple of 1010.