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Q5:
Check whether $6^n$ can end with the digit 0 for any natural number $n$.
Solution :
Given: A number of the form $6^n$, where $n$ is a natural number ($n \in \mathbb{N}$).
To Find: Whether there exists any natural number $n$ such that $6^n$ ends with the digit $0$.
Step 1: Understanding the condition for a number to end with the digit 0.
Any positive integer that ends with the digit $0$ must be divisible by $10$. In terms of prime factorization, if a number is divisible by $10$, it must also be divisible by the prime factors of $10$. Since $10 = 2 \times 5$, any number ending in $0$ must have both $2$ and $5$ as prime factors in its prime factorization.
Step 2: Prime factorization of the base.
The given number is $6^n$. First, we find the prime factorization of the base, which is $6$.
$6 = 2 \times 3$.
Step 3: Expressing $6^n$ in terms of its prime factors.
Substituting the prime factorization of $6$ into the expression $6^n$:
$6^n = (2 \times 3)^n$
Using the exponent rule $(a \times b)^n = a^n \times b^n$, we get:
$6^n = 2^n \times 3^n$
Step 4: Analyzing the Fundamental Theorem of Arithmetic.
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.
In the prime factorization of $6^n = 2^n \times 3^n$, the only prime factors present are $2$ and $3$.
Step 5: Conclusion based on the analysis.
For $6^n$ to end with the digit $0$, its prime factorization must contain the prime factor $5$. However, we have shown that the prime factorization of $6^n$ is $2^n \times 3^n$. Since $5$ is not a prime factor of $6^n$ for any value of $n$, it is impossible for $6^n$ to be divisible by $10$.
Therefore, there is no natural number $n$ for which $6^n$ ends with the digit $0$.
Final Answer: No, $6^n$ cannot end with the digit 0 for any natural number $n$ because its prime factorization does not contain the prime factor 5.
More Questions from Class 10 Mathematics Real numbers EXERCISE 1.1
- Q1(i): Express each number as a product of its prime factors: (i) 140
- Q1(ii): Express each number as a product of its prime factors: (ii) 156
- Q1(iii): Express each number as a product of its prime factors: (iii) 3825
- Q1(iv): Express each number as a product of its prime factors: (iv) 5005
- Q1(v): Express each number as a product of its prime factors: (v) 7429
- Q2(i): Find the LCM and HCF of the following pairs of integers and verify that LCM $\times$ HCF = product of the two numbers. (i) 26 and 91
- Q2(ii): Find the LCM and HCF of the following pairs of integers and verify that LCM $\times$ HCF = product of the two numbers. (ii) 510 and 92
- Q2(iii): Find the LCM and HCF of the following pairs of integers and verify that LCM $\times$ HCF = product of the two numbers. (iii) 336 and 54
- Q3(i): Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21
- Q3(ii): Find the LCM and HCF of the following integers by applying the prime factorisation method. (ii) 17, 23 and 29
- Q3(iii): Find the LCM and HCF of the following integers by applying the prime factorisation method. (iii) 8, 9 and 25
- Q4: Given that HCF (306, 657) = 9, find LCM (306, 657).
- Q6: Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers.
- Q7: There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
CBSE Solutions for Class 10 Mathematics Real numbers
Chapters in CBSE - Class 10 Mathematics
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