NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 5
NCERT Class X Chapter 1: Real Numbers Exercise 1.1 Question 5
Question:
Check whether \( 6^n \) can end with the digit 0 for any natural number \( n \).
Given:
The expression \( 6^n \), where \( n \) is a natural number.
To Find:
Whether \( 6^n \) can end with the digit 0 for any natural number \( n \).
Formula:
To end with 0, a number must be divisible by 10.
\( 10 = 2 \times 5 \)
Solution:
Step 1: For a number to end with 0, it must be divisible by 10.
The prime factorization of 10 is \( 10 = 2 \times 5 \).
Step 2: Write 6 as a product of its prime factors.
$$ 6 = 2 \times 3 $$Step 3: Raise both sides to the power \( n \).
$$ 6^n = (2 \times 3)^n $$Step 4: Apply the exponent rule: \( (a \times b)^n = a^n \times b^n \).
$$ 6^n = 2^n \times 3^n $$Step 5: Observe the prime factors of \( 6^n \).
The only prime factors are 2 and 3 (each raised to the power \( n \)).
There is no factor of 5 in \( 6^n \) for any natural number \( n \).
Step 6: Since \( 6^n \) does not contain a factor of 5, it cannot be divisible by 10.
Therefore, \( 6^n \) cannot end with the digit 0 for any natural number \( n \).
Result:
No, \( 6^n \) cannot end with the digit 0 for any natural number \( n \).
Reason: Its prime factorization is \( 2^n \times 3^n \), which never includes a factor of 5, so it cannot be divisible by 10.
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