Question 22 of 26intermediate🔧 ApplyLong Answer3 marks

Use proof by contradiction to show that there is no value of nn for which 6n6^n ends with the digit zero.

Correct Answer

Let us assume that there exists a natural number nn for which 6n6^n ends with the digit zero.

If 6n6^n ends with the digit zero, then 6n6^n must be divisible by 10. For a number to be divisible by 10, it must have both 2 and 5 as prime factors (since 10=2×510 = 2 \times 5).

Now,

6n=(2×3)n=2n×3n6^n = (2 \times 3)^n = 2^n \times 3^n

The prime factorization of 6n6^n contains only the pr \times 2 and 3. It does not contain 5 as a prime factor. This is a contradiction, because we assumed 6n6^n ends with zero, but 6n6^n lacks the prime factor 5 which is necessary for divisibility by 10. The contradiction has arisen because we assumed that there exists a value of nn for which 6n6^n ends with the digit zero. Hence, there is no value of nn for which 6n6^n ends with the digit zero.

Exercise: EXERCISE A1.6 | Q: 5 | (Chapter: 21)
For More Understanding

Explanation

The question requires proof by contradiction. The context shows the structure of proof by contradiction: assume the negation, derive a logical deduction, reach a contradiction, and conclude. Following this pattern, we assume 6n6^n ends with zero, which means divisible by 10, requiring factor 5. But 6n=2n×3n6^n = 2^n \times 3^n has no factor 5, creating the contradiction.

Solution Steps

  1. Step 1: Assume there exists nn such that 6n6^n ends with digit zero

  2. Step 2: If 6n6^n ends with zero, it must be divisible by 10

  3. Step 3: For divisibility by 10, both 2 and 5 must be prime factors

  4. Step 4: Show 6n=2n×3n6^n = 2^n \times 3^n has no factor 5

  5. Step 5: State the contradiction and conclude

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