Question 5 of 10intermediate🔍 AnalyzeLong Answer3 marks

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

Correct Answer

No, 6n6^n cannot end with the digit 00 for any natural number nn.

Reason:

For a number to end with the digit 00, it must be divisible by 1010.

Since 10=2×510 = 2 \times 5, the number must have both 22 and 55 as prime factors.

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

The prime factorisation of 6n6^n contains only the primes 22 and 33.

There is no prime factor 55 in the factorisation of 6n6^n.

By the Fundamental Theorem of Arithmetic, the prime factorisation of every composite number is unique.

Since 55 is not a factor of 6n6^n, the number 6n6^n cannot be divisible by 1010.

∴ 6n6^n cannot end with the digit 00 for any natural number nn.

Exercise: Exercise 1.1 | Q: 5 | (Page Page 5)
For More Understanding

Explanation

The chapter explains that for a number to end with digit 00, it must be divisible by both 22 and 55 (since 10=2×510 = 2 \times 5). The Fundamental Theorem of Arithmetic guarantees unique prime factorisation.

Since 6n=2n×3n6^n = 2^n \times 3^n contains only primes 22 and 33, and no prime factor 55, it cannot end with digit 00.

This is similar to Example 1 in the chapter which shows that 4n4^n cannot end with digit 00.

Solution Steps

  1. Step 1: A number ending with digit 00 must be divisible by 10=2×510 = 2 \times 5

  2. Step 2: Express 6n=(2×3)n=2n×3n6^n = (2 \times 3)^n = 2^n \times 3^n

  3. Step 3: Observe that prime factorisation of 6n6^n contains only primes 22 and 33

  4. Step 4: Since prime factor 55 is absent, 6n6^n is not divisible by 1010

  5. Step 5: Conclude that 6n6^n cannot end with digit 00

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