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HomeClass 10mathematicsChapter 1Exercise 1.1Q5
| Exercise 1.1

Question 5

Check whether 6n6^n can end with the digit 0 for any natural number nn.

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SOLUTION

UNDERSTAND THE QUESTION

The problem asks whether any power of 6 can have 0 as its last decimal digit. An integer ends with 0 exactly when it is divisible by 10 (=2·5). Thus we must examine the prime factors of 6ⁿ and see if a factor 5 ever appears.

STEP 1 — FACTORISATION OF 6ⁿ

6n=(2⋅3)n=2n 3n6^{n}= (2\cdot3)^{n}=2^{n}\,3^{n}

STEP 2 — DIVISIBILITY CONDITION FOR A TRAILING 0

A decimal integer ends with the digit 00 iff it is a multiple of 10=2⋅510=2\cdot5. Hence 6n6^{n} would need at least one factor 55 in its prime decomposition.

STEP 3 — SEARCH FOR A FACTOR 5

The prime factorisation of 6n6^{n} is 2n3n2^{n}3^{n}, containing only the primes 22 and 33. No factor 55 occurs for any natural nn, so 6n6^{n} is never divisible by 55.

STEP 4 — CONCLUSION

Since 6n6^{n} lacks a factor 55, it cannot be a multiple of 1010 and therefore can never end with the digit 00.

ANSWER

No, 6n6^{n} never ends with the digit 00 for any natural number nn.

COMMON MISTAKES

Assuming that because 66 is even it automatically provides the factor 55 needed for a trailing zero; overlooking that the prime factor 55 must appear explicitly in the factorisation.

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