Appendix 1: Proofs in Mathematics | A1.6

Question 5

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

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Solution
Understand the Question
  • In a proof by contradiction, we assume the opposite of what we want to prove and demonstrate that this assumption leads to an impossible statement.
  • For a number to end in the digit 00, it must be divisible by 10=2×510 = 2 \times 5, which requires both 22 and 55 in its prime factorisation.
  • By the Fundamental Theorem of Arithmetic, prime factorisation is unique. We inspect the prime factors of 6n=(2×3)n6^n = (2 \times 3)^n to verify whether 55 can ever appear as a factor.

Step 1 · Assume the Opposite

Assume that for some natural number nn, 6n6^n ends with the digit 00.

If a number ends with 00, it must be divisible by 1010. Therefore, for some integer kk: 6n=10k6^n = 10kDiagram 1

Step 2 · Find Prime Factorisations

Prime factorise 6n6^n:

6=2×36n=2n×3n\begin{aligned} 6 &= 2 \times 3 \\ 6^n &= 2^n \times 3^n \end{aligned}

Prime factorise 10k10k:

10=2×510k=2×5×k\begin{aligned} 10 &= 2 \times 5 \\ 10k &= 2 \times 5 \times k \end{aligned}

Step 3 · Arrive at a Contradiction

Equating both sides from the initial assumption: 2n×3n=2×5×k2^n \times 3^n = 2 \times 5 \times k

  • The right-hand side is divisible by 55.
  • By the Fundamental Theorem of Arithmetic, the prime factorisation of 6n6^n is unique and contains only the primes 22 and 33.
  • Since 55 is not a prime factor of 6n6^n, 6n6^n cannot be divisible by 55 or 1010.

This creates a contradiction. Hence, the initial assumption is false.

Answer

There is no value of nn for which 6n6^n ends with the digit zero.

Common Mistakes
  • Empirical Testing vs. General Proof: Checking a few values (e.g., 61=6,62=36,63=2166^1 = 6, 6^2 = 36, 6^3 = 216) only illustrates the pattern—it does not constitute a valid proof for all natural numbers nn.
  • Omitting Uniqueness of Primes: Failing to cite the Fundamental Theorem of Arithmetic leaves the proof incomplete, as uniqueness guarantees no other prime factors (like 55) can exist.

More questions in A1.6

Q1

Suppose a+b=c+da + b = c + d, and a<ca < c. Use proof by contradiction to show b>db > d.

Q2

Let rr be a rational number and xx be an irrational number. Use proof by contradiction to show that r+xr + x is an irrational number.

Q3

Use proof by contradiction to prove that if for an integer aa, a2a^2 is even, then so is aa.

[Hint : Assume aa is not even, that is, it is of the form 2n+12n + 1, for some integer nn, and then proceed.]

Q4

Use proof by contradiction to prove that if for an integer aa, a2a^2 is divisible by 3, then aa is divisible by 3.

Q5

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

Q6

Prove by contradiction that two distinct lines in a plane cannot intersect in more than one point.

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