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# Basic Maths

Show that the number 111...11 with 3^n digits is divisible by 3^n.

Note by Biswaroop Roy
3 years, 8 months ago

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We prove this by induction:

Base case:

When n = 0, it is obviously true that 1 is divisible by 1.

When n = 1:

The number is 111 which is divisible by 3.

Inductive Case: Suppose the number 1111.....111 (which has 3^n digits) is divisible by 3^n.

For 3^(n+1) digits of 1, The sum of digits for this case is always a power of 3. Hence, the assertion is true.

Numerical Proof: 3^(n+1) divides 111....111 (which has 3^(n+1) digits)

(3^n)(3) divides 111....111 (which has (3^n)(3) = 3^n + 3^n + 3^n digits)

Remembering the case for n: the number 1111.....111 (which has 3^n digits) is divisible by 3^n and the base case, the proof is done. · 3 years, 8 months ago

Just to clarify your induction, as it is a bit unclear, $$111\cdots 1(3^{n+1}$$ digits $$)=111\cdots 1(3^n$$ digits $$)\times (10^{2n}+10^n+1)$$. Since $$(10^{2n}+10^n+1)$$ is divisible by 3, by the inductive hypothesis we finish the induction. · 3 years, 8 months ago

Nicely expressed. This also shows that $$3^n$$ is the largest power of 3 that divides $$111\ldots 1$$.

As a side note, another way of stating the question is that $$3^{n+2} \mid 10^{3^n} - 1$$. Staff · 3 years, 8 months ago

Yeah that's true. Or even better as you mentioned, $$v_3(10^{3^n}-1)=n+2$$ or $$3^{n+2}||10^{3^n}-1$$. · 3 years, 8 months ago

Thanks for clarifying! I also didn't notice that... · 3 years, 8 months ago

I just solved this same problem using algebra.Would you like to try? · 3 years, 8 months ago

Sure thing.. · 3 years, 8 months ago

Just a hint,you might find geometric progression an useful tool:) · 3 years, 8 months ago