Basic Maths

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

#NumberTheory #MathProblem #Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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.

John Ashley Capellan - 7 years, 6 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.

Yong See Foo - 7 years, 6 months ago

Thank you very much,both of you. I should have solved myself. Thanks again:)

Biswaroop Roy - 7 years, 6 months ago

Thanks for clarifying! I also didn't notice that...

John Ashley Capellan - 7 years, 6 months ago

@John Ashley Capellan – I just solved this same problem using algebra.Would you like to try?

Biswaroop Roy - 7 years, 6 months ago

@Biswaroop Roy – Sure thing..

John Ashley Capellan - 7 years, 6 months ago

@John Ashley Capellan – Just a hint,you might find geometric progression an useful tool:)

Biswaroop Roy - 7 years, 6 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 $3^{n+2} \mid 10^{3^n} - 1$ .

Calvin Lin Staff - 7 years, 6 months ago

@Calvin Lin – Yeah that's true. Or even better as you mentioned, $v_3(10^{3^n}-1)=n+2$ or $3^{n+2}||10^{3^n}-1$ .

Yong See Foo - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$