Interesting problem Prove that every positive integer having 3^m equal digits is divisible by 3^m

Please provide me a solution...

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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

first of all m>or= 1, Now let there be a no. with 3^m equal digits and he digit be k ; k = {1,2,3,4,5,6,7,8,9} A no. is divisible by 3 if the sum of it's digits is divisible by 3. therefore the sum of digits of the no. is (3^m)k. on dividing by 3 we get the sum of digits must be [3^(m-1)]k, which itself is divisible by 3. and by induction we conclude that the no. is divisible by 3. However a case may arise when [3^(m-1)]k turns out to be a single digit no., while the actual no. on dividing by 3 and calculating the sum gives a different result, say n- digit. however notice a no. is divisible by 3 if the sum of it's digits is divisible by 3. However this implies if the sum of digits is a multiple of 3, thus sum of its own digits must be a multiple of 3. Thus when you take the sum of that n-digit no.'s digit and go on doing so until you get a single digit you find that the value is equal to [3^(m-1)]k.

Ekdeep SIngh Lubana - 8 years, 3 months ago

It's a tough explanation although, but try with some examples and you'll get what I'm saying.

Ekdeep SIngh Lubana - 8 years, 3 months ago

Can you provide me your gmail or any other working id....

alpha beta - 8 years, 3 months ago

@Alpha Beta – [email protected]

Ekdeep SIngh Lubana - 8 years, 3 months ago

You can show this easily by induction. Any number consisting of 3^m equal digits can be written as $k(111111\ldots111)$ where $k$ is the digit. We can factor the sting of 1s as into a string of 1s with one third the length, multiplied by a number consisting of three 1s and many 0s in the middle. (the number of 1s and 0s can be made more exact in terms of $m$ , if you want to be more rigorous). By induction, the smaller string of 1s is divisible by $3^{m-1}$ and the number with three 1s is clearly divisible by 3, so the original number is divisible by $3^m$ .

Lino Demasi - 8 years, 3 months ago

Prove that there are infinetly many primes of the form 6n-1.

Aritri Samaddar - 5 years, 2 months ago

Just a quick thought... One can say tat if there are limited primes of the form of 6n-1, then all the remaining primes after the largest 6n-1 prime will be of the form of 6n+1... But we know that not all the primes are of the form of 6n+1, this can lead to a rigorous proof via contradiction... you can try it... seems easy EDIT- I forgot to mention that all primes are of the form 6n+1 and 6n-1 only

Ekdeep SIngh Lubana - 5 years, 1 month 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}`	$\boxed{123}$