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 \(3^m\).

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.

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

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TopNewestYou 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 \(3^m\).

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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.

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

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

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

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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

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