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

Note by Alpha Beta
5 years, 3 months ago

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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 $$3^m$$.

- 5 years, 3 months ago

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.

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

ekdeeplubana@gmail.com

- 5 years, 3 months ago

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

- 2 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

- 2 years, 1 month ago