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

Note by Alpha Beta
4 years 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 $$3^m$$. · 4 years 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. · 4 years ago

It's a tough explanation although, but try with some examples and you'll get what I'm saying. · 4 years ago

Can you provide me your gmail or any other working id.... · 4 years ago

ekdeeplubana@gmail.com · 4 years ago

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