Number Theory

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korean

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Let N be a number whose decimal expansion consists of 3^n identical digits.

Show by INDUCTION that 3^n | N. (3^n divides N)

so i get what this is saying and it makes since to me...in that if you have 3^1 it should divided any identical (3) digits and so on and so forth depending on what power you raise it to...for example 3^1 | 222 or 3^1 | 333 etc. and 3^2 should divide any identical (9) digits.

but i have no idea on how to prove it beside getting the first part of the induction. which i got as....

n=1....then 3^1 | (any three identical digits)....but i have no idea on how to write it... can someone help me?
any help is appreciated.
 
In the textbook, look up the divisibility rules.
If the sum of the digits of a number is divisible by 3 then the number is also.
See if that applies to powers of three.
 
Let N be a number whose decimal expansion consists of 3^n identical digits.

Show by INDUCTION that 3^n | N. (3^n divides N)

so i get what this is saying and it makes since to me...in that if you have 3^1 it should divided any identical (3) digits and so on and so forth depending on what power you raise it to...for example 3^1 | 222 or 3^1 | 333 etc. and 3^2 should divide any identical (9) digits.

but i have no idea on how to prove it beside getting the first part of the induction. which i got as....

n=1....then 3^1 | (any three identical digits)....but i have no idea on how to write it... can someone help me?
any help is appreciated.korean
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The induction is on N, so the base case is n=1. If you have 3^1 identical digits, then the number has the form DDD, where D is your digit. But

DDD really = 100D + 10D + D = D(111) = D(3)(37), so 3 | DDD.

Now assume that any number of the form N1 = DDD(k times) is div by 3.
To prove that N2 = DDD(k+1 times) is div by 3.
N2 = DDD followed by DDD(k times)

N2 = DDD(10^(3k) + N1

N2 = (100D + 10D + D)(10^(3k) + N1

Now the first term of that is (111D)(10^(3k) which is easily shown divisible by 3, and the second term is N1, which is div by 3, BY ASSUMPtION.

The rest is easy, I think.
 
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