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I recently saw a question Is there any perfect square in the sequence $12,123,1234,12345,...$?

This led to thinking about a new question.


Consider this sequence that is the sequence formed by concatenating next digit at right-hand end (where the next digit after 9 is again 0).

ie $1, 12, 123, 1234, 12345, 123456, 1234567, … 123456789012 …$

Is there any perfect square in the sequence?

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NO


Computational check for numbers mod powers of 10.

  1. Mod 10: 𝑛∈{0,1,4,5,6,9} perfect squares can end only with these digits
  2. Mod 100: 𝑛∈{01, 56, 89} They are the only pairs with consecutive numbers
  3. Mod 1000: 𝑛∈{456} Using computation, 901 and 789 are eliminated. No perfect square can end with 789 or 901.
  4. Mod 10000: 𝑛∈{3456}, Mod 100000: 𝑛∈{23456} and so on continues

Let us assume that the number is (1234567890)..k..123456

Test of divisibility by 3: 1+2+3+4+5+6+7+8+9+0 = 45 1+2+3+4+5+6 = 21 The sum of all digits of a number is 45*k+21 which is divisible by 3

Test of divisibility by 9: 45*k+21 can never be divisible by 9

A perfect square divisible by 3 but not divisible by 9 cannot exist.

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