Square Number from Consecutive Digits?

Number Theory Level pending

Consider all 45-digit positive integers with one digit that is a 1, two digits that are 2s, three digits that are 3s, and so on up to nine digits that are 9s. Does there exist at least one integer of this form that is a perfect square?

Yes No

Steven Yuan by Steven Yuan , 20, USA

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1 solution

Steven Yuan
Jan 6, 2018

A perfect square must be congruent to one of 0, 1, 4, or 7 modulo 9. Since a number is congruent to its digit sum modulo 9, any number of the form described in the problem is congruent to

1 ( 1 ) + 2 ( 2 ) + 3 ( 3 ) + ⋯ + 9 ( 9 ) ≡ 9 ( 10 ) ( 19 ) 6 ≡ 15 ( 19 ) ≡ 6 ⁣ ⁣ ⁣ ⁣ ( m o d 9 ) . 1(1) + 2(2) + 3(3) + \cdots + 9(9) \equiv \dfrac{9(10)(19)}{6} \equiv 15(19) \equiv 6 \! \! \! \! \pmod{9}.

We conclude that no , no such integer can be a perfect square.

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