Large square number

Find the largest possible 3-digit positive integer N N such that N 2 N^2 is of the form A B C , A B D ABC, ABD , where A A , B B , C C , and D D are (not necessarily distinct) digits and D = C + 1 D = C+1 .

(An example of a number of this form is 123,124 or 998,999.)


The answer is 846.

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1 solution

Patrick Corn
Nov 24, 2017

Let M M be the three-digit number A B C . ABC. Then we want N 2 = 1000 M + M + 1 = 1001 M + 1. N^2 = 1000M + M + 1 = 1001M + 1. So N 2 1 N^2 \equiv 1 mod 1001. 1001. Since 1001 = 7 11 13 , 1001 = 7\cdot 11\cdot 13, this has eight solutions between 1 1 and 1000 1000 by the Chinese remainder theorem , by picking various signs in the congruences N ± 1 ( m o d 7 ) N ± 1 ( m o d 11 ) N ± 1 ( m o d 13 ) \begin{aligned} N &\equiv \pm 1 \pmod{7} \\ N &\equiv \pm 1 \pmod{11} \\ N &\equiv \pm 1 \pmod{13} \end{aligned} The eight solutions are N = 1 , 1000 ; 428 , 573 ; 274 , 727 ; 155 , 846. N = 1,1000; 428, 573; 274, 727; 155, 846. So the largest three-digit one is 846 . \fbox{846}.

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