Concealed Square

Find the unique positive integer whose square has the form 1_2_3_4_5_6_7_8_9_0, where each “_” is a single digit.

This problem is challenging because it limits the number of possibilities to a certain pattern. The only instances where a squared integer ends in 0 is when the integer ends with 0. Let’s establish the following:

1) The square root of an integer ending in 0 must also end in 0.
2) Thus, the last “0” in the digit and its square are satisfied by the last 0 in the integer and its square respectively.
3) The second last digit of the square is formed by the square of the second last digit of the integer itself.

This implies that the integer we’re looking for must also end with a zero, meaning the last two characters of the unique integer are 00 (because, if the square ends in a “0”, then the integer also ends in a “0”).

So, let’s format the number to only have the variables we will be calculating for in the following manner 1_2_3_4_5_6_7_8_900.

Moving on, we recognize everything from 3^2 to 9^2 ends with a digit other than 0, 1, and 4. As such, we know that _8 must be 00, _6 must be 40, _4 must be 60, and _2 must be 80. This completely determines _4,_6, and _8, which leaves _0 to fill using: 0, 1, 4, and 9.

Now, the constructed integer would look like this 1_200_300_400_500_600_700_800_900

At this point, given that the square root of a five-digit number is a three-digit number, and the square root of a six-digit number is a four-digit number, we know that the _1 and _3 might be a two-digit number and _5 and _7 might be a three-digit number. This simply implies that _1 and _3 can only take 0, while _5 and _7 can take both 0 and 1. To check for these, we only need to check the last digit of each square which gives the following conditions:

1) _7 must end with 0 because the square of the last digit must end with 1.
2) _5 must end with 0 because the square of the last digit must end with 4.
3) _3 can end with 0 or 1 because the square of the last digit can end with 0 or 9.
4) _1 can end with 0 because the square of the last digit can end with 1.

In checking the roots manually, we find that the root solution for 1_200_300_400_500_600_700_800_900 is 1929394959697989900. Hence, the solution for the original problem, 1_2_3_4_5_6_7_8_9_0, is represented by the last two digits removed from the root solution, which is 19293949596979899.

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