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444...444 (Posted on 2019-08-02) Difficulty: 3 of 5
Find the smallest perfect square whose decimal representation begins and ends with three 4's.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution; spreadsheet assist Comment 5 of 5 |
Note that only 4 3-digit "winning" numbers have squares that end in 444:
038^2 = 1,444
538^2 = 289,444
462^2 = 213,444
962^2 = 925,444

So, a low effort brute force attack is to take 444...444 starting with 0 intervening digits, then one intervening digit, then two etc, both for all 0's and for all 9's; take the square roots, and by inspection see if there is a number between those 2 square roots that matches one of our 4 three-digit winners above.

Example:
sqrt of 444000444 is 21,071.3180 etc
sqrt of 444999444 is 21,095.0099 etc
of the 23 numbers between 21,071 and 21,095, none are "winners"

But when there are 6 intervening numbers (444000000444 and 444999999444) we get two square roots that are
666,333.25 etc and 667,083.20 etc

The smallest three-digit winner (which is larger than 333) is 462.
So our square root is 666,462.
Squaring this gives 444,171,597,444

  Posted by Larry on 2019-11-02 14:58:55
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