Find the smallest perfect square whose decimal representation begins and ends with three 4's.
Note that only 4 3digit "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 threedigit 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 threedigit 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 20191102 14:58:55 