Find the smallest perfect square whose decimal representation begins and ends with three 4's.
(In reply to
solution by Daniel)
taking modular square root we get n=538 mod 1000
x^2=y*10^3+444, {x = 500n + 38,x = 500n + 462} with possible values ...038, ...462, ...538, ...962.
This suggests a somewhat easier mode of attack, because we can also say something about the first 3 digits of n:
n^2=444, n=2 sqrt(111) or nā21.071
n^2=4440, n = 2 sqrt(1110) or nā66.633
and these are the only two variants.
If n had less than 6 digits, the first 3 above would overlap with the known and different value of the last 3. So the first 3 digits of n are 210 or 666.
21038^2 = 442597444
210038^2 = 44115961444
210462^2 = 44294253444
210538^2 = 44326249444
210962^2 = 44504965444
666038^2 = 443606617444
not enough 4s at the start.
Therefore, 666462^2 = 444171597444 must be the smallest possible solution.
Edited on August 3, 2019, 12:16 am
|
|
Posted by broll
on 2019-08-02 23:11:12 |
re: solution
