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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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re: solution | Comment 4 of 5 |
(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 nā‰ˆ21.071
n^2=4440, n = 2 sqrt(1110) or nā‰ˆ66.633
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

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