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Point to 2012 (Posted on 2012-04-04) Difficulty: 3 of 5
Determine the minimum value of a positive integer N such that the four digits immediately following the decimal point in the base ten expansion of √N is 2012.

*** For an extra challenge, solve this puzzle without using a computer program.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Non computer solution (pretty much) | Comment 3 of 7 |
We seek N such that for some integer a
(a+.2012)^2 < N < (a+.2013)^2

For simplicity lets look for N ≈ (a+.20125)^2
N ≈ (a + 161/800)^2 = a^2 + 161a/400 + 25921/600000
Since a^2 is already an integer we need
161a/400 + 25921/640000 ≈ b for some integer b
Solving for a gives
a ≈ 400b/161 - 161/1600

So what is sought is a value of b that makes a close to an integer. 
A quick chart shows
b = 23, a=57.042 which is close to 57
57.20125^2 = 3271.983
√3272 = 57.201398 close but not quite

b = 25, a=62.011 which is even closer to 62
62.20125^2 = 3868.99955
√3869 = 62.20128616 bingo!

The solution is N = 3869

  Posted by Jer on 2012-04-04 13:38:30
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