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Rooting for 2015 (Posted on 2015-08-14) Difficulty: 3 of 5
Determine the smallest value of a positive integer N such that the digit immediately preceding the decimal point in the expansion of √N is 2 and the three digits immediately following the decimal point is 015 (in this order)

*** Extra Challenge: A non-computer program assisted method.

No Solution Yet Submitted by K Sengupta    
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Solution Likely Solution | Comment 1 of 3
√n = __2.015__
√n = 10x + 2.015 (±small amount) where x is an integer
n ≈ 100x²+40.3x+4.060225
by inspection if x=3 we will be close to an integer
100*3²+40.3*3+4.060225=1024.960225
try n=1025
√1025=32.01562119__



  Posted by Jer on 2015-08-14 09:35:22
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