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'Two' many roots (Posted on 2019-03-14) Difficulty: 3 of 5
Find the largest positive integer k possible in the following expression, n >= 0

((√2)1+(√2)2)*((√2)3+(√2)4)*...*((√2)2019+(√2)2020) = 2k + n

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Solution Comment 2 of 2 |
Each of the 1010 terms of the product can be expressed as sqrt(2)^(2x-1) + sqrt(2)^(2x) for x=1 to 1010.

The expression can be simplified to 2^(x-1) * (sqrt(2) + 2).

Then for all 1010 terms the product is (2^0 * 2^1 * ... * 2^1009) * (sqrt(2) + 2)^1010 = 2^509545 * sqrt(2)^1010 * (1+sqrt(2))^1010

Some logs yields 1+sqrt(2) = 2^1.27155.  Then the product simplifies to 2^(509545 + 505 + 1284.26884) = 2^511334.26884.  Then the value k sought is 511334

  Posted by Brian Smith on 2019-03-15 01:52:46
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