Find the required smallest positive odd integer n?

Algebra Level 3

Question Question

11 9 17 7

Navneet Kumar by Navneet Kumar , 18, India

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2 solutions

2^10 = 1024 > 1000 so the exponents should add up to almost 10.

n=7: (1+3+5+7+9+11+13+15)/7=64/7 = 9 1/7 is much less than 10

n=8: (1+3+5+7+9+11+13+15+17)/7= 81/7 = 11 4/7 > 10, however, n must be odd, so n must be at least 9

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Yvonne Killian
Nov 13, 2018

Two solutions.

Solution 1:

  • x = 2^(1/7) * 2^(3/7) * 2^(5/7) * ... * 2^((2n+1)/7) = 2^(1/7+3/7+...+(2n+1)/7) = 2^(((n+1)^2)/7)
  • Find the smallest odd value for n for which x > 1000
  • As 2^10 = 1024, ((n+1)^2)/7 is probably close to 10, so n is probably 7 or 9
  • For n = 7, x = 2^(((7+1)^2)/7) = 2^(64/7) = (2^9) * (2^(1/7)) = 512 * 1.1... < 1000
  • For n = 9, x = 2^(((9+1)^2)/7) = 2^(100/7) = (2^14) * (2^(2/7) ) = 16384 * 1.5... > 1000
  • So the answer is 9

Solution 2:

  • y = 1/7+3/7+5/7+...+(2n+1)/7 = ((n+1)^2)/7
  • Find the smallest odd value for n for which 2^y > 1000
  • For 2^y = 1000, y = ln 1000 / ln 2 = 9.965...
  • As 2^10 = 1024, ((n+1)^2)/7 is probably close to 10, so n is probably 7 or 9
  • For n = 7, y = ((7+1)^2)/7 = 64/7 = 9.1...< 9.965... so 2^y < 1000
  • For n = 9, y = ((9+1)^2)/7 = 100/7 =14.3... > 9.965... so 2^y > 1000
  • So the answer is 9
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