In this question x^y stands for x raised to the power y .for example 2^3=8 and 4^1.5=8

Find the number of positive integers n>2000 which can be expressed as n=2^m+2^n where m and n are integers (for example, 33=2^0+2^5)

Answer: 65

• I think this exercise is incorrect because n is much more < than 2^n. Thus, there is no n
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• If n
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• In this question x^y stands for x raised to the power y .for example 2^3=8 and 4^1.5=8 Find the number of positive integers n < 2000 which can be expressed as n=2^m+2^n where m and n are integers
  
  Number of positive integers < 2000
  
  2^m & 2ⁿ can be
  
  2⁰ = 1
  
  2¹ = 2
  
  2² = 4
  
  2³ = 8
  
  2⁴ = 16
  
  2⁵ = 32
  
  2⁶ = 64
  
  2⁷ = 128
  
  2⁸ = 256
  
  2⁹ = 512
  
  2¹⁰ = 1024
  
  with 2⁰ = 1  all 11 numbers can be added
  
  with 2¹ = 2   10 numbers can be added except 2⁰ = 1 which is already considered
  
  Similarly
  
  till 2⁹ = 512   , 2 numbers can be added
  
  for 2¹⁰ = 1024 , no numbers can be added as it will make number > 2000
  
  11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 +
  
  Sum = (10/2)(11 + 2)
  
  = 5 * 13
  
  = 65
• 4 years agoHelpfull: Yes(0) No(0)