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Numerical Ability
Trigonometry
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
Read Solution (Total 3)
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- I think this exercise is incorrect because n is much more < than 2^n. Thus, there is no n
- 5 years agoHelpfull: Yes(0) No(1)
- If n
- 5 years agoHelpfull: Yes(0) No(0)
- 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 - 5 years agoHelpfull: Yes(0) No(0)
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