Solve 81^x = 27^x+2 X=1 X=2 X=5 X=6
2 answers:
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Answer:
= - 3n
Step-by-step explanation:
There is a common difference between consecutive terms, that is
- 6 - (- 3) = - 9 - (- 6) = - 12 - (- 9) = - 3
This indicate the sequence is arithmetic with n th term
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Here a₁ = - 3 and d = - 3 , thus
= - 3 - 3(n - 1) = - 3 - 3n + 3 = - 3n
The n th term is - 3n
This is a simple problem based on combinatorics which can be easily tackled by using inclusion-exclusion principle.
We are asked to find number of positive integers less than 1,000,000 that are not divisible by 6 or 4.
let n be the number of positive integers.
∴ 1≤n≤999,999
Let c₁ be the set of numbers divisible by 6 and c₂ be the set of numbers divisible by 4.
Let N(c₁) be the number of elements in set c₁ and N(c₂) be the number of elements in set c₂.
∴N(c₁) =
N(c₂) =
∴N(c₁c₂) =
∴ Number of positive integers that are not divisible by 4 or 6,
N(c₁`c₂`) = 999,999 - (166666+250000) + 41667 = 625000
Therefore, 625000 integers are not divisible by 6 or 4
We are asked to find number of positive integers less than 1,000,000 that are not divisible by 6 or 4.
let n be the number of positive integers.
∴ 1≤n≤999,999
Let c₁ be the set of numbers divisible by 6 and c₂ be the set of numbers divisible by 4.
Let N(c₁) be the number of elements in set c₁ and N(c₂) be the number of elements in set c₂.
∴N(c₁) =
N(c₂) =
∴N(c₁c₂) =
∴ Number of positive integers that are not divisible by 4 or 6,
N(c₁`c₂`) = 999,999 - (166666+250000) + 41667 = 625000
Therefore, 625000 integers are not divisible by 6 or 4