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Alexus [3.1K]
3 years ago
13

The product of two consecutive whole numbers is less than the sum of the square of the smaller number and 13

Mathematics
1 answer:
Norma-Jean [14]3 years ago
8 0

Answer: a<13

Step-by-step explanation:

a*(a+1)<a^2+13

a^2+a<a^2+13

subtract a^2 from both sides:

a<13

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Use the principle of inclusion and exclusion to find the number of positive integers less than 1,000,000 that are not divisable
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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.
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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) = \frac{999,999}{6} = 166666
N(c₂) = \frac{999,999}{4} = 250000
∴N(c₁c₂) = \frac{999,999}{24} = 41667
∴ 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
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