Total number of integers ' n ' such that 2 ≤ n ≤ 2000 and H.C.F of ' n ' and 36 is one, is equal to

36 − 2 2 × 3 2 From 2 to 2000 number of multiples of 2 are [ 2 2000 ] = 1000 From 2 to 2000 number of multiples of 3 are [ 3 2 c 00 ] − 66 From 2 to 2000 number of multiples of 6 are [ 5 2 C 00 ] = 333 ∴ Number of possible ′ n 'are = 1999 − [ 1000 + 666 − 333 ]

Mathematics - Permutation & Combination Question with Solution | TestHub

MathematicsPermutation & CombinationFundamental PrincipleMedium2 minQB

MathematicsMediumsingle choice

Total number of integers ' $n$ ' such that $2 \leq n \leq 2000$ and H.C.F of ' $n$ ' and 36 is one, is equal to

Options:

Answer:

Solution:

$36 - 2^{2} \times 3^{2}$

From 2 to 2000 number of multiples of 2 are $[\frac{2000}{2}] = 1000$

From 2 to 2000 number of multiples of 3 are $[\frac{2 c 00}{3}] - 66$

From 2 to 2000 number of multiples of 6 are $[\frac{2 C 00}{5}] = 333$

∴ Number of possible $^{'} n$ 'are $= 1999 - [1000 + 666 - 333]$

Stream:JEESubject:MathematicsTopic:Permutation & CombinationSubtopic:Fundamental Principle

⏱ 2mℹ️ Source: QB

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