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2 Answers

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1 votes

Answer: $\mathbf A$

Explanation:

This can be solved using the inclusion-exclusion principle and Demorgan's Law:

$|\bar {A}| \cap \bar {|B|}\cap \bar {|C|} = 1000 - |A\cap B\cap C|= 1000 - (|A| + |B|+|C|) + (|A|\cap |
B|) + (|B|\cap |
C|) + |C|\cap |A|) - |A\cap B\cap C|\tag{1}$

$|A| = \Big\lfloor {\frac{1000}{17}}\Big \rfloor = 58 $

$|B| = \Big\lfloor {\frac{1000}{19}}\Big \rfloor = 52$

$|C| = \Big\lfloor {\frac{1000}{23}}\Big \rfloor = 43$

 
$|A\cap B| = \Big\lfloor {\frac{1000}{17.19}}\Big \rfloor = 3 $

$|B\cap C| = \Big\lfloor {\frac{1000}{19.23}}\Big \rfloor = 2$

$|C\cap A| = \Big\lfloor {\frac{1000}{23.17}}\Big \rfloor = 2$

$|A\cap B \cap C| = \Big\lfloor {\frac{1000}{17.19.23}}\Big \rfloor = 0$

On substituting these values in $\mathrm {{eq}^n} \; (1)$, we get:

Answer $=1000 -( 58+52+43)+3+2+2 = 854$

$\therefore \mathbf A$ is the correct option.

edited by
0 votes
0 votes

$854 =2\times 7\times 61$

Hence, is the correct answer.

 

Note: Numbers smaller than 1000 are easy to factorize.

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