2 votes 2 votes The number of positive integers which are less than or equal to $1000$ and are divisible by none of $17$, $19$ and $23$ equals $854$ $153$ $160$ none of the above Quantitative Aptitude isi2015-mma quantitative-aptitude number-system remainder-theorem + – Arjun asked Sep 23, 2019 retagged Nov 16, 2019 by Lakshman Bhaiya Arjun 1.1k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 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. `JEET answered Sep 26, 2019 edited Nov 10, 2019 by `JEET `JEET comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes $854 =2\times 7\times 61$ Hence, A is the correct answer. Note: Numbers smaller than 1000 are easy to factorize. NastyBall answered Jun 17, 2021 NastyBall comment Share Follow See all 0 reply Please log in or register to add a comment.