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Question Number 2381 by Yozzi last updated on 18/Nov/15

Of the numbers 1, 2, 3, ... , 6000,  how many are not multiples of 2, 3 or 5?

$${Of}\:{the}\:{numbers}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:...\:,\:\mathrm{6000}, \\ $$$${how}\:{many}\:{are}\:{not}\:{multiples}\:{of}\:\mathrm{2},\:\mathrm{3}\:{or}\:\mathrm{5}? \\ $$

Commented by 123456 last updated on 18/Nov/15

N=N_2 +N_3 +N_5 −N_(2,3) −N_(2,5) −N_(3,5) +N_(2,3,5)   M=X−N

$$\mathrm{N}=\mathrm{N}_{\mathrm{2}} +\mathrm{N}_{\mathrm{3}} +\mathrm{N}_{\mathrm{5}} −\mathrm{N}_{\mathrm{2},\mathrm{3}} −\mathrm{N}_{\mathrm{2},\mathrm{5}} −\mathrm{N}_{\mathrm{3},\mathrm{5}} +\mathrm{N}_{\mathrm{2},\mathrm{3},\mathrm{5}} \\ $$$$\mathrm{M}=\mathrm{X}−\mathrm{N} \\ $$

Commented by prakash jain last updated on 18/Nov/15

n(A∪B∪C)=n(A)+n(B)+n(C)                                 −n(A∩B)−n(A∩C)−n(B∩C)                               +n(A∩B∩C)

$${n}\left(\mathrm{A}\cup\mathrm{B}\cup\mathrm{C}\right)={n}\left(\mathrm{A}\right)+{n}\left(\mathrm{B}\right)+{n}\left(\mathrm{C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{n}\left(\mathrm{A}\cap\mathrm{B}\right)−{n}\left(\mathrm{A}\cap\mathrm{C}\right)−{n}\left(\mathrm{B}\cap\mathrm{C}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{n}\left(\mathrm{A}\cap\mathrm{B}\cap\mathrm{C}\right) \\ $$

Answered by 123456 last updated on 18/Nov/15

first we have that 1,2,...,6000 have  6000 numer (X=6000)  lets count the number of multiples of  2,3,5, for this, we only need to count  the number of multiply of these and  subtract tese that are multply of two  and sum the multiple of tree of these  ps:−⌈(1/a)⌉+1=0, a∈N^∗   N_2 =⌊((6000)/2)⌋=3000  N_3 =⌊((6000)/3)⌋=2000  N_5 =⌊((6000)/5)⌋=1200  N_(2,3) =⌊((6000)/(2×3))⌋=⌊((6000)/6)⌋=1000  N_(2,5) =⌊((6000)/(2×5))⌋=⌊((6000)/(10))⌋=600  N_(3,5) =⌊((6000)/(3×5))⌋=⌊((6000)/(15))⌋=400  N_(2,3,5) =⌊((6000)/(2×3×5))⌋=⌊((6000)/(30))⌋=200  N=3000+2000+1200−1000−600−400+200  N=4400  so, the number of no multiple is the  total minus the number of multiples  M=6000−4400=1600

$$\mathrm{first}\:\mathrm{we}\:\mathrm{have}\:\mathrm{that}\:\mathrm{1},\mathrm{2},...,\mathrm{6000}\:\mathrm{have} \\ $$$$\mathrm{6000}\:\mathrm{numer}\:\left(\mathrm{X}=\mathrm{6000}\right) \\ $$$$\mathrm{lets}\:\mathrm{count}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{multiples}\:\mathrm{of} \\ $$$$\mathrm{2},\mathrm{3},\mathrm{5},\:\mathrm{for}\:\mathrm{this},\:\mathrm{we}\:\mathrm{only}\:\mathrm{need}\:\mathrm{to}\:\mathrm{count} \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{multiply}\:\mathrm{of}\:\mathrm{these}\:\mathrm{and} \\ $$$$\mathrm{subtract}\:\mathrm{tese}\:\mathrm{that}\:\mathrm{are}\:\mathrm{multply}\:\mathrm{of}\:\mathrm{two} \\ $$$$\mathrm{and}\:\mathrm{sum}\:\mathrm{the}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{tree}\:\mathrm{of}\:\mathrm{these} \\ $$$$\mathrm{ps}:−\lceil\frac{\mathrm{1}}{{a}}\rceil+\mathrm{1}=\mathrm{0},\:{a}\in\mathbb{N}^{\ast} \\ $$$$\mathrm{N}_{\mathrm{2}} =\lfloor\frac{\mathrm{6000}}{\mathrm{2}}\rfloor=\mathrm{3000} \\ $$$$\mathrm{N}_{\mathrm{3}} =\lfloor\frac{\mathrm{6000}}{\mathrm{3}}\rfloor=\mathrm{2000} \\ $$$$\mathrm{N}_{\mathrm{5}} =\lfloor\frac{\mathrm{6000}}{\mathrm{5}}\rfloor=\mathrm{1200} \\ $$$$\mathrm{N}_{\mathrm{2},\mathrm{3}} =\lfloor\frac{\mathrm{6000}}{\mathrm{2}×\mathrm{3}}\rfloor=\lfloor\frac{\mathrm{6000}}{\mathrm{6}}\rfloor=\mathrm{1000} \\ $$$$\mathrm{N}_{\mathrm{2},\mathrm{5}} =\lfloor\frac{\mathrm{6000}}{\mathrm{2}×\mathrm{5}}\rfloor=\lfloor\frac{\mathrm{6000}}{\mathrm{10}}\rfloor=\mathrm{600} \\ $$$$\mathrm{N}_{\mathrm{3},\mathrm{5}} =\lfloor\frac{\mathrm{6000}}{\mathrm{3}×\mathrm{5}}\rfloor=\lfloor\frac{\mathrm{6000}}{\mathrm{15}}\rfloor=\mathrm{400} \\ $$$$\mathrm{N}_{\mathrm{2},\mathrm{3},\mathrm{5}} =\lfloor\frac{\mathrm{6000}}{\mathrm{2}×\mathrm{3}×\mathrm{5}}\rfloor=\lfloor\frac{\mathrm{6000}}{\mathrm{30}}\rfloor=\mathrm{200} \\ $$$$\mathrm{N}=\mathrm{3000}+\mathrm{2000}+\mathrm{1200}−\mathrm{1000}−\mathrm{600}−\mathrm{400}+\mathrm{200} \\ $$$$\mathrm{N}=\mathrm{4400} \\ $$$$\mathrm{so},\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{no}\:\mathrm{multiple}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{total}\:\mathrm{minus}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{multiples} \\ $$$$\mathrm{M}=\mathrm{6000}−\mathrm{4400}=\mathrm{1600} \\ $$

Commented by Yozzi last updated on 18/Nov/15

Thanks. I got that.

$${Thanks}.\:{I}\:{got}\:{that}. \\ $$

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