How-many-natural-number-from-1-to-900-which-are-not-divisible-2-3-and-5-

Question Number 155856 by peter frank last updated on 05/Oct/21

$$\mathrm{How}\:\mathrm{many}\:\mathrm{natural}\:\mathrm{number}\:\mathrm{from} \\ $$$$\mathrm{1}\:\mathrm{to}\:\mathrm{900}\:\mathrm{which}\:\mathrm{are}\:\mathrm{not}\:\mathrm{divisible} \\ $$$$\mathrm{2},\mathrm{3}\:\mathrm{and}\:\mathrm{5} \\ $$

Answered by TheSupreme last updated on 05/Oct/21

2,3 and m.c.m(2,3 and 5)=30 D_(30=) ⌊900/30⌋=30 2,3 or 5 D_2 =⌊900/2⌋=450 D_3 =⌊900/3⌋=300 D_5 =⌊900/5⌋=180 D_6 =⌊900/6⌋=150 D_(10) =...=90 D_(15) =80 D_(30) =30 D_(2,3 or 5) =450+300+180−150−90−80+30= =640

$$\mathrm{2},\mathrm{3}\:{and} \\ $$$${m}.{c}.{m}\left(\mathrm{2},\mathrm{3}\:{and}\:\mathrm{5}\right)=\mathrm{30} \\ $$$${D}_{\mathrm{30}=} \lfloor\mathrm{900}/\mathrm{30}\rfloor=\mathrm{30} \\ $$$$ \\ $$$$\mathrm{2},\mathrm{3}\:{or}\:\mathrm{5} \\ $$$${D}_{\mathrm{2}} =\lfloor\mathrm{900}/\mathrm{2}\rfloor=\mathrm{450} \\ $$$${D}_{\mathrm{3}} =\lfloor\mathrm{900}/\mathrm{3}\rfloor=\mathrm{300} \\ $$$${D}_{\mathrm{5}} =\lfloor\mathrm{900}/\mathrm{5}\rfloor=\mathrm{180} \\ $$$${D}_{\mathrm{6}} =\lfloor\mathrm{900}/\mathrm{6}\rfloor=\mathrm{150} \\ $$$${D}_{\mathrm{10}} =…=\mathrm{90} \\ $$$${D}_{\mathrm{15}} =\mathrm{80} \\ $$$${D}_{\mathrm{30}} =\mathrm{30} \\ $$$${D}_{\mathrm{2},\mathrm{3}\:{or}\:\mathrm{5}} =\mathrm{450}+\mathrm{300}+\mathrm{180}−\mathrm{150}−\mathrm{90}−\mathrm{80}+\mathrm{30}= \\ $$$$=\mathrm{640} \\ $$

Commented by puissant last updated on 05/Oct/21

$${Sir}\:{you}\:{use}\:{which}\:{formula}.? \\ $$

Answered by mr W last updated on 05/Oct/21

numbers divisible by 30: ⌊((900)/(30))⌋=30 numbers divisible by 6: ⌊((900)/6)⌋−30=120 numbers divisible by 10: ⌊((900)/(10))⌋−30=60 numbers divisible by 15: ⌊((900)/(15))⌋−30=30 numbers divisible by 2: ⌊((900)/2)⌋−30−120−60=240 numbers divisible by 3: ⌊((900)/3)⌋−30−120−30=120 numbers divisible by 5: ⌊((900)/5)⌋−30−60−30=60 total numbers divisible by 2,3,5: 30+120+60+30+240+120+60=660 numbers not divisible by 2,3,5: 900−660=240

$${numbers}\:{divisible}\:{by}\:\mathrm{30}: \\ $$$$\lfloor\frac{\mathrm{900}}{\mathrm{30}}\rfloor=\mathrm{30} \\ $$$${numbers}\:{divisible}\:{by}\:\mathrm{6}: \\ $$$$\lfloor\frac{\mathrm{900}}{\mathrm{6}}\rfloor−\mathrm{30}=\mathrm{120} \\ $$$${numbers}\:{divisible}\:{by}\:\mathrm{10}: \\ $$$$\lfloor\frac{\mathrm{900}}{\mathrm{10}}\rfloor−\mathrm{30}=\mathrm{60} \\ $$$${numbers}\:{divisible}\:{by}\:\mathrm{15}: \\ $$$$\lfloor\frac{\mathrm{900}}{\mathrm{15}}\rfloor−\mathrm{30}=\mathrm{30} \\ $$$${numbers}\:{divisible}\:{by}\:\mathrm{2}: \\ $$$$\lfloor\frac{\mathrm{900}}{\mathrm{2}}\rfloor−\mathrm{30}−\mathrm{120}−\mathrm{60}=\mathrm{240} \\ $$$${numbers}\:{divisible}\:{by}\:\mathrm{3}: \\ $$$$\lfloor\frac{\mathrm{900}}{\mathrm{3}}\rfloor−\mathrm{30}−\mathrm{120}−\mathrm{30}=\mathrm{120} \\ $$$${numbers}\:{divisible}\:{by}\:\mathrm{5}: \\ $$$$\lfloor\frac{\mathrm{900}}{\mathrm{5}}\rfloor−\mathrm{30}−\mathrm{60}−\mathrm{30}=\mathrm{60} \\ $$$${total}\:{numbers}\:{divisible}\:{by}\:\mathrm{2},\mathrm{3},\mathrm{5}: \\ $$$$\mathrm{30}+\mathrm{120}+\mathrm{60}+\mathrm{30}+\mathrm{240}+\mathrm{120}+\mathrm{60}=\mathrm{660} \\ $$$$ \\ $$$${numbers}\:{not}\:{divisible}\:{by}\:\mathrm{2},\mathrm{3},\mathrm{5}: \\ $$$$\mathrm{900}−\mathrm{660}=\mathrm{240} \\ $$

Commented by mr W last updated on 05/Oct/21

⌊((900)/7)⌋=128 means there are 128 numbers from 1 to 900 which are divisible by 7.

$$\lfloor\frac{\mathrm{900}}{\mathrm{7}}\rfloor=\mathrm{128}\:{means}\:{there}\:{are}\:\mathrm{128} \\ $$$${numbers}\:{from}\:\mathrm{1}\:{to}\:\mathrm{900}\:{which}\:{are} \\ $$$${divisible}\:{by}\:\mathrm{7}. \\ $$

Commented by puissant last updated on 05/Oct/21