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Question Number 11754 by Joel576 last updated on 31/Mar/17
How many numbers between 1 − 2017 that aren′t divisible by 5, 6, 7, 8 ?
$$\mathrm{How}\:\mathrm{many}\:\mathrm{numbers}\:\mathrm{between}\:\mathrm{1}\:−\:\mathrm{2017} \\ $$$$\mathrm{that}\:\mathrm{aren}'\mathrm{t}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{5},\:\mathrm{6},\:\mathrm{7},\:\mathrm{8}\:? \\ $$
Answered by mrW1 last updated on 31/Mar/17
let a=count of numbers which are multiple of 5 let b=count of numbers which are multiple of 6 let c=count of numbers which are multiple of 7 let d=count of numbers which are multiple of 8 a=2017/5=403 a_1 =2017/(5×6)=67 (also in b) a_2 =2017/(5×7)=57 (also in c) a_3 =2017/(5×8)=50 (also in d) b=2017/6=336 b_1 =2017/(6×7)=48 (also in c) b_2 =2017/(6×8)=42 (also in d) c=2017/7=288 c_1 =2017/(7×8)=36 (also in d) d=2017/8=252 a_1 ,a_2 ,a_3 ,b_1 ,b_2 ,c_1 are count of numbers which are double counted. count of numbers which are divisible by 5,6,7,8: a+b+c+d−a_1 −a_2 −a_3 −b_1 −b_2 −c_1 =403+336+288+252−67−57−50−48−42−36 =979 count of numbers which are not divisible by 5,6,7,8: 2017−979=1038
$${let}\:{a}={count}\:{of}\:{numbers}\:{which}\:{are}\:{multiple}\:{of}\:\mathrm{5} \\ $$$${let}\:{b}={count}\:{of}\:{numbers}\:{which}\:{are}\:{multiple}\:{of}\:\mathrm{6} \\ $$$${let}\:{c}={count}\:{of}\:{numbers}\:{which}\:{are}\:{multiple}\:{of}\:\mathrm{7} \\ $$$${let}\:{d}={count}\:{of}\:{numbers}\:{which}\:{are}\:{multiple}\:{of}\:\mathrm{8} \\ $$$$ \\ $$$${a}=\mathrm{2017}/\mathrm{5}=\mathrm{403} \\ $$$${a}_{\mathrm{1}} =\mathrm{2017}/\left(\mathrm{5}×\mathrm{6}\right)=\mathrm{67}\:\left({also}\:{in}\:{b}\right) \\ $$$${a}_{\mathrm{2}} =\mathrm{2017}/\left(\mathrm{5}×\mathrm{7}\right)=\mathrm{57}\:\left({also}\:{in}\:{c}\right) \\ $$$${a}_{\mathrm{3}} =\mathrm{2017}/\left(\mathrm{5}×\mathrm{8}\right)=\mathrm{50}\:\left({also}\:{in}\:{d}\right) \\ $$$$ \\ $$$${b}=\mathrm{2017}/\mathrm{6}=\mathrm{336} \\ $$$${b}_{\mathrm{1}} =\mathrm{2017}/\left(\mathrm{6}×\mathrm{7}\right)=\mathrm{48}\:\left({also}\:{in}\:{c}\right) \\ $$$${b}_{\mathrm{2}} =\mathrm{2017}/\left(\mathrm{6}×\mathrm{8}\right)=\mathrm{42}\:\left({also}\:{in}\:{d}\right) \\ $$$$ \\ $$$${c}=\mathrm{2017}/\mathrm{7}=\mathrm{288} \\ $$$${c}_{\mathrm{1}} =\mathrm{2017}/\left(\mathrm{7}×\mathrm{8}\right)=\mathrm{36}\:\left({also}\:{in}\:{d}\right) \\ $$$$ \\ $$$${d}=\mathrm{2017}/\mathrm{8}=\mathrm{252} \\ $$$$ \\ $$$${a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,{a}_{\mathrm{3}} ,{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,{c}_{\mathrm{1}} \:{are}\:{count}\:{of}\:{numbers} \\ $$$${which}\:{are}\:{double}\:{counted}. \\ $$$$ \\ $$$${count}\:{of}\:{numbers}\:{which}\:{are}\:{divisible} \\ $$$${by}\:\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8}: \\ $$$${a}+{b}+{c}+{d}−{a}_{\mathrm{1}} −{a}_{\mathrm{2}} −{a}_{\mathrm{3}} −{b}_{\mathrm{1}} −{b}_{\mathrm{2}} −{c}_{\mathrm{1}} \\ $$$$=\mathrm{403}+\mathrm{336}+\mathrm{288}+\mathrm{252}−\mathrm{67}−\mathrm{57}−\mathrm{50}−\mathrm{48}−\mathrm{42}−\mathrm{36} \\ $$$$=\mathrm{979} \\ $$$$ \\ $$$${count}\:{of}\:{numbers}\:{which}\:{are}\:{not}\:{divisible} \\ $$$${by}\:\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8}: \\ $$$$\mathrm{2017}−\mathrm{979}=\mathrm{1038} \\ $$
Commented by Joel576 last updated on 01/Apr/17
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$

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