Q-How-many-common-three-digit-numbers-are-there-in-the-following-two-sequences-a-n-1-5-9-13-b-m-4-7-10-13-

Question Number 173902 by mnjuly1970 last updated on 20/Jul/22

$$ \\ $$$$\:\:{Q}:\:{How}\:{many}\:{common}\:{three}−{digit}\:{numbers} \\ $$$$\:\:\:\:{are}\:{there}\:{in}\:{the}\:{following} \\ $$$$\:\:\:\:\:{two}\:{sequences}? \\ $$$$\:\:\:\:\:\begin{cases}{\:\:{a}_{{n}} \:=\:\mathrm{1}\:\:,\:\mathrm{5}\:,\:\mathrm{9}\:,\mathrm{13}\:,\:…}\\{\:\:{b}_{\:{m}} \:=\:\mathrm{4}\:,\:\mathrm{7}\:,\:\mathrm{10}\:,\:\mathrm{13}\:,…}\end{cases} \\ $$

Answered by mr W last updated on 21/Jul/22

a_n =1+4(n−1) b_m =4+3(m−1) a_n =b_m 1+4(n−1)=4+3(m−1) 4(n−1)=3m ⇒n−1=3k, m=4k ⇒a_n =1+12k=b_m 1+12k≥100 ⇒k≥((99)/(12))≥8 1+12k≤999 ⇒k≤((998)/(12))≤83 ⇒83−8+1=76 numbers with 3 digits are common in sequences a_n and b_n .

$${a}_{{n}} =\mathrm{1}+\mathrm{4}\left({n}−\mathrm{1}\right) \\ $$$${b}_{{m}} =\mathrm{4}+\mathrm{3}\left({m}−\mathrm{1}\right) \\ $$$${a}_{{n}} ={b}_{{m}} \\ $$$$\mathrm{1}+\mathrm{4}\left({n}−\mathrm{1}\right)=\mathrm{4}+\mathrm{3}\left({m}−\mathrm{1}\right) \\ $$$$\mathrm{4}\left({n}−\mathrm{1}\right)=\mathrm{3}{m} \\ $$$$\Rightarrow{n}−\mathrm{1}=\mathrm{3}{k},\:{m}=\mathrm{4}{k} \\ $$$$\Rightarrow{a}_{{n}} =\mathrm{1}+\mathrm{12}{k}={b}_{{m}} \\ $$$$\mathrm{1}+\mathrm{12}{k}\geqslant\mathrm{100}\:\Rightarrow{k}\geqslant\frac{\mathrm{99}}{\mathrm{12}}\geqslant\mathrm{8} \\ $$$$\mathrm{1}+\mathrm{12}{k}\leqslant\mathrm{999}\:\Rightarrow{k}\leqslant\frac{\mathrm{998}}{\mathrm{12}}\leqslant\mathrm{83} \\ $$$$\Rightarrow\mathrm{83}−\mathrm{8}+\mathrm{1}=\mathrm{76}\:{numbers}\:{with}\:\mathrm{3}\:{digits}\: \\ $$$${are}\:{common}\:{in}\:{sequences}\:{a}_{{n}} \:{and}\:{b}_{{n}} . \\ $$

Commented by Rasheed.Sindhi last updated on 21/Jul/22