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The-first-23-natural-numbers-are-written-in-an-increasing-order-beside-each-other-to-form-a-single-number-What-is-the-remainder-when-this-number-is-divided-by-18-

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Question Number 113000 by Aina Samuel Temidayo last updated on 10/Sep/20
The first 23 natural numbers are written in an increasing order beside each other to form a single number. What is the remainder when this number is divided by 18?
$$\mathrm{The}\:\mathrm{first}\:\mathrm{23}\:\mathrm{natural}\:\mathrm{numbers} \\ $$$$\mathrm{are}\:\mathrm{written}\:\mathrm{in}\:\mathrm{an}\:\mathrm{increasing}\:\mathrm{order} \\ $$$$\mathrm{beside}\:\mathrm{each}\:\mathrm{other}\:\mathrm{to}\:\mathrm{form}\:\mathrm{a}\:\mathrm{single} \\ $$$$\mathrm{number}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when} \\ $$$$\mathrm{this}\:\mathrm{number}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{18}? \\ $$
Answered by floor(10²Eta[1]) last updated on 11/Sep/20
N=12345...212223(mod18) 18=9.2 the solution to N(mod 18) is equivalent to find the solution to the system: { ((N≡x(mod2))),((N≡y(mod9))) :} I: since N is odd so N≡1(mod 2) II: any number N when divided by 9 leaves the same remainder that the number formed by the sum of the digits of N N≡S(N)(mod 9), where S(N) is the sum of the digits of N. N≡1+2+...+1+0+1+1+...(mod9) (1+...+9)+(1+0+...+1+9)+(2+0+...+2+3) =45+(10.1+(0+1+2+...+9))+(2.4+(0+1+2+3)) =45+10+45+8+6=114≡6(mod9) { ((N≡1(mod 2))),((N≡6(mod 9))) :} ⇒N=9k+6≡1(mod2)∴k≡1(mod2) k=2t+1⇒N=18t+15 N leaves remainder 15 when divided by 18.
$$\mathrm{N}=\mathrm{12345}…\mathrm{212223}\left(\mathrm{mod18}\right) \\ $$$$\mathrm{18}=\mathrm{9}.\mathrm{2} \\ $$$$\mathrm{the}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{N}\left(\mathrm{mod}\:\mathrm{18}\right)\:\mathrm{is}\:\mathrm{equivalent} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{system}: \\ $$$$\begin{cases}{\mathrm{N}\equiv\mathrm{x}\left(\mathrm{mod2}\right)}\\{\mathrm{N}\equiv\mathrm{y}\left(\mathrm{mod9}\right)}\end{cases} \\ $$$$\mathrm{I}: \\ $$$$\mathrm{since}\:\mathrm{N}\:\mathrm{is}\:\mathrm{odd}\:\mathrm{so}\:\mathrm{N}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right) \\ $$$$\mathrm{II}: \\ $$$$\mathrm{any}\:\mathrm{number}\:\mathrm{N}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{9}\:\mathrm{leaves} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{remainder}\:\mathrm{that}\:\mathrm{the}\:\mathrm{number}\: \\ $$$$\mathrm{formed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{N} \\ $$$$\mathrm{N}\equiv\mathrm{S}\left(\mathrm{N}\right)\left(\mathrm{mod}\:\mathrm{9}\right),\:\mathrm{where}\:\mathrm{S}\left(\mathrm{N}\right)\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{N}. \\ $$$$\mathrm{N}\equiv\mathrm{1}+\mathrm{2}+…+\mathrm{1}+\mathrm{0}+\mathrm{1}+\mathrm{1}+…\left(\mathrm{mod9}\right) \\ $$$$\left(\mathrm{1}+…+\mathrm{9}\right)+\left(\mathrm{1}+\mathrm{0}+…+\mathrm{1}+\mathrm{9}\right)+\left(\mathrm{2}+\mathrm{0}+…+\mathrm{2}+\mathrm{3}\right) \\ $$$$=\mathrm{45}+\left(\mathrm{10}.\mathrm{1}+\left(\mathrm{0}+\mathrm{1}+\mathrm{2}+…+\mathrm{9}\right)\right)+\left(\mathrm{2}.\mathrm{4}+\left(\mathrm{0}+\mathrm{1}+\mathrm{2}+\mathrm{3}\right)\right) \\ $$$$=\mathrm{45}+\mathrm{10}+\mathrm{45}+\mathrm{8}+\mathrm{6}=\mathrm{114}\equiv\mathrm{6}\left(\mathrm{mod9}\right) \\ $$$$\begin{cases}{\mathrm{N}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{2}\right)}\\{\mathrm{N}\equiv\mathrm{6}\left(\mathrm{mod}\:\mathrm{9}\right)}\end{cases} \\ $$$$\Rightarrow\mathrm{N}=\mathrm{9k}+\mathrm{6}\equiv\mathrm{1}\left(\mathrm{mod2}\right)\therefore\mathrm{k}\equiv\mathrm{1}\left(\mathrm{mod2}\right) \\ $$$$\mathrm{k}=\mathrm{2t}+\mathrm{1}\Rightarrow\mathrm{N}=\mathrm{18t}+\mathrm{15} \\ $$$$\mathrm{N}\:\mathrm{leaves}\:\mathrm{remainder}\:\mathrm{15}\:\mathrm{when}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{18}. \\ $$
Commented by Aina Samuel Temidayo last updated on 11/Sep/20
Thanks.
$$\mathrm{Thanks}. \\ $$

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