Question Number 152139 by Ar Brandon last updated on 26/Aug/21
![How many numbers x are found between 9000 and 11000 and verify x=12[19], x=5[13], x=9[11] ?](Q152139.png)
$$\mathrm{How}\:\mathrm{many}\:\mathrm{numbers}\:{x}\:\mathrm{are}\:\mathrm{found}\:\mathrm{between}\:\mathrm{9000}\:\mathrm{and}\:\mathrm{11000} \\ $$$$\mathrm{and}\:\mathrm{verify}\:{x}=\mathrm{12}\left[\mathrm{19}\right],\:{x}=\mathrm{5}\left[\mathrm{13}\right],\:{x}=\mathrm{9}\left[\mathrm{11}\right]\:? \\ $$
Commented by Rasheed.Sindhi last updated on 26/Aug/21
$$\mathrm{One}\:\left({x}=\mathrm{10899}\right) \\ $$
Answered by Rasheed.Sindhi last updated on 26/Aug/21
![x≡12[19], x≡5[13], x≡9[11] x≡12+19[19], x≡5+13[13], x≡9+11[11] x≡31[19], x≡18[13], x≡20[11] x≡31+19[19], x≡18+13[13], x≡20+11[11] x≡50[19], x≡3113], x≡31[11] Smallest value of x: 31 which satisfy all three congruence. next such value 31+LCM(11,13,19) =31+2717=2748 In general 31+nLCM(11,13,19) To find least value of n n=⌊((9000)/(2717))⌋=3 or n=⌊((9000)/(2717))⌋+1=4 31+3×2717=8182<9000 (×) 31+4×2717=10899 (✓) 31+5×2727=13616>11000 (×) ONE ONLY](Q152202.png)
$${x}\equiv\mathrm{12}\left[\mathrm{19}\right],\:{x}\equiv\mathrm{5}\left[\mathrm{13}\right],\:{x}\equiv\mathrm{9}\left[\mathrm{11}\right] \\ $$$${x}\equiv\mathrm{12}+\mathrm{19}\left[\mathrm{19}\right],\:{x}\equiv\mathrm{5}+\mathrm{13}\left[\mathrm{13}\right],\:{x}\equiv\mathrm{9}+\mathrm{11}\left[\mathrm{11}\right] \\ $$$${x}\equiv\mathrm{31}\left[\mathrm{19}\right],\:{x}\equiv\mathrm{18}\left[\mathrm{13}\right],\:{x}\equiv\mathrm{20}\left[\mathrm{11}\right] \\ $$$${x}\equiv\mathrm{31}+\mathrm{19}\left[\mathrm{19}\right],\:{x}\equiv\mathrm{18}+\mathrm{13}\left[\mathrm{13}\right],\:{x}\equiv\mathrm{20}+\mathrm{11}\left[\mathrm{11}\right] \\ $$$$\left.{x}\equiv\mathrm{50}\left[\mathrm{19}\right],\:{x}\equiv\mathrm{3113}\right],\:{x}\equiv\mathrm{31}\left[\mathrm{11}\right] \\ $$$${Smallest}\:{value}\:{of}\:{x}:\:\mathrm{31}\:{which}\:{satisfy} \\ $$$${all}\:{three}\:{congruence}. \\ $$$${next}\:{such}\:{value}\:\mathrm{31}+\mathrm{LCM}\left(\mathrm{11},\mathrm{13},\mathrm{19}\right) \\ $$$$=\mathrm{31}+\mathrm{2717}=\mathrm{2748} \\ $$$${In}\:{general}\:\mathrm{31}+{n}\mathrm{LCM}\left(\mathrm{11},\mathrm{13},\mathrm{19}\right) \\ $$$${To}\:{find}\:\mathrm{least}\:{value}\:{of}\:{n} \\ $$$${n}=\lfloor\frac{\mathrm{9000}}{\mathrm{2717}}\rfloor=\mathrm{3}\:\mathrm{or}\:\mathrm{n}=\lfloor\frac{\mathrm{9000}}{\mathrm{2717}}\rfloor+\mathrm{1}=\mathrm{4} \\ $$$$\mathrm{31}+\mathrm{3}×\mathrm{2717}=\mathrm{8182}<\mathrm{9000}\:\left(×\right) \\ $$$$\mathrm{31}+\mathrm{4}×\mathrm{2717}=\mathrm{10899}\:\left(\checkmark\right) \\ $$$$\mathrm{31}+\mathrm{5}×\mathrm{2727}=\mathrm{13616}>\mathrm{11000}\:\left(×\right) \\ $$$$\mathrm{ONE}\:\:\mathrm{ONLY} \\ $$
Commented by Ar Brandon last updated on 26/Aug/21
$$\mathrm{OK}\:\mathrm{Sir}\:\mathrm{Thanks} \\ $$