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Question Number 188196 by cortano12 last updated on 26/Feb/23

(1)solve Diopthantine equation 754x+221y=13 (2) find the number abcd such that 4×(abcd)=dcba

$$\left(\mathrm{1}\right)\mathrm{solve}\:\mathrm{Diopthantine}\:\mathrm{equation} \\ $$$$\:\:\:\:\:\mathrm{754x}+\mathrm{221y}=\mathrm{13} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{abcd}\: \\ $$$$\:\:\:\mathrm{such}\:\mathrm{that}\:\mathrm{4}×\left(\mathrm{abcd}\right)=\mathrm{dcba} \\ $$

Commented by ARUNG_Brandon_MBU last updated on 26/Feb/23

#include <iostream> using namespace std; int main(void) { for (int i=1000; i<2500; i++) { for(int j=1001; j<10000;j++) { if (4*i == j) if (i%10 == j/1000) if (i%100/10 == j%1000/100) if (i%1000/100 == j%100/10) if (i/1000 == j%10) cout << "4*" << i <<"=" << j; } } return 0; } // output: 4*2178=8712

Commented by ARUNG_Brandon_MBU last updated on 26/Feb/23

a=2, b=1, c=7, d=8

$${a}=\mathrm{2},\:{b}=\mathrm{1},\:{c}=\mathrm{7},\:{d}=\mathrm{8} \\ $$

Answered by ARUNG_Brandon_MBU last updated on 26/Feb/23

754x+221y=13 754=3(221)+91 221=2(91)+39 91=2(39)+13 ⇒13=91−2(39) =91−2[221−2(91)] =−2(221)+5(91) =−2(221)+5[754−3(221)] =5(754)−17(221) ⇒754(5)+221(−17)=13 ⇒x=5, y=−17

$$\mathrm{754}{x}+\mathrm{221y}=\mathrm{13} \\ $$$$\mathrm{754}=\mathrm{3}\left(\mathrm{221}\right)+\mathrm{91} \\ $$$$\mathrm{221}=\mathrm{2}\left(\mathrm{91}\right)+\mathrm{39} \\ $$$$\mathrm{91}=\mathrm{2}\left(\mathrm{39}\right)+\mathrm{13} \\ $$$$ \\ $$$$\Rightarrow\mathrm{13}=\mathrm{91}−\mathrm{2}\left(\mathrm{39}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{91}−\mathrm{2}\left[\mathrm{221}−\mathrm{2}\left(\mathrm{91}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\mathrm{2}\left(\mathrm{221}\right)+\mathrm{5}\left(\mathrm{91}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\mathrm{2}\left(\mathrm{221}\right)+\mathrm{5}\left[\mathrm{754}−\mathrm{3}\left(\mathrm{221}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{5}\left(\mathrm{754}\right)−\mathrm{17}\left(\mathrm{221}\right) \\ $$$$\Rightarrow\mathrm{754}\left(\mathrm{5}\right)+\mathrm{221}\left(−\mathrm{17}\right)=\mathrm{13} \\ $$$$\Rightarrow{x}=\mathrm{5},\:\mathrm{y}=−\mathrm{17} \\ $$

Answered by mr W last updated on 26/Feb/23

(2) 4×(abcd)=(dcba) abcd × 4 dcba a=1 (rejected) or 2 ⇒a=2 ✓ 4×d=12 ⇒d=3 (rejected) 4×d=32 ⇒d=8 ✓ 2bc8 × 4 8cb2 (2000+100b+10c+8)×4=8000+100c+10b+2 13b+1=2c ⇒b=1, c=7 so the number is 2178 with 2178×4=8712

$$\left(\mathrm{2}\right) \\ $$$$\mathrm{4}×\left({abcd}\right)=\left({dcba}\right) \\ $$$$\:\:\:\:{abcd} \\ $$$$\underline{×\:\:\:\:\:\:\:\mathrm{4}} \\ $$$$\:\:\:\:{dcba} \\ $$$${a}=\mathrm{1}\:\left({rejected}\right)\:{or}\:\mathrm{2}\:\Rightarrow{a}=\mathrm{2}\:\checkmark \\ $$$$\mathrm{4}×{d}=\mathrm{12}\:\Rightarrow{d}=\mathrm{3}\:\left({rejected}\right) \\ $$$$\mathrm{4}×{d}=\mathrm{32}\:\Rightarrow{d}=\mathrm{8}\:\checkmark \\ $$$$\:\:\:\:\mathrm{2}{bc}\mathrm{8} \\ $$$$\underline{×\:\:\:\:\:\:\mathrm{4}} \\ $$$$\:\:\:\:\mathrm{8}{cb}\mathrm{2} \\ $$$$\left(\mathrm{2000}+\mathrm{100}{b}+\mathrm{10}{c}+\mathrm{8}\right)×\mathrm{4}=\mathrm{8000}+\mathrm{100}{c}+\mathrm{10}{b}+\mathrm{2} \\ $$$$\mathrm{13}{b}+\mathrm{1}=\mathrm{2}{c} \\ $$$$\Rightarrow{b}=\mathrm{1},\:{c}=\mathrm{7} \\ $$$${so}\:{the}\:{number}\:{is}\:\mathrm{2178}\:{with} \\ $$$$\mathrm{2178}×\mathrm{4}=\mathrm{8712} \\ $$

Answered by manxsol last updated on 27/Feb/23

determinant ((d,c,b,a,,a,b,c,d),(4,4,1,2,,6,9,4,4),(5,5,1,1,,4,7,6,6),(6,6,1,1,,4,7,6,6),(7,7,1,1,,8,1,7,7),(8,8,2,2,,2,5,8,8),(9,9,2,2,,2,5,8,8),(9,9,2,3,,6,9,9,9),(,,,,,,,,),(8,7,1,2,,2,1,7,8))

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