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Question Number 205175 by BaliramKumar last updated on 12/Mar/24
Answered by Rasheed.Sindhi last updated on 12/Mar/24
413283P759387 •Divide the number in groups of 3 from right side 4,132,83P,759,387 •Apply subtraction and addition alternatively between the groups: 387−759+83P−132+4 The number is divisible by 13 if the result is 0 or a multiple of 13 ∵ 13 ∣ (387−759+83P−132+4) or13 ∣ (387−759+(830+P)−132+4) or 13 ∣ 330+P or 13 ∣ 13×25+5+P or 13 ∣ 5+P Recall that 0≤P≤9 also. ∴ P=8
$$\mathrm{413283P759387} \\ $$$$\bullet{Divide}\:{the}\:{number}\:{in}\:{groups}\:{of}\:\mathrm{3} \\ $$$${from}\:{right}\:{side} \\ $$$$\mathrm{4},\mathrm{132},\mathrm{83P},\mathrm{759},\mathrm{387} \\ $$$$\bullet{Apply}\:{subtraction}\:{and}\:{addition} \\ $$$${alternatively}\:{between}\:{the}\:{groups}: \\ $$$$\mathrm{387}−\mathrm{759}+\mathrm{83P}−\mathrm{132}+\mathrm{4} \\ $$$${The}\:{number}\:{is}\:{divisible}\:{by}\:\mathrm{13}\:{if} \\ $$$${the}\:{result}\:{is}\:\mathrm{0}\:{or}\:{a}\:{multiple}\:{of}\:\mathrm{13} \\ $$$$\because\:\mathrm{13}\:\mid\:\left(\mathrm{387}−\mathrm{759}+\mathrm{83P}−\mathrm{132}+\mathrm{4}\right) \\ $$$$\:\:\:{or}\mathrm{13}\:\mid\:\left(\mathrm{387}−\mathrm{759}+\left(\mathrm{830}+\mathrm{P}\right)−\mathrm{132}+\mathrm{4}\right) \\ $$$$\:\:\:{or}\:\mathrm{13}\:\mid\:\mathrm{330}+\mathrm{P} \\ $$$$\:\:\:{or}\:\mathrm{13}\:\mid\:\mathrm{13}×\mathrm{25}+\mathrm{5}+\mathrm{P} \\ $$$$\:\:\:{or}\:\mathrm{13}\:\mid\:\mathrm{5}+\mathrm{P} \\ $$$${Recall}\:{that}\:\mathrm{0}\leqslant\mathrm{P}\leqslant\mathrm{9}\:{also}. \\ $$$$\therefore\:\mathrm{P}=\mathrm{8} \\ $$
Commented by BaliramKumar last updated on 12/Mar/24
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$
Answered by A5T last updated on 12/Mar/24
[3^(3k+1) ≡^(13) 3;3^(3k+2) ≡^(13) ;3^(3k) ≡^(13) 1] 4(10^(12) )+1(10^(11) )+3(10^(10) )+...+3(10^2 )+8(10)+7 ≡^(13) 4(−3)^(12) +(−3)^(11) +3(−3)^(10) +...+3(−3)^2 +8(−3)+7 ≡4(1)+1(−9)+3(3)+2(−1)+8(9)+3(−3)+P +7(−9)+5(3)+9(−1)+3(9)+8(−3)+7=P+18 ≡^(13) 0⇒P≡−18≡−5≡8(mod 13)⇒P=8
$$\left[\mathrm{3}^{\mathrm{3}{k}+\mathrm{1}} \overset{\mathrm{13}} {\equiv}\mathrm{3};\mathrm{3}^{\mathrm{3}{k}+\mathrm{2}} \overset{\mathrm{13}} {\equiv};\mathrm{3}^{\mathrm{3}{k}} \overset{\mathrm{13}} {\equiv}\mathrm{1}\right] \\ $$$$\mathrm{4}\left(\mathrm{10}^{\mathrm{12}} \right)+\mathrm{1}\left(\mathrm{10}^{\mathrm{11}} \right)+\mathrm{3}\left(\mathrm{10}^{\mathrm{10}} \right)+…+\mathrm{3}\left(\mathrm{10}^{\mathrm{2}} \right)+\mathrm{8}\left(\mathrm{10}\right)+\mathrm{7} \\ $$$$\overset{\mathrm{13}} {\equiv}\mathrm{4}\left(−\mathrm{3}\right)^{\mathrm{12}} +\left(−\mathrm{3}\right)^{\mathrm{11}} +\mathrm{3}\left(−\mathrm{3}\right)^{\mathrm{10}} +…+\mathrm{3}\left(−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{8}\left(−\mathrm{3}\right)+\mathrm{7} \\ $$$$\equiv\mathrm{4}\left(\mathrm{1}\right)+\mathrm{1}\left(−\mathrm{9}\right)+\mathrm{3}\left(\mathrm{3}\right)+\mathrm{2}\left(−\mathrm{1}\right)+\mathrm{8}\left(\mathrm{9}\right)+\mathrm{3}\left(−\mathrm{3}\right)+{P} \\ $$$$+\mathrm{7}\left(−\mathrm{9}\right)+\mathrm{5}\left(\mathrm{3}\right)+\mathrm{9}\left(−\mathrm{1}\right)+\mathrm{3}\left(\mathrm{9}\right)+\mathrm{8}\left(−\mathrm{3}\right)+\mathrm{7}={P}+\mathrm{18} \\ $$$$\overset{\mathrm{13}} {\equiv}\mathrm{0}\Rightarrow{P}\equiv−\mathrm{18}\equiv−\mathrm{5}\equiv\mathrm{8}\left({mod}\:\mathrm{13}\right)\Rightarrow{P}=\mathrm{8}\: \\ $$
Commented by BaliramKumar last updated on 12/Mar/24
thanks sir what is osculator how it works
$$\mathrm{thanks}\:\mathrm{sir} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{osculator}\:\mathrm{how}\:\mathrm{it}\:\mathrm{works} \\ $$$$ \\ $$
Commented by A5T last updated on 12/Mar/24
If you want to find the osculator of a number that is co-prime with 10,say 17 or 19, find the smallest multiple of the number, say 17:(51) or 19:(19), that is +_− 1 mod 10. If the smallest multiple that is +_− 1mod 10 is actually +1(mod 10) as in the case of 17, the first digit of that multiple is called the “negative” osculator of the number. So, if you want to check the divisibility of a number by 17,say 3281, then you “subtract 5 times” the last digit,1, from the remaining part: 328−5×1 =323→^(same_(operation) ) 32−3×5=17. Since 17 is divisible by 17, 3281 must also be divisible by 17. For a number like 19≡−1(mod 10); the “positive” osculator is the first digit of 19+1,which is 2, so instead of “subtracting” the last digit, you add it. For example,3876: 387+6×2=399 →39+9×2=57=19×3(divisible by 19)⇒19∣3876
$${If}\:{you}\:{want}\:{to}\:{find}\:{the}\:{osculator}\:{of}\:{a}\:{number} \\ $$$${that}\:{is}\:{co}-{prime}\:{with}\:\mathrm{10},{say}\:\mathrm{17}\:{or}\:\mathrm{19},\:{find}\:{the} \\ $$$${smallest}\:{multiple}\:{of}\:{the}\:{number},\:{say}\:\mathrm{17}:\left(\mathrm{51}\right) \\ $$$${or}\:\mathrm{19}:\left(\mathrm{19}\right),\:{that}\:{is}\:\underset{−} {+}\mathrm{1}\:{mod}\:\mathrm{10}.\: \\ $$$${If}\:{the}\:{smallest}\:{multiple}\:{that}\:{is}\:\underset{−} {+}\mathrm{1}{mod}\:\mathrm{10}\:{is}\: \\ $$$${actually}\:+\mathrm{1}\left({mod}\:\mathrm{10}\right)\:{as}\:{in}\:{the}\:{case}\:{of}\:\mathrm{17},\:{the}\:{first}\: \\ $$$${digit}\:{of}\:{that}\:{multiple}\:{is}\:{called}\:{the}\:“{negative}''\: \\ $$$${osculator}\:{of}\:{the}\:{number}. \\ $$$${So},\:{if}\:{you}\:{want}\:{to}\:{check}\:{the}\:{divisibility}\:{of}\:{a}\:{number} \\ $$$${by}\:\mathrm{17},{say}\:\mathrm{3281},\:{then}\:{you}\:“{subtract}\:\mathrm{5}\:{times}''\:{the}\: \\ $$$${last}\:{digit},\mathrm{1},\:{from}\:{the}\:{remaining}\:{part}:\:\mathrm{328}−\mathrm{5}×\mathrm{1} \\ $$$$=\mathrm{323}\overset{\underset{{operation}} {{same}}} {\rightarrow}\mathrm{32}−\mathrm{3}×\mathrm{5}=\mathrm{17}.\:{Since}\:\mathrm{17}\:{is}\:{divisible}\:{by}\:\mathrm{17}, \\ $$$$\mathrm{3281}\:{must}\:{also}\:{be}\:{divisible}\:{by}\:\mathrm{17}. \\ $$$$ \\ $$$${For}\:{a}\:{number}\:{like}\:\mathrm{19}\equiv−\mathrm{1}\left({mod}\:\mathrm{10}\right);\:{the}\:“{positive}'' \\ $$$${osculator}\:{is}\:{the}\:{first}\:{digit}\:{of}\:\mathrm{19}+\mathrm{1},{which}\:{is}\:\mathrm{2}, \\ $$$${so}\:{instead}\:{of}\:“{subtracting}''\:{the}\:{last}\:{digit},\:{you} \\ $$$${add}\:{it}.\:{For}\:{example},\mathrm{3876}:\:\mathrm{387}+\mathrm{6}×\mathrm{2}=\mathrm{399} \\ $$$$\rightarrow\mathrm{39}+\mathrm{9}×\mathrm{2}=\mathrm{57}=\mathrm{19}×\mathrm{3}\left({divisible}\:{by}\:\mathrm{19}\right)\Rightarrow\mathrm{19}\mid\mathrm{3876} \\ $$
Commented by A5T last updated on 12/Mar/24
The negative osculator of 7 is 2 First multiples of 7: 7,14,21: +1(mod 10) So,osculator is first digit of 21−1 The positive osculator of 13 is 4 First multiples of 13: 13,26,39: −1(mod 10) So,osculator is first digit of 39+1
$${The}\:{negative}\:{osculator}\:{of}\:\mathrm{7}\:{is}\:\mathrm{2} \\ $$$${First}\:{multiples}\:{of}\:\mathrm{7}:\:\mathrm{7},\mathrm{14},\mathrm{21}:\:+\mathrm{1}\left({mod}\:\mathrm{10}\right) \\ $$$${So},{osculator}\:{is}\:{first}\:{digit}\:{of}\:\mathrm{21}−\mathrm{1} \\ $$$${The}\:{positive}\:{osculator}\:{of}\:\mathrm{13}\:{is}\:\mathrm{4} \\ $$$${First}\:{multiples}\:{of}\:\mathrm{13}:\:\mathrm{13},\mathrm{26},\mathrm{39}:\:−\mathrm{1}\left({mod}\:\mathrm{10}\right) \\ $$$${So},{osculator}\:{is}\:{first}\:{digit}\:{of}\:\mathrm{39}+\mathrm{1} \\ $$
Commented by BaliramKumar last updated on 12/Mar/24
thanks sir
$$\mathrm{thanks}\:\mathrm{sir} \\ $$
Answered by A5T last updated on 12/Mar/24
[10^3 ≡^(13) −1;10^6 ≡^(13) 1] 413283P759387=387+759×1000+83P×10^6 +132×10^9 +4×10^(12) ≡387−759+83P−132+4 ≡^(13) 0⇒83P−500≡^(13) 0⇒830+P≡500≡6(mod 13) 11+P≡^(13) 6⇒P≡−5≡^(13) 8
$$\left[\mathrm{10}^{\mathrm{3}} \overset{\mathrm{13}} {\equiv}−\mathrm{1};\mathrm{10}^{\mathrm{6}} \overset{\mathrm{13}} {\equiv}\mathrm{1}\right] \\ $$$$\mathrm{413283}{P}\mathrm{759387}=\mathrm{387}+\mathrm{759}×\mathrm{1000}+\mathrm{83}{P}×\mathrm{10}^{\mathrm{6}} \\ $$$$+\mathrm{132}×\mathrm{10}^{\mathrm{9}} +\mathrm{4}×\mathrm{10}^{\mathrm{12}} \equiv\mathrm{387}−\mathrm{759}+\mathrm{83}{P}−\mathrm{132}+\mathrm{4} \\ $$$$\overset{\mathrm{13}} {\equiv}\mathrm{0}\Rightarrow\mathrm{83}{P}−\mathrm{500}\overset{\mathrm{13}} {\equiv}\mathrm{0}\Rightarrow\mathrm{830}+{P}\equiv\mathrm{500}\equiv\mathrm{6}\left({mod}\:\mathrm{13}\right) \\ $$$$\mathrm{11}+{P}\overset{\mathrm{13}} {\equiv}\mathrm{6}\Rightarrow{P}\equiv−\mathrm{5}\overset{\mathrm{13}} {\equiv}\mathrm{8} \\ $$
Answered by Rasheed.Sindhi last updated on 13/Mar/24
413283P759387 =4132830000000+P759387 Osculator of 13=4 P759387 determinant ((P,7,5,9,3,8,7),((4×4+2+P),(4×4+1+7),(4×2+1+5),(4×0+3+9),(4×6+3+3),(4×7+8), ),((18+P),(24),(14),(12),(30),(36), )) 4132830000000 determinant ((4,1,3,2,8,3,0,(...),0),((4×8+3+4),(4×9+1+1),(4×4+3),(4×0+2+2),(4×3+8),(4×0+3),(4×0+0), , ),((39),(38),(19),4,(20),3,0, , )) 13∣(18+P+39) 13∣(5+P) P=8
$$\mathrm{413283P759387} \\ $$$$=\mathrm{4132830000000}+\mathrm{P759387} \\ $$$${Osculator}\:{of}\:\mathrm{13}=\mathrm{4} \\ $$$$\mathrm{P759387} \\ $$$$\begin{array}{|c|c|c|}{\mathrm{P}}&\hline{\mathrm{7}}&\hline{\mathrm{5}}&\hline{\mathrm{9}}&\hline{\mathrm{3}}&\hline{\mathrm{8}}&\hline{\mathrm{7}}\\{\mathrm{4}×\mathrm{4}+\mathrm{2}+\mathrm{P}}&\hline{\mathrm{4}×\mathrm{4}+\mathrm{1}+\mathrm{7}}&\hline{\mathrm{4}×\mathrm{2}+\mathrm{1}+\mathrm{5}}&\hline{\mathrm{4}×\mathrm{0}+\mathrm{3}+\mathrm{9}}&\hline{\mathrm{4}×\mathrm{6}+\mathrm{3}+\mathrm{3}}&\hline{\mathrm{4}×\mathrm{7}+\mathrm{8}}&\hline{\:}\\{\mathrm{18}+\mathrm{P}}&\hline{\mathrm{24}}&\hline{\mathrm{14}}&\hline{\mathrm{12}}&\hline{\mathrm{30}}&\hline{\mathrm{36}}&\hline{\:}\\\hline\end{array} \\ $$$$\mathrm{4132830000000} \\ $$$$\begin{array}{|c|c|c|}{\mathrm{4}}&\hline{\mathrm{1}}&\hline{\mathrm{3}}&\hline{\mathrm{2}}&\hline{\mathrm{8}}&\hline{\mathrm{3}}&\hline{\mathrm{0}}&\hline{…}&\hline{\mathrm{0}}\\{\mathrm{4}×\mathrm{8}+\mathrm{3}+\mathrm{4}}&\hline{\mathrm{4}×\mathrm{9}+\mathrm{1}+\mathrm{1}}&\hline{\mathrm{4}×\mathrm{4}+\mathrm{3}}&\hline{\mathrm{4}×\mathrm{0}+\mathrm{2}+\mathrm{2}}&\hline{\mathrm{4}×\mathrm{3}+\mathrm{8}}&\hline{\mathrm{4}×\mathrm{0}+\mathrm{3}}&\hline{\mathrm{4}×\mathrm{0}+\mathrm{0}}&\hline{\:}&\hline{\:}\\{\mathrm{39}}&\hline{\mathrm{38}}&\hline{\mathrm{19}}&\hline{\mathrm{4}}&\hline{\mathrm{20}}&\hline{\mathrm{3}}&\hline{\mathrm{0}}&\hline{\:}&\hline{\:}\\\hline\end{array} \\ $$$$\: \\ $$$$\mathrm{13}\mid\left(\mathrm{18}+\mathrm{P}+\mathrm{39}\right) \\ $$$$\mathrm{13}\mid\left(\mathrm{5}+\mathrm{P}\right) \\ $$$$\mathrm{P}=\mathrm{8} \\ $$

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