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Find-A-and-prove-that-2021-A-if-abcd-A-a-d-1-c-b-c-a-b-b-c-

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Question Number 150571 by mathdanisur last updated on 13/Aug/21
Find A and prove that 2021∈A if abcd^(−) ∈A, (a/(d + 1)) = ((c - b)/c) = ((a + b)/(b + c))
FindAandprovethat2021∈Aifabcd―∈A,ad+1=c−bc=a+bb+c
Answered by Rasheed.Sindhi last updated on 15/Aug/21
abcd^(−) ∈A; (a/(d + 1)) = ((c - b)/c) = ((a + b)/(b + c));A=? 0≤a,b,c,d≤9; a,c≠0 ((c - b)/c) = ((a + b)/(b + c))⇒ (c−b)c+b)=c(a+b) c^2 −b^2 −ac−bc=0 c^2 −c(a+b)−b^2 =0..........A D=(a+b)^2 −4(−b^2 ) =a^2 +5b^2 −2ab a^2 −2a+5⇒a=1 (a/(d + 1)) = ((c - b)/c)⇒ d=((ac)/(c−b))−1.......................B b=0:^• (c−0)(c+0)=c(a+0)⇒c=a d=(a^2 /(a−0))−1=a−1=c−1 a=c=u, the required numbers will be: determinant ((a,b,c,( d)),(u,0,u,(u−1))) u∈{1,2,3,...,9} 1010,2021,3032,4043,5054,6065,7076,8087,9098 b=1:^• c^2 −c(a+1)−1^2 =0 c^2 −c(a+1)−1=0 b=2:^• c^2 −c(a+2)−2^2 =0 b=3:^• c^2 −c(a+3)−3^2 =0 b=4:^• c^2 −c(a+4)−4^2 =0 b=5:^• c^2 −c(a+5)−5^2 =0 b=6:^• c^2 −c(a+6)−6^2 =0 b=7:^• c^2 −c(a+7)−7^2 =0 b=8:^• c^2 −c(a+8)−8^2 =0 b=9:^• c^2 −c(a+9)−9^2 =0
abcd―∈A;ad+1=c−bc=a+bb+c;A=?0⩽a,b,c,d⩽9;a,c≠0c−bc=a+bb+c⇒(c−b)c+b)=c(a+b)c2−b2−ac−bc=0c2−c(a+b)−b2=0……….AD=(a+b)2−4(−b2)=a2+5b2−2aba2−2a+5⇒a=1ad+1=c−bc⇒d=acc−b−1…………………..Bb=0:∙(c−0)(c+0)=c(a+0)⇒c=ad=a2a−0−1=a−1=c−1a=c=u,therequirednumberswillbe:abcdu0uu−1u∈{1,2,3,…,9}1010,2021,3032,4043,5054,6065,7076,8087,9098b=1:∙c2−c(a+1)−12=0c2−c(a+1)−1=0b=2:∙c2−c(a+2)−22=0b=3:∙c2−c(a+3)−32=0b=4:∙c2−c(a+4)−42=0b=5:∙c2−c(a+5)−52=0b=6:∙c2−c(a+6)−62=0b=7:∙c2−c(a+7)−72=0b=8:∙c2−c(a+8)−82=0b=9:∙c2−c(a+9)−92=0
Answered by Rasheed.Sindhi last updated on 14/Aug/21
abcd^(−) ∈A, (a/(d + 1)) = ((c - b)/c) = ((a + b)/(b + c));A=? 0≤a,b,c,d≤9;a,c≠0 determinant (((b=0)))(a/(d+1))=((c−0)/c)=((a+0)/(0+c)) a=d+1 ∧ a=c a=u(say),b=0,c=a=u,d=a−1=u−1 (a/u)(b/0)(c/u)(d/( u-1)); u=1,2,3,...,9 1010,2021,3032,4043,5054,6065, 7076,8087,9098 b=1: (a/(d + 1)) = ((c - b)/c) = ((a + b)/(b + c)) ⇒ (a/(d + 1)) = ((c - 1)/c) = ((a + 1)/(1+ c)) a=((c^2 −1)/c)−1 c divides c^2 −1 only in case c=1 a=((1^2 −1)/1)−1=−1(×) Case b=1 form no numbers b=2: (a/(d + 1)) = ((c - b)/c) = ((a + b)/(b + c)) (a/(d + 1)) = ((c - 2)/c) = ((a + 2)/(2 + c)) a=((c^2 - 4)/c) −2⇒c=4⇒a=1 (a/(d + 1)) = ((c - b)/c)⇒(1/(d+1))=((4−2)/4)=(1/2)⇒d=1 1241 b=3: ((c - b)/c)=((a + b)/(b + c))⇒((c - 3)/c)=((a + 3)/(3 + c)) a=((c^2 −9)/c)−3⇒c=9⇒a=5 d=((ac)/(c−b))−1=((45)/(9−3))−1 b=4: ((c - b)/c)=((a + b)/(b + c))⇒((c−4)/c)=((a+4)/(c+4)) a=((c^2 −16)/c)−4⇒c=8⇒a=2 d=((ac)/(c−b))−1=((16)/(8−4))−1=3⇒d=3 2483 ..... ...
abcd―∈A,ad+1=c−bc=a+bb+c;A=?0⩽a,b,c,d⩽9;a,c≠0b=0ad+1=c−0c=a+00+ca=d+1∧a=ca=u(say),b=0,c=a=u,d=a−1=u−1aub0cudu−1;u=1,2,3,…,91010,2021,3032,4043,5054,6065,7076,8087,9098b=1:ad+1=c−bc=a+bb+c⇒ad+1=c−1c=a+11+ca=c2−1c−1cdividesc2−1onlyincasec=1a=12−11−1=−1(×)Caseb=1formnonumbersb=2:ad+1=c−bc=a+bb+cad+1=c−2c=a+22+ca=c2−4c−2⇒c=4⇒a=1ad+1=c−bc⇒1d+1=4−24=12⇒d=11241b=3:c−bc=a+bb+c⇒c−3c=a+33+ca=c2−9c−3⇒c=9⇒a=5d=acc−b−1=459−3−1b=4:c−bc=a+bb+c⇒c−4c=a+4c+4a=c2−16c−4⇒c=8⇒a=2d=acc−b−1=168−4−1=3⇒d=32483…..…―
Answered by Rasheed.Sindhi last updated on 14/Aug/21
abcd^(−) ∈A, (a/(d + 1)) = ((c - b)/c) = ((a + b)/(b + c));A=? 0≤abcd≤9; c≠0,a≠0(∵ abcd^(−) is 4-digit nmbr) ((c - b)/c) = ((a + b)/(c + b)) c^2 −b^2 =ac+bc b^2 +bc+ac−c^2 =0...........(i) D=c^2 −4(ac−c^2 ) D =5c^2 −4ac (must be perfect square)....(ii) (a/(d + 1)) = ((c - b)/c)⇒((d+1)/a)=(c/(c−b)) ⇒d=((ac)/(c−b))−1.................(iii) c=1:(iii)⇒D=5−4a_((perfect square)) ⇒a=1 (i)⇒b^2 +b(1)+(1)(1)−(1)^2 =0 b(b+1)=0⇒b=0 ∨ b=−1(rejected) (iii)⇒d=(((1)(1))/(1−0))−1=0⇒d=0 abcd^(−) =1010 c=2:(ii)⇒D=5(2)^2 −4a(2)=20−8a⇒a=2 (i)⇒b^2 +b(2)+(2)(2)−(2)^2 =0 b(b+2)=0⇒b=0∨b=−2(rejected) (iii)⇒d=(((2)(2))/(2−0))−1⇒d=1 abcd^(−) =2021 c=3:D=5(3)^2 −4a(3)=45−12a⇒a=3 (i)⇒b^2 +bc+ac−c^2 =0 b^2 +b(3)+(3)(3)−(3)^2 =0 b(b+3)=0⇒b=0∨b=−3(rejected) (iii)⇒d=(((3)(3))/(3−0))−1⇒d=2 abcd^(−) =3032 c=4:D=5c^2 −4ac=5(4)^2 −4a(4) =80−16a⇒a=1,4,5 a=1: b^2 +b(4)+(1)(4)−(4)^2 =0 b^2 +4b−12=0⇒(b+6)(b−2)=0⇒b=2 d=((ac)/(c−b))−1=((4×1)/(4−2))−1⇒d=1 abcd^(−) =1241 a=4 b^2 +b(4)+(4)(4)−(4)^2 =0 b(b+4)=0⇒b=0 d=((4×4)/(4−0))−1=3⇒d=3 abcd^(−) =4043 a=5 b^2 +b(4)+(5)(4)−(4)^2 =0 b^2 +4b+4=0 (b+2)^2 =0⇒b=−2(rejected) Rejected case c=5:D =5c^2 −4ac=5(5)^2 −4a(5) =125−20a⇒a=5 b^2 +bc+ac−c^2 =0 b^2 +5b+25−25=0 b(b+5)=0⇒b=0,−5 (×) d=((ac)/(c−b))−1=((25)/(5−0))−1=4⇒d=4 abcd^(−) =5054 c=6:D =5c^2 −4ac=5(6)^2 −4a(6) =180−24a⇒a=6 b^2 +bc+ac−c^2 =0 b^2 +6b+36−36=0⇒b(b+6)=0⇒b=0,−6(×) od=((ac)/(c−b))−1=((36)/(6−0))−1⇒d=5 abcd^(−) =6065 c=7:D =5c^2 −4ac=5(7)^2 −4a(7) =245−28a⇒a=7 b^2 +bc+ac−c^2 =0 b^2 +7b+49−49=0⇒b(b+7)=0 b=0,−7(×) d=((ac)/(c−b))−1=((49)/(7−0))−1=6⇒d=6 abcd^(−) =7076 c=8:D =5c^2 −4ac=5(8)^2 −4a(8) =320−32a⇒a=2,8 a=2: b^2 +bc+ac−c^2 =0⇒b^2 +8b+16−64=0 b^2 +8b−48=0⇒(b+12)(b−4)=0⇒b=4,−12(×) d=((ac)/(c−b))−1=((16)/(8−4))−1=3⇒d=3 abcd^(−) =2483 a=8 b^2 +bc+ac−c^2 =0⇒b^2 +8b+64−64=0 b(b+8)=0⇒b=0,−8(×) d=((ac)/(c−b))−1⇒((64)/(8−0))−1=7⇒d=7 abcd^(−) =8087 c=9: D =5c^2 −4ac=5(9)^2 −4a(9)=405−36a ⇒a=5,9 a=5:b^2 +bc+ac−c^2 =0⇒b^2 +9b+45−81=0 b^2 +9b−36=0⇒(b+12)(b−3)=0⇒b=3,−12(×) d=((ac)/(c−b))−1=((45)/(9−3))−1(×) a=9:b^2 +bc+ac−c^2 =0⇒b^2 +9b+81−81=0 ⇒b(b+9)=0⇒b=0,−9(×) d=((ac)/(c−b))−1=((81)/(9−0))−1=8⇒d=8 abcd^(−) =9098
abcd―∈A,ad+1=c−bc=a+bb+c;A=?0⩽abcd⩽9;c≠0,a≠0(∵abcd―is4−digitnmbr)c−bc=a+bc+bc2−b2=ac+bcb2+bc+ac−c2=0………..(i)D=c2−4(ac−c2)D=5c2−4ac(mustbeperfectsquare)….(ii)ad+1=c−bc⇒d+1a=cc−b⇒d=acc−b−1……………..(iii)c=1:(iii)⇒D=5−4a(perfectsquare)⇒a=1(i)⇒b2+b(1)+(1)(1)−(1)2=0b(b+1)=0⇒b=0∨b=−1(rejected)(iii)⇒d=(1)(1)1−0−1=0⇒d=0abcd―=1010c=2:(ii)⇒D=5(2)2−4a(2)=20−8a⇒a=2(i)⇒b2+b(2)+(2)(2)−(2)2=0b(b+2)=0⇒b=0∨b=−2(rejected)(iii)⇒d=(2)(2)2−0−1⇒d=1abcd―=2021c=3:D=5(3)2−4a(3)=45−12a⇒a=3(i)⇒b2+bc+ac−c2=0b2+b(3)+(3)(3)−(3)2=0b(b+3)=0⇒b=0∨b=−3(rejected)(iii)⇒d=(3)(3)3−0−1⇒d=2abcd―=3032c=4:D=5c2−4ac=5(4)2−4a(4)=80−16a⇒a=1,4,5a=1:b2+b(4)+(1)(4)−(4)2=0b2+4b−12=0⇒(b+6)(b−2)=0⇒b=2d=acc−b−1=4×14−2−1⇒d=1abcd―=1241a=4b2+b(4)+(4)(4)−(4)2=0b(b+4)=0⇒b=0d=4×44−0−1=3⇒d=3abcd―=4043a=5b2+b(4)+(5)(4)−(4)2=0b2+4b+4=0(b+2)2=0⇒b=−2(rejected)Rejectedcasec=5:D=5c2−4ac=5(5)2−4a(5)=125−20a⇒a=5b2+bc+ac−c2=0b2+5b+25−25=0b(b+5)=0⇒b=0,−5(×)d=acc−b−1=255−0−1=4⇒d=4abcd―=5054c=6:D=5c2−4ac=5(6)2−4a(6)=180−24a⇒a=6b2+bc+ac−c2=0b2+6b+36−36=0⇒b(b+6)=0⇒b=0,−6(×)od=acc−b−1=366−0−1⇒d=5abcd―=6065c=7:D=5c2−4ac=5(7)2−4a(7)=245−28a⇒a=7b2+bc+ac−c2=0b2+7b+49−49=0⇒b(b+7)=0b=0,−7(×)d=acc−b−1=497−0−1=6⇒d=6abcd―=7076c=8:D=5c2−4ac=5(8)2−4a(8)=320−32a⇒a=2,8a=2:b2+bc+ac−c2=0⇒b2+8b+16−64=0b2+8b−48=0⇒(b+12)(b−4)=0⇒b=4,−12(×)d=acc−b−1=168−4−1=3⇒d=3abcd―=2483a=8b2+bc+ac−c2=0⇒b2+8b+64−64=0b(b+8)=0⇒b=0,−8(×)d=acc−b−1⇒648−0−1=7⇒d=7abcd―=8087c=9:D=5c2−4ac=5(9)2−4a(9)=405−36a⇒a=5,9a=5:b2+bc+ac−c2=0⇒b2+9b+45−81=0b2+9b−36=0⇒(b+12)(b−3)=0⇒b=3,−12(×)d=acc−b−1=459−3−1(×)a=9:b2+bc+ac−c2=0⇒b2+9b+81−81=0⇒b(b+9)=0⇒b=0,−9(×)d=acc−b−1=819−0−1=8⇒d=8abcd―=9098
Commented by Rasheed.Sindhi last updated on 14/Aug/21
1010,2021,3032,1241,4043,5054, 6065,7076,2483,8087,9098 (11 solutions)
1010,2021,3032,1241,4043,5054,6065,7076,2483,8087,9098(11solutions)
Commented by mathdanisur last updated on 14/Aug/21
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