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w-x-y-z-are-digits-in-respective-base-system-and-a-b-are-bases-Find-out-an-example-examples-which-satisfy-the-following-wxyz-a-wxyz-b-wxyz-a-b-




Question Number 7249 by Rasheed Soomro last updated on 19/Aug/16
w,x,y,z are digits in respective base system and  a,b are bases.  Find out an example/examples which satisfy  the following                 wxyz_a +wxyz_b =wxyz_(a+b)
w,x,y,zaredigitsinrespectivebasesystemanda,barebases.Findoutanexample/exampleswhichsatisfythefollowingwxyza+wxyzb=wxyza+b
Commented by Yozzia last updated on 19/Aug/16
2^1 +2^1 =(2+2)^1 =4^1   {1×2+0×2^0 }+{1×2+0×2^0 }=1×4+0×4^0   or (10)_2 +(10)_2 =(10)_4
21+21=(2+2)1=41{1×2+0×20}+{1×2+0×20}=1×4+0×40or(10)2+(10)2=(10)4
Commented by Rasheed Soomro last updated on 19/Aug/16
Four digit number is required.
Fourdigitnumberisrequired.
Commented by Yozzia last updated on 19/Aug/16
wxyz_a +wxyz_b =wxyz_(a+b)     (∗)  Suppose wxyz is an integer identical on both sides of (∗)  e.g  1234_a +1234_b =1234_(a+b)    for some given a,b∈N.  [wa^3 +xa^2 +ya+z]+[wb^3 +xb^2 +yb+z]=w(a+b)^3 +x(a+b)^2 +y(a+b)+z  w{a^3 +b^3 −(a+b)^3 }+x{a^2 +b^2 −(a+b)^2 }+y(a+b−a−b)+2z−z=0  w{−3a^2 b−3ab^2 }+x{−2ab}+z=0  w{3a^2 b+3ab^2 }+2xab−z=0    The form of the above equation is   independent of the value of y∈N.  z=ab(2x+3w(a+b))  w,x∈Z^≥ , a,b∈N−{1} and ab∣z. If x>0 or w>0  then 2x+3w(a+b)>1⇒z>ab>a and z>b.  However, 0≤z≤min(a−1,b−1)<min(a,b)  for z in wxyz_a  and wxyz_b , a contradiction.  Hence, we must have x=w=0 ⇒ z=0.  ∴ we get 00y0_a +00y0_b =00y0_(a+b)   or y0_a +y0_b =y0_(a+b)   or ya+yb=y(a+b) for any y∈N ,0≤y<a,y<b  when a and b are known bases.  No four digit integer wxyz satisfies  wxyz_a +wxyz_b =wxyz_(a+b) .
wxyza+wxyzb=wxyza+b()Supposewxyzisanintegeridenticalonbothsidesof()e.g1234a+1234b=1234a+bforsomegivena,bN.[wa3+xa2+ya+z]+[wb3+xb2+yb+z]=w(a+b)3+x(a+b)2+y(a+b)+zw{a3+b3(a+b)3}+x{a2+b2(a+b)2}+y(a+bab)+2zz=0w{3a2b3ab2}+x{2ab}+z=0w{3a2b+3ab2}+2xabz=0TheformoftheaboveequationisindependentofthevalueofyN.z=ab(2x+3w(a+b))w,xZ,a,bN{1}andabz.Ifx>0orw>0then2x+3w(a+b)>1z>ab>aandz>b.However,0zmin(a1,b1)<min(a,b)forzinwxyzaandwxyzb,acontradiction.Hence,wemusthavex=w=0z=0.weget00y0a+00y0b=00y0a+bory0a+y0b=y0a+borya+yb=y(a+b)foranyyN,0y<a,y<bwhenaandbareknownbases.Nofourdigitintegerwxyzsatisfieswxyza+wxyzb=wxyza+b.
Commented by Rasheed Soomro last updated on 20/Aug/16
V. NicE!  What is meant by :     w,x∈Z^≥    ?
V.NicE!Whatismeantby:w,xZ?
Commented by Yozzia last updated on 20/Aug/16
Z^≥ ={0,1,2,3,4,...}  w,x∈Z^≥ ≡x,w are non−negative integers.
Z={0,1,2,3,4,}w,xZx,warenonnegativeintegers.
Commented by Rasheed Soomro last updated on 20/Aug/16
ThαnkS!
ThαnkS!
Commented by Yozzia last updated on 20/Aug/16
Let q be a number of digit length n≥3,  and c,b∈[N−{1}]. Write k_m  for the  number k corresponding to base m≥2.  Suppose that q_c +q_b =q_(c+b) .   Then, no q exists satisfying this equation  for n≥3.  −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−  PROOF:  Looking at the equation q_c +q_b =q_(c+b) ,  q has a digit representation of the form  q=a_0 ^� a_1 ^� a_2 ^� a_3 ^� ...a_(n−1) ^�  where 0≤a_i ≤min(c−1,b−1)  for i=1 , 2 , ... , n−1, and 1≤a_0 ≤min(c−1,b−1).  Therefore q_c =Σ_(i=0) ^(n−1) a_i c^(n−1−i)  , q_b =Σ_(i=1) ^(n−1) a_i b^(n−1−i)   and q_(c+b) =Σ_(i=0) ^(n−1) a_i (b+c)^(n−1−i) .   Σ_(i=0) ^(n−1) a_i c^(n−1−i) +Σ_(i=0) ^(n−1) a_i b^(n−1−i) =Σ_(i=0) ^(n−1) a_i (b+c)^(n−1−i)   Σ_(i=0) ^(n−1) {a_i c^(n−1−i) +a_i b^(n−1−i) −a_i (b+c)^(n−1−i) }=0  Σ_(i=0) ^(n−1) [a_i {c^(n−1−i) +b^(n−1−i) −(b+c)^(n−1−i) }]=0  Σ_(i=0) ^(n−3) [a_i {c^(n−1−i) +b^(n−1−i) −(b+c)^(n−1−i) }]+a_(n−2) {c^(n−1−n+2) +b^(n−1−n+2) −(b+c)^(n−1−n+2) }+a_(n−1) {c^(n−1−n+1) +b^(n−1−n+1) −(b+c)^(n−1−n+1) }=0  Σ_(i=0) ^(n−3) [a_i {c^(n−1−i) +b^(n−1−i) −(b+c)^(n−1−i) }]+a_(n−2) {c+b−(b+c)}+a_(n−1) {1+1−1}=0  Σ_(i=0) ^(n−3) [a_i {c^(n−1−i) +b^(n−1−i) −(b+c)^(n−1−i) }]+a_(n−1) =0  a_(n−1) =Σ_(i=0) ^(n−3) [a_i {(b+c)^(n−1−i) −c^(n−1−i) −b^(n−1−i) }]    According to the Binomial Theorem,  (b+c)^(n−1−i) =Σ_(k=0) ^(n−1−i)  (((n−1−i)),(k) ) b^(n−1−i−k) c^k   or (b+c)^(n−1−i) =b^(n−1−i) +c^(n−1−i) +Σ_(k=1) ^(n−2−i)  (((n−1−i)),(k) ) b^(n−1−i−k) c^k .  ∴ (b+c)^(n−1−i) −b^(n−1−i) −c^(n−1−i) =Σ_(k=1) ^(n−2−i)  (((n−1−i)),(k) ) b^(n−1−i−k) c^k .  a_(n−1) =Σ_(i=0) ^(n−3) [a_i {Σ_(k=1) ^(n−2−i)  (((n−1−i)),(k) ) b^(n−1−i−k) c^k }]   a_(n−1) =a_0 {Σ_(k=1) ^(n−2)  (((n−1)),(k) ) b^(n−1−k) c^k }+a_1 {Σ_(k=1) ^(n−3)  (((n−2)),(k) ) b^(n−2−k) c^k }                +a_2 {Σ_(k=1) ^(n−4)  (((n−3)),(k) ) b^(n−3−k) c^k }+...+a_(n−4) {Σ_(k=1) ^2  ((3),(k) ) b^(3−k) c^k }+a_(n−3) {Σ_(k=1) ^1  ((2),(k) ) b^(2−k) c^k }  Observe that the form of the above equation  is indepedent of a_(n−2) ; so a_(n−2)  is a free  variable such that 0≤a_(n−2) ≤min(c−1,b−1).  −−−−−−−−−−−−−−−−−−−−−−−−−−−−−  Proposition: For any n≥3, bc divides Σ_(k=1) ^(n−2)  (((n−1)),(k) ) b^(n−1−k) c^k .  Proof (induction): Base case n=3: Σ_(k=1) ^(3−2)  (((3−1)),(k) ) b^(3−1−k) c^k = ((2),(1) ) bc  ⇒bc divides Σ_(k=1) ^(n−2)  (((n−1)),(k) ) b^(n−1−k) c^k  for n=3.    Inductive step: Suppose bc divides Σ_(k=1) ^(j−2)  (((j−1)),(k) ) b^(j−1−k) c^k  for n=j.  i.e Σ_(k=1) ^(j−2)  (((j−1)),(k) ) b^(j−1−k) c^k =rbc for some r∈N.  For n=j+1, we have that Σ_(k=1) ^(j−1)  ((j),(k) ) b^(j−k) c^k =Σ_(k=1) ^(j−1) { (((j−1)),(k) ) + (((j−1)),((k−1)) )}b^(j−k) c^k
Letqbeanumberofdigitlengthn3,andc,b[N{1}].Writekmforthenumberkcorrespondingtobasem2.Supposethatqc+qb=qc+b.Then,noqexistssatisfyingthisequationforn3.PROOF:Lookingattheequationqc+qb=qc+b,qhasadigitrepresentationoftheformq=a¯0a¯1a¯2a¯3a¯n1where0aimin(c1,b1)fori=1,2,,n1,and1a0min(c1,b1).Thereforeqc=n1i=0aicn1i,qb=n1i=1aibn1iandqc+b=n1i=0ai(b+c)n1i.n1i=0aicn1i+n1i=0aibn1i=n1i=0ai(b+c)n1in1i=0{aicn1i+aibn1iai(b+c)n1i}=0n1i=0[ai{cn1i+bn1i(b+c)n1i}]=0n3i=0[ai{cn1i+bn1i(b+c)n1i}]+an2{cn1n+2+bn1n+2(b+c)n1n+2}+an1{cn1n+1+bn1n+1(b+c)n1n+1}=0n3i=0[ai{cn1i+bn1i(b+c)n1i}]+an2{c+b(b+c)}+an1{1+11}=0n3i=0[ai{cn1i+bn1i(b+c)n1i}]+an1=0an1=n3i=0[ai{(b+c)n1icn1ibn1i}]AccordingtotheBinomialTheorem,(b+c)n1i=n1ik=0(n1ik)bn1ikckor(b+c)n1i=bn1i+cn1i+n2ik=1(n1ik)bn1ikck.(b+c)n1ibn1icn1i=n2ik=1(n1ik)bn1ikck.an1=n3i=0[ai{n2ik=1(n1ik)bn1ikck}]an1=a0{n2k=1(n1k)bn1kck}+a1{n3k=1(n2k)bn2kck}+a2{n4k=1(n3k)bn3kck}++an4{2k=1(3k)b3kck}+an3{1k=1(2k)b2kck}Observethattheformoftheaboveequationisindepedentofan2;soan2isafreevariablesuchthat0an2min(c1,b1).\boldsymbolProposition:Foranyn3,bcdividesn2k=1(n1k)bn1kck.Proof(induction):Basecasen=3:32k=1(31k)b31kck=(21)bcbcdividesn2k=1(n1k)bn1kckforn=3.Inductivestep:Supposebcdividesj2k=1(j1k)bj1kckforn=j.i.ej2k=1(j1k)bj1kck=rbcforsomerN.Forn=j+1,wehavethatj1k=1(jk)bjkck=j1k=1{(j1k)+(j1k1)}bjkck
Commented by Yozzia last updated on 20/Aug/16
Let t(n)=Σ_(k=1) ^(n−2)  (((n−1)),(k) ) b^(n−1−k) c^k , n≥3.  Since bc divides t(n) for n≥3,then  bc divides a_(n−1)  since a_(n−1) =a_0 t(n)+a_1 t(n−1)+a_2 t(n−2)+a_3 t(n−3)+...+a_(n−3) t(3).  Now, b>1, c>1 and all a_i   (i=0,1,2,...)  are non−negative integers, where min(a_0 )=1.  ∴a_(n−1) =bcla_0 +... (l∈N). But, if min(a_0 )=1⇒a_(n−1) >b and a_(n−1) >c,  while 0≤a_(n−1) ≤min(c−1,b−1)<min(c,b).  This contradiction indicates that q  cannot be a number having digit length  n≥3, and q_c +q_b =q_(b+c) .                        ////
Lett(n)=n2k=1(n1k)bn1kck,n3.Sincebcdividest(n)forn3,thenbcdividesan1sincean1=a0t(n)+a1t(n1)+a2t(n2)+a3t(n3)++an3t(3).Now,b>1,c>1andallai(i=0,1,2,)arenonnegativeintegers,wheremin(a0)=1.an1=bcla0+(lN).But,ifmin(a0)=1an1>bandan1>c,while0an1min(c1,b1)<min(c,b).Thiscontradictionindicatesthatqcannotbeanumberhavingdigitlengthn3,andqc+qb=qb+c.////
Commented by Yozzia last updated on 20/Aug/16
Σ_(k=1) ^(j−1)  ((j),(k) ) b^(j−k) c^k =b{Σ_(k=1) ^(j−2)  (((j−1)),(k) ) b^(j−1−k) c^k + (((j−1)),((j−1)) ) b^0 c}+Σ_(k=1) ^(j−1)  (((j−1)),((k−1)) ) b^(j−k) c^k   Σ_(k=1) ^(j−1)  ((j),(k) ) b^(j−k) c^k =b×rbc+bc+Σ_(k=1) ^(j−1)  (((j−1)),((k−1)) ) b^(j−k) c^k   Σ_(k=1) ^(j−1)  ((j),(k) ) b^(j−k) c^k =bc(br+1)+{Σ_(k=1) ^(j−2)  (((j−1)),((k−1)) ) b^(j−k) c^k + (((j−1)),((j−2)) ) b^(j−j+1) c^(j−1) }  Σ_(k=1) ^(j−1)  ((j),(k) ) b^(j−k) c^k =bc(br+1)+(j−1)bc^(j−1) +{Σ_(k=1) ^(j−2)  (((j−1)),((k−1)) ) b^(j−k) c^k }  Σ_(k=1) ^(j−1)  ((j),(k) ) b^(j−k) c^k =bc(br+1)+(j−1)bc^(j−1) +b^(j−1) c+{c (((j−1)),(1) ) b^(j−2) c+c (((j−1)),(2) ) b^(j−3) c^2 +...+c (((j−1)),((j−3)) ) b^2 c^(j−3) + (((j−1)),((j−2)) ) bc^(j−2) − (((j−1)),((j−2)) ) bc^(j−2) }  Σ_(k=1) ^(j−1)  ((j),(k) ) b^(j−k) c^k =bc(br+1)+(j−1)bc^(j−1) +b^(j−1) c+c{ (((j−1)),(1) ) b^(j−2) c+ (((j−1)),(2) ) b^(j−3) c^2 +...+ (((j−1)),((j−3)) ) b^2 c^(j−3) + (((j−1)),((j−2)) ) bc^(j−2) }− (((j−1)),((j−2)) ) bc^(j−1)   Σ_(k=1) ^(j−1)  ((j),(k) ) b^(j−k) c^k =bc(br+1)+(j−1)bc^(j−1) +b^(j−1) c+cΣ_(k=1) ^(j−2)  (((j−1)),(k) ) b^(j−1−k) c^k −(j−1) bc^(j−1)   Σ_(k=1) ^(j−1)  ((j),(k) ) b^(j−k) c^k =bc(br+1+cr)+b^(j−1) c  Since j≥3⇒ j−1>1 ⇒bc divides b^(j−1) c.  Therefore, bc divides Σ_(k=1) ^(j−1)  ((j),(k) ) b^(j−k) c^k , if bc divides Σ_(k=1) ^(j−2)  (((j−1)),(k) ) b^(j−1−k) c^k .  Hence, by P.M.I, the proposition is true.  −−−−−−−−−−−−−−−−−−−−−−−−−−−
j1k=1(jk)bjkck=b{j2k=1(j1k)bj1kck+(j1j1)b0c}+j1k=1(j1k1)bjkckj1k=1(jk)bjkck=b×rbc+bc+j1k=1(j1k1)bjkckj1k=1(jk)bjkck=bc(br+1)+{j2k=1(j1k1)bjkck+(j1j2)bjj+1cj1}j1k=1(jk)bjkck=bc(br+1)+(j1)bcj1+{j2k=1(j1k1)bjkck}j1k=1(jk)bjkck=bc(br+1)+(j1)bcj1+bj1c+{c(j11)bj2c+c(j12)bj3c2++c(j1j3)b2cj3+(j1j2)bcj2(j1j2)bcj2}j1k=1(jk)bjkck=bc(br+1)+(j1)bcj1+bj1c+c{(j11)bj2c+(j12)bj3c2++(j1j3)b2cj3+(j1j2)bcj2}(j1j2)bcj1j1k=1(jk)bjkck=bc(br+1)+(j1)bcj1+bj1c+cj2k=1(j1k)bj1kck(j1)bcj1j1k=1(jk)bjkck=bc(br+1+cr)+bj1cSincej3j1>1bcdividesbj1c.Therefore,bcdividesj1k=1(jk)bjkck,ifbcdividesj2k=1(j1k)bj1kck.Hence,byP.M.I,thepropositionistrue.
Commented by Rasheed Soomro last updated on 20/Aug/16
Wow! What a depth you have in mathematics!  Good wishes for you and your future!
Wow!Whatadepthyouhaveinmathematics!Goodwishesforyouandyourfuture!
Commented by Yozzia last updated on 20/Aug/16
Thanks for your kind words!
Thanksforyourkindwords!Thanksforyourkindwords!

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