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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-

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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,b∈N.[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+b−a−b)+2z−z=0w{−3a2b−3ab2}+x{−2ab}+z=0w{3a2b+3ab2}+2xab−z=0Theformoftheaboveequationisindependentofthevalueofy∈N.z=ab(2x+3w(a+b))w,x∈Z⩾,a,b∈N−{1}andab∣z.Ifx>0orw>0then2x+3w(a+b)>1⇒z>ab>aandz>b.However,0⩽z⩽min(a−1,b−1)<min(a,b)forzinwxyzaandwxyzb,acontradiction.Hence,wemusthavex=w=0⇒z=0.∴weget00y0a+00y0b=00y0a+bory0a+y0b=y0a+borya+yb=y(a+b)foranyy∈N,0⩽y<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,x∈Z⩾?
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,x∈Z⩾≡x,warenon−negativeintegers.
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
Letqbeanumberofdigitlengthn⩾3,andc,b∈[N−{1}].Writekmforthenumberkcorrespondingtobasem⩾2.Supposethatqc+qb=qc+b.Then,noqexistssatisfyingthisequationforn⩾3.−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−PROOF:Lookingattheequationqc+qb=qc+b,qhasadigitrepresentationoftheformq=a¯0a¯1a¯2a¯3…a¯n−1where0⩽ai⩽min(c−1,b−1)fori=1,2,…,n−1,and1⩽a0⩽min(c−1,b−1).Thereforeqc=∑n−1i=0aicn−1−i,qb=∑n−1i=1aibn−1−iandqc+b=∑n−1i=0ai(b+c)n−1−i.∑n−1i=0aicn−1−i+∑n−1i=0aibn−1−i=∑n−1i=0ai(b+c)n−1−i∑n−1i=0{aicn−1−i+aibn−1−i−ai(b+c)n−1−i}=0∑n−1i=0[ai{cn−1−i+bn−1−i−(b+c)n−1−i}]=0∑n−3i=0[ai{cn−1−i+bn−1−i−(b+c)n−1−i}]+an−2{cn−1−n+2+bn−1−n+2−(b+c)n−1−n+2}+an−1{cn−1−n+1+bn−1−n+1−(b+c)n−1−n+1}=0∑n−3i=0[ai{cn−1−i+bn−1−i−(b+c)n−1−i}]+an−2{c+b−(b+c)}+an−1{1+1−1}=0∑n−3i=0[ai{cn−1−i+bn−1−i−(b+c)n−1−i}]+an−1=0an−1=∑n−3i=0[ai{(b+c)n−1−i−cn−1−i−bn−1−i}]AccordingtotheBinomialTheorem,(b+c)n−1−i=∑n−1−ik=0(n−1−ik)bn−1−i−kckor(b+c)n−1−i=bn−1−i+cn−1−i+∑n−2−ik=1(n−1−ik)bn−1−i−kck.∴(b+c)n−1−i−bn−1−i−cn−1−i=∑n−2−ik=1(n−1−ik)bn−1−i−kck.an−1=∑n−3i=0[ai{∑n−2−ik=1(n−1−ik)bn−1−i−kck}]an−1=a0{∑n−2k=1(n−1k)bn−1−kck}+a1{∑n−3k=1(n−2k)bn−2−kck}+a2{∑n−4k=1(n−3k)bn−3−kck}+…+an−4{∑2k=1(3k)b3−kck}+an−3{∑1k=1(2k)b2−kck}Observethattheformoftheaboveequationisindepedentofan−2;soan−2isafreevariablesuchthat0⩽an−2⩽min(c−1,b−1).−−−−−−−−−−−−−−−−−−−−−−−−−−−−−Proposition:Foranyn⩾3,bcdivides∑n−2k=1(n−1k)bn−1−kck.Proof(induction):Basecasen=3:∑3−2k=1(3−1k)b3−1−kck=(21)bc⇒bcdivides∑n−2k=1(n−1k)bn−1−kckforn=3.Inductivestep:Supposebcdivides∑j−2k=1(j−1k)bj−1−kckforn=j.i.e∑j−2k=1(j−1k)bj−1−kck=rbcforsomer∈N.Forn=j+1,wehavethat∑j−1k=1(jk)bj−kck=∑j−1k=1{(j−1k)+(j−1k−1)}bj−kck
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)=∑n−2k=1(n−1k)bn−1−kck,n⩾3.Sincebcdividest(n)forn⩾3,thenbcdividesan−1sincean−1=a0t(n)+a1t(n−1)+a2t(n−2)+a3t(n−3)+…+an−3t(3).Now,b>1,c>1andallai(i=0,1,2,…)arenon−negativeintegers,wheremin(a0)=1.∴an−1=bcla0+…(l∈N).But,ifmin(a0)=1⇒an−1>bandan−1>c,while0⩽an−1⩽min(c−1,b−1)<min(c,b).Thiscontradictionindicatesthatqcannotbeanumberhavingdigitlengthn⩾3,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. −−−−−−−−−−−−−−−−−−−−−−−−−−−
∑j−1k=1(jk)bj−kck=b{∑j−2k=1(j−1k)bj−1−kck+(j−1j−1)b0c}+∑j−1k=1(j−1k−1)bj−kck∑j−1k=1(jk)bj−kck=b×rbc+bc+∑j−1k=1(j−1k−1)bj−kck∑j−1k=1(jk)bj−kck=bc(br+1)+{∑j−2k=1(j−1k−1)bj−kck+(j−1j−2)bj−j+1cj−1}∑j−1k=1(jk)bj−kck=bc(br+1)+(j−1)bcj−1+{∑j−2k=1(j−1k−1)bj−kck}∑j−1k=1(jk)bj−kck=bc(br+1)+(j−1)bcj−1+bj−1c+{c(j−11)bj−2c+c(j−12)bj−3c2+…+c(j−1j−3)b2cj−3+(j−1j−2)bcj−2−(j−1j−2)bcj−2}∑j−1k=1(jk)bj−kck=bc(br+1)+(j−1)bcj−1+bj−1c+c{(j−11)bj−2c+(j−12)bj−3c2+…+(j−1j−3)b2cj−3+(j−1j−2)bcj−2}−(j−1j−2)bcj−1∑j−1k=1(jk)bj−kck=bc(br+1)+(j−1)bcj−1+bj−1c+c∑j−2k=1(j−1k)bj−1−kck−(j−1)bcj−1∑j−1k=1(jk)bj−kck=bc(br+1+cr)+bj−1cSincej⩾3⇒j−1>1⇒bcdividesbj−1c.Therefore,bcdivides∑j−1k=1(jk)bj−kck,ifbcdivides∑j−2k=1(j−1k)bj−1−kck.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!

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