Question Number 103888 by mr W last updated on 18/Jul/20

find all such numbers:  if we make its last digit, say k, as its  first digit, the number becomes k  times large as before.  (□□...□k)→(k□□...□)=k×(□□...□k)

Answered by MAB last updated on 18/Jul/20

x=Σ_(i=0) ^n 10^i a_i    where a_i ∈{0,1...9} and a_0 =k  x′=10^n k+Σ_(i=0) ^(n−1) 10^i a_(i+1) =kx  hence  10^n k+Σ_(i=0) ^(n−1) 10^i a_(i+1) =k^2 +Σ_(i=0) ^(n−1) 10^(i+1) a_(i+1)   (10^n −k)k=9Σ_(i=0) ^(n−1) 10^i a_(i+1)   (10^n −k)k=9kx  10^n −k=9x  k=1 and x=111...1

Commented bymr W last updated on 18/Jul/20

thanks sir!

Answered by mr W last updated on 18/Jul/20

let x=□□...□□_(↽ n digits⇁)  ≤10^n −1  N=□□...□□_(↽ n digits⇁) k=10x+k  k□□...□□_(↽ n digits⇁) =10^n k+x  10^n k+x=k(10x+k)  ⇒x=((k(10^n −k))/(10k−1)) ≤10^n −1    case k=1:  x=((10^n −1)/9)=111..111_(↽ n digits ⇁)   x=((10^n −1)/9)=111..111_(↽ n digits ⇁)   ⇒N=111..111_(↽ n digits ⇁) 1 with n∈N    case k=2:  x=((2(10^n −2))/(19))  we can find  n=18m−1 with m∈N  example n=17:  x=10526315789473784  ⇒N=105263157894737842  check:  2×105263157894737842     =210526315789473784  example n=35:  x=10526315789473784210526315789473784  ⇒N=105263157894737842105263157894737842  check:  2×105263157894737842105263157894737842     =210526315789473784210526315789473784    (to be continued)

Commented bymr W last updated on 18/Jul/20

case k=3:  here we′ll try an other way to find  the number x.  note: in following the digits of the  number are displayed inversely, i.e.  the last digit is the left most.            39731427155698602685728443013  ×3   97314271556986026857284430139            ∣← last digit              first digit →∣  ⇒ x=103448275862068965517241379  ⇒N=1034482758620689655172413793  check:  3×1034482758620689655172413793     =3103448275862068965517241379  we see the digits of number N may be  repeated any times. so we have  infinite numbers which satisfy the  requirement.

Commented bymr W last updated on 18/Jul/20

case k=4:  using the same method as before,           4652014  ×4  6520146  ⇒ x=10256  ⇒N=102564  check:  4×102564     =410256  similarly the digits of N may be  repeated any times. for example  N=102564102564102564

Commented bymr W last updated on 18/Jul/20

case k=5:  x=((k(10^n −k))/(10k−1))=((5(10^n −5))/(7×7))  it seems to have no solution.

Commented bymr W last updated on 19/Jul/20

case k=6:  x=((k(10^n −k))/(10k−1))=((6(10^n −6))/(59))  it seems to have no solution.