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.