Cycling Digits

Cycling Digits

 

I have in mind a number which when you remove the units digit and place it at the front, gives the same result as multiplying the original number by 2. Am I telling the truth?

Choose {a, b, c, d, e} each elements of the set (0, 1, 2, 3, 4, 5, 6, 7, 8, 9), forming numbers of the type with:

one digit that is a = 2a             | Implies a = 0 the only solution

two digit that is ba = 2(ab) or            | converting to units

10b + a = 2(ab) = 20a + 2b            |simplify

8b = 19a                    | Must belong to the set above,

| if a =8 then b =19 and it can not

ba = ab + 9(b-a)                    | reversing is 9 times difference of b and a

b=a gives the two digit Palindromes 11, 22, ...    | This does not give two times original number! False.

 

Three digit number of the form 2(cba) = acb    | converting to units

200c +20b + 2a = 100a + 10c + b            | simplifying

190c + 19b = 98a which is 19(10c+b)=98a    | to be integer solution c=9, b=8 and a=19 False.

 

cba = acb                     |work on solution

100c+ 10b+ a = 100a + 10c+ b            | simplify

90c + 9b = 99a    which is 10c + b = 11a        | a=b=c set of (0, 1, ..., 8, 9) three digit Palindromes

 

Perhaps ba = n(ab)                | Something to try!

10b + a = n(ab) = na + nb | convert to units

(10-n)b = (n-1)a | simplify

This problem has potential for more investigation when time permits or with your help.