Cycling Digits

Cycling Digits from "Thinking Mathematically" page 165

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) | Here is the solution and it is not two

| times the reverse number. False.

| See solution on left above. The reverse

| of ba is ab plus 9 times difference of b

|and a, therefor not equal to twice ab.

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

|number! False.

Three digit numbers 2(cba) = acb | converting to units

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

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

cba = acb + 9( combination of 21 for #s 1 apart?) |work on solution as they are not equal

cba = acb + n (189= 21*9) | the RHS is acb plus or minus some

cba = acb + n (189= 21*9) |multiple of 9, sign depending order of -

|Again not equal to 2acb.

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 111, 222, ..., 888, 999

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

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

Sorry about the image quality tha is why I left the type, but the alignment is ???