Solve this simple Maths?

Question

Find a number so that when you move the first digit to the last position, the result is 1.5 times the original number.

All other positions are constant (e.g. 1234 becomes 2341). The number of digits is not specified. I was told that this can be solved with grade 8 Maths (which I assume is algebra).

Answer 1

1176470588235294

Method: Suppose the number has n+1 digits. Let a be the first digit and b denote the other n digits. Then the number is 10^n*a+b and the number obtained by moving the first digit to the last position is 10b+a. Then we have:

10b+a = 3/2 (10^n*a+b)

Solving this equation for b:

20b+2a = 3*10^n*a+3b
17b=(3*10^n-2)a
b=(3*10^n-2)a/17

Now, since b is an integer, (3*10^n-2)a must be divisible by 17. However, a cannot be divisible by 17, since it is only a single digit, so we must have:

3*10^n-2≡0 mod 17

Solving this congruence:

3*10^n≡2 mod 17
18*10^n≡12 mod 17
10^n≡12 mod 17

Now, I know from Fermat's little theorem that 10^16 ≡ 1 mod 17, and I also know 12*10≡1 mod 17. Thus 10^16≡12*10 mod 17 and thus 10^15≡12 mod 17. Thus n must be 15. (this could also have been found by exhaustive search -- there are only 15 possibilities to check). So now substituting this into the original equation:

b=(3*10^15-2)a/17

Choosing the simplest value for a, namely a=1, yields:

b=(3*10^15-2)/17 = 176470588235294

Thus the number we are looking for is 1176470588235294

Check: if we move the first digit to the last position, we obtain 1764705882352941, which is indeed 1.5 times the original number.

Answer 2

how many digits are there in the number??

Answer 3

are other all positions constant ?
while placing first digit to last, do last digit becomes second last ?

please put clear question.