What's the smallest number with first digit 1 that triples when this digit is moved to the end? There's this homework question I have, and while I know people generally don't like these, I would like a hint on how to get started please.
A positive integer begins with the digit 1 when written in decimal. When this digit is transferred to the end of the number, the number is tripled. Find the smallest number that has this property.
I know it has something to do with divisibility by 3.
How should I start? Could someone please give an insightful comment that might let me solve it (without feeling like cheating)?
Edit: I suppose that transfer means that 12345 would become 23451...
 A: Here's a slightly different solution:
We must have
$$
10x + 1 = 3(x+10^k)
$$
for some non-negative integers $x$ and $k$.
Simplifying, we get
$$
7x = 3\times 10^k - 1
$$
so
$$
3\times 10^k \equiv 1 (\mod 7)
$$ 
or 
$$
3 ^ {k+1} \equiv 1 (\mod 7)
$$
Working mod 7, we have $3^2\equiv2$, $3^3\equiv6$, $3^4\equiv4$, $3^5\equiv5$ and $3^6\equiv1$, so the minimal $k$ is 5. 
Thus 
$$
7x = 3\times10^5-1 = 299,999
$$
and $x=42,857$; the numbers are $428,571 = 3 \times 142,857$.
A: $x=1\cdot 10^n + t$, with $t < 10^n$
$y=10t+1 \implies y=3x$
$10t+1 = 3 \cdot 10^n + 3t \implies 7t=3 \cdot 10^n-1$
So the question is: what is the smallest $n$ such that $7$ divides $3 \cdot 10^n-1$ ?
Another way to express this is: how long is the period in $3/7$ when expressed in decimal?
This will give you the smallest $x$.
A: I guess the last digit is a 7 because only $7 \times 3 =21$ will produce a number which ends with 1. I would start by showing that a two digit number is impossible (17) and then try $1a7$ or $1ab7$ writing the equations in the form (e.g. if it was a 4 digit number):
$$
 3 \times(1ab7) = ab71
$$
From this you can extract information by writing out the decimal expansion, for instance:
$$1ab7 = 1000 + 100 \times a + 10 \times b + 7$$
and try to figure out how $a$ and $b$ are related to 
A: Here is an alternative more elementary solution (expanding on Rogelios answer):
Let's call the number with the $1$ in the front $a$ and the other one $b$. Then $a \times 3 = b$.
We know that $b$ has a $1$ in its last place so $a$ must have a $7$ in its last place because $7 \times 3 = 21$ is the only product of three which has a $1$ in its last place. So
$a = 7$ and $b = 7 \times 3 =21$
Now the last digit of both numbers is correct. However, if $a$ ends with $7$ then $b$ needs to end with $71$ instead of $21$. So we lack another $50$. How can we get another $50$ into $b$? By adding $50$ to $a$! Because $5 \times 3 = 15$ is the only product of three which has a $5$ in its last place. So
$a = 57$ and $b = 57 \times 3 = 171$
Now the last two digits are correct. However, as above, if $a$ ends with $57$ then $b$ needs to end with $571$ instead of $171$ so we need another $400$. How can we get another 400 into $b$? By adding $800$ to $a$! Because $8 \times 3 = 24$ is the only product of three which has a $4$ in its last place. So
$a = 857$ and $b = 857 \times 3 = 2571$
Now the last three digits are correct. Can you take it from there?
A: Hints: last digit should be 7.
Number is 142857
