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...

  • 1
    $\begingroup$ What does "transferred to the end of the number" mean? Can you give an example? $\endgroup$ – tangrs Jun 16 '15 at 2:55
  • $\begingroup$ How about writing a normal title? Whom will this randomness help? $\endgroup$ – Salvador Dali Jun 16 '15 at 4:26
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    $\begingroup$ For completeness, I don't think the community on the whole dislikes homework questions; if you're asking a homework question, we just prefer that you show some degree of effort first. (This is just a general assessment; I'm not making a judgment on this particular post.) $\endgroup$ – Ken Jun 16 '15 at 4:32

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?


$x=1\cdot 10^n + t$, 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 $x$.

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    $\begingroup$ Decimal expansions of fractions of the form $a/7$ are fun. $\endgroup$ – lhf Jun 16 '15 at 3:02
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    $\begingroup$ Yeah, given my knowledge of $142857$ and its properties, my personal answer to the OP was "OK $142857$ is an answer, is there a smaller one?" $\endgroup$ – Mark Hurd Jun 17 '15 at 1:30
  • $\begingroup$ No, the other way to express this is: how long is the digit sequence in $3/7$ after the decimal point before periodic repetition begins? $t=\frac{3}{7}\cdot 10^n-\frac{1}{7}$ and $\frac{3}{7}=0.42857\overline{142857}$ and $\frac{1}{7}=0.\overline{142857}$, so the least $n$ is $|\{4,2,8,5,7\}|=5$. All integer solutions are given when $n=6k+5,\, k\in\Bbb Z_{\ge 0}$. $\endgroup$ – user26486 Jun 17 '15 at 8:13

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

$$ 7x = 3\times 10^k - 1 $$


$$ 3\times 10^k \equiv 1 (\mod 7) $$


$$ 3 ^ {k+1} \equiv 1 (\mod 7) $$

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.


$$ 7x = 3\times10^5-1 = 299,999 $$

and $x=42,857$; the numbers are $428,571 = 3 \times 142,857$.

  • $\begingroup$ Ah, I guess this is the same as @lhf above, but presented slightly differently at the end. $\endgroup$ – james_d Jun 16 '15 at 14:05

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

  • $\begingroup$ Ok, thanks about that! I completely forgot about the 7's. $\endgroup$ – Kbot Jun 16 '15 at 2:58
  • $\begingroup$ Good approach, but I think you have this backwards. Shouldn't it be 3*(1ab7) = ab71 ? $\endgroup$ – dberm22 Jun 16 '15 at 12:08
  • $\begingroup$ I am sorry, indeed it is backwards! I will fix it. $\endgroup$ – Rogelio Molina Jun 16 '15 at 16:58

Hints: last digit should be 7. Number is 142857

  • $\begingroup$ If one takes all other digits as a number say x then equation would look like: 100x + 71 = 3*(10 * 10^floor(logx) + 10x +7). further simplification gives (7x + 5)/3 should be powers of 10. that means 7x can take values 25, 295, 2995, 29995 an so on. since we have to get least possible one, we can try divisibility by 7 on these numbers. the least one would be 29995. so x = 29995/7 that is 4285. hence the number is 142857. $\endgroup$ – Amitesh Singh Jun 16 '15 at 9:29
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    $\begingroup$ You know, you can edit your question, if you feel like adding details. $\endgroup$ – user228113 Jun 16 '15 at 9:30
  • $\begingroup$ sorry for delay in giving details. $\endgroup$ – Amitesh Singh Jun 16 '15 at 9:31

protected by Community Jun 17 '15 at 10:08

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