All credit for this problem goes to Dr. J. Bronowski.

Find the least positive integer such that moving the leading digit to the end produces a new integer one and a half times the original.

  • $\begingroup$ For the record, does "moving the leading digit to the rear" mean 123 => 312 or is it 123 => 231? $\endgroup$ Commented May 27, 2015 at 20:39
  • $\begingroup$ Find $123$ such as $231 = 123\times 1.5$ $\endgroup$
    – Masclins
    Commented May 27, 2015 at 20:41
  • $\begingroup$ Forgive my ignorance, but are negative numbers still classified as integers? (Not that it matters much, since the further you go negative the further you get away from the 1.5 margin) $\endgroup$ Commented May 27, 2015 at 20:47
  • $\begingroup$ @BradChristie negative numbers are classified as integers, yes. $\endgroup$
    – Bailey M
    Commented May 27, 2015 at 20:48
  • $\begingroup$ I simply copied the problem as posed by Bronowski, but it's intended to be positive, so I edited the question $\endgroup$
    – Masclins
    Commented May 27, 2015 at 20:54

3 Answers 3


The smallest number I could find is


My process for finding it:

Start with 1 as the number. To find more digits, multiply the current number by 1.5 and use the first case where the nth digit of the result is not the same as the n+1th digit of the number. Set the n+1th digit of your number to be that nth digit. Repeat until we find a number that works.

An example of my process:

number is 1
1.0 * 1.5 = 1.5
- ^ ^ <- 0 doesn't match 1
number is now 11
11.0 * 1.5 = 16.5
- ^ ^ <- 0 doesn't match 6
number is now 116
116.0 * 1.5 = 174
- ^ ^ <- 6 doesn't match 7
number is now 117
117.0 * 1.5 = 175.5
- ^ ^ <- 0 doesn't match 5
number is now 1175


number is now 1176470588235293
1176470588235293 * 1.5 = 1764705882352939.5

11764705882352939 * 1.5 = 17647058823529408.5

1176470588235294 * 1.5 = 1764705882352941 <- FOUND!


Here's a number that is smaller that almost works:

588235294117647.0 * 1.5 = 882352941176470.5

It doesn't quite work because the number produced is not an integer.

I'm confident my number is the smallest number that works. I tried my same process

starting with 2, 3, 4, and 5

and found that the resulting numbers were all larger than what I found. Something else I realized is that the are no numbers (of any size) that work that start with 6, 7, 8, or 9. This is because the number * 1.5 needs to have the same number of digits as the original number, and the product for numbers starting with 6-9 have an extra digit.

  • $\begingroup$ I really love your alternative solution. Yes, your finding is the smallest number. There are ways to solve it without that many tries, in a more elegant way, but still you managed to solved it in a practical way. $\endgroup$
    – Masclins
    Commented May 27, 2015 at 22:06
  • $\begingroup$ "I'm not sure that it counts as it requires moving the decimal point." -- I think it doesn't count as the result is not an integer. The question doesn't just specify that the initial number is an integer, it also specifies that the digit move "produces a new integer one and a half times the original". But good show of creativity anyway. :) $\endgroup$
    – hvd
    Commented May 28, 2015 at 7:25

Here is a fun approach to the same answer:

Let's call $X$ the number you start with, and $Y$ the number you get after moving the leftmost digit of $X$ to the right -- so the problem requires that $Y = 1.5X$. Now imagine the infinite repeating decimals $x=0.XXXXXX\ldots$ and $y=0.YYYYYY\ldots$ -- for example if $X = 123$ then $x=0.123123123\ldots$.

Now the point is:

We have two relations between $x$ and $y$. On the one hand the problem requires that $y=1.5x$. On the other hand the relationship between the digits implies that $10x = a+y$, where $a$ is the leading digit of $X$, an integer between $1$ and $9$. Hence we find that $x = 2a/17$!

Now we're basically done, let's just wrap up:

If you compute the decimal expansions of $1/17, 2/17, \ldots$, you will find that they all are repeating decimals, with repeating sequence of length $16$, so $X$ must be at least a $16$-digit number. ($X$ could also be a $32$-digit number, a $48$-digit number, etc., containing several repeated units.) But anyway the smallest $X$ will have $16$ digits and it's pretty clear that the smallest $X$ comes from the decimal expansion of $2/17$.

And the answer:

The decimal expansion of $2/17$ is the repeating decimal $0.1176470588235294\ 1176470588235294\ 1176470588235294 \ldots$.

  • $\begingroup$ I noticed that the ones I found were 16 digits. It's cool to have an explanation for why that is the case! $\endgroup$
    – Rob Watts
    Commented May 28, 2015 at 3:10

If we let the leading digit of the number be $a$ and the rest of the number be $b$, where $b$ has $n$ digits, our number is $a\cdot 10^n+b$ We are requiring that $1.5(a\cdot 10^n+b)=10b+a$ or $(3\cdot 10^n-2)a=17b$
A quick Excel search finds that $3\cdot 10^{15}-2$ is a multiple of $17$, leading to $a=1, b=176470588235294$ and the number is $1176470588235294$


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