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Knowing that $m$ and $n$ are two positive integers, find all the solutions for the equation:

$\frac{m}{n} = n.m$

For example, the pair $(m=5, n=2)$ is a solution because $\frac{5}{2} = 2.5$ while $(m=294, n=17)$ comes close to a solution but it is not because $\frac{294}{17} = 17.29411...$ .

Don
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Robert Werner
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3 Answers3

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We begin by rewriting $m/n=n.m$ as

$${m\over n}=n+{m\over10^k}\quad\text{where }k=1+\lfloor\log m\rfloor$$

This can be rewritten as

$$(10^k-n)m=10^kn^2$$

which, since the right hand side is clearly positive, implies $10^k\gt n$. Hold that thought.

Next, solving for $m$, we see

$$m=n^2+{n^3\over10^k-n}$$

Since we already know $10^k-n$ is positive, this implies $m\gt n^2$. Hold that thought too.

Notice next that if $p$ is a prime divisor of $10^k-n$, then $p$ must also divide $n^3$, which is to say $p$ divides $n$, in which case $p$ also divides $(10^k-n)+n=10^k$, which is to say $p$ can only be $2$ or $5$. If we now write

$$10^k-n=2^\sigma\cdot5^\tau\cdot10^h\quad\text{with }0\le\sigma+\tau\le1$$

(and all exponents understood to be nonnegative integers), we see that $n\gt0$ implies $k\gt h$. Using the fact that $2^\sigma\cdot5^\tau$ is either $1$, $2$, or $5$, we have

$$2^\sigma\cdot5^\tau\cdot10^h\le5\cdot10^{k-1}\lt9\cdot10^{k-1}=10^k-10^{k-1}$$ so that $10^{k-1}\lt10^k-2^\sigma\cdot5^\tau\cdot10^h=n\lt10^k$, which implies

$$\lfloor\log n\rfloor=k-1$$

But now the inequality $m\gt n^2$ implies

$$k-1=\lfloor\log m\rfloor\ge\lfloor2\log n\rfloor\ge2\lfloor\log n\rfloor=2(k-1)$$

Thus $k=1$, so that $m$ is a one-digit number. The inequality $9\ge m\gt n^2$ now limits $n$ to either $1$ or $2$. The first of these would give $m=10/9$, so we are left with $n=2$ and $m=4+8/(10-2)=5$ as the only solution.

Barry Cipra
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$m/n = n.m= n + m/10^k$ so $10^kn^2 + mn - 10^km$

So by quadratic equation $n = \frac{-m + \sqrt{m^2 + 4m*10^{2k}}}{2*10^k} $

Notice:

m has k digits.

So n = $\frac{-m + \sqrt{m^2 + 4m*10^{2k}}}{2*10^k}$ has $\lceil k/2 \rceil$ digits. But m/n $\approx$ n then has ${\lceil k/2 \rceil}$ digits, so $m$ has between $\lfloor k/2 \rfloor$ and ${\lceil k/2 \rceil}$ digits. But m has $k $ digits. So k = 1 or 2 and m has one or 2 digits and n has 1 digit. $nm$ is a multiple of $10^k$ but m/n has only 1 digit so k = 1 and m has 1 digit.

$n = \frac{-m + \sqrt{m^2 + 400}}{20} $

So $m^2 + 400$ is a perfect square. Only possibility is m =5 so n = 2.

fleablood
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    That only solves the one-digit example. What about $m/n=n+m/100$, when $10\leq m<100$, for example. – Thomas Andrews Nov 11 '15 at 00:14
  • Good point. I was assuming m.n represented single digis. If not then it's a power of 10. Still must have m (m+4.10^2k) being a perfect square which severely limits the results. – fleablood Nov 11 '15 at 00:27
  • I don't understand the justification for the statement "So n, in that equation, has only 1 digit." (I'm not even sure which equation is being referred to.) – Barry Cipra Nov 11 '15 at 17:20
  • n = quadratic expression. -m + square root of m^2 + 10^2k dividide by 2*10^k will have 1 digit. – fleablood Nov 11 '15 at 17:22
  • @fleablood, but you're taking the square root of $m^2+4m\cdot10^{2k}$, not $m^2+10^{2k}$. The leaves the possibility that $n\approx10^{k/2}$. – Barry Cipra Nov 11 '15 at 17:28
  • Yeah, your right. But as nm is a multiple of 10^k m has both approximately k/2 digits and $ k/2 \approx k$ so k = 1. – fleablood Nov 11 '15 at 17:38
  • Now I don't understand the justification for the assertion "so $m$ has at most $\lceil k/2\rceil$ digit[s]." But maybe I'm just being dense. – Barry Cipra Nov 11 '15 at 17:47
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$$m/n=n + 10^{-k}m$$

Or $m = n^2 + 10^{-k}mn$.

Or $$10^kn^2+mn -10^km =0$$

So $$n = \frac{-m\pm \sqrt{m^2+4\cdot 10^{2k}m}}{2\cdot 10^k}.$$

So, you need $m\left(m+4\cdot 10^{2k}\right)$ to be a perfect square, and you need its square root congruent to $m\pmod {2\cdot 10^k}$.

So Let $$m^2+4m\cdot 10^{2k} = \left(m+2r\cdot 10^k\right)^2 = m^2 +4mr \cdot 10^k +4\cdot 10^{2k} r^2$$

So $m=\frac{10^kr^2}{10^k-r}$. So Letting $s=10^k-r$ then $$m=\frac{10^k(10^k-s)^2}{s}=\frac{10^{3k}}{s} -2\cdot 10^{2k} +10^ks.$$

So $s$ needs to be a divisor of $10^{3k}$ and $m<10^k$.

For example, in your case, $k=1,s=8, m=5$.

Now, for $m<10^k$, you need $(10^k-s)^2<s$ or $s^2-(2\cdot 10^k +1)s + 10^{2k}<0$.

This means that you need $s$ in between the roots:

$$\frac{2\cdot 10^k+1\pm \sqrt{4\cdot 10^k +1}}{2}$$

Take for example, $k=2$. then you need a divisor of $10^6$ that is in the range:

$$\frac{201\pm\sqrt{401}}{2}$$

which, for integers, means $[91,111]$. The only $s$ that we can get is $s=100$.

In general, we'd need a factor of $10^{3k}$ that is within $1+\sqrt{10^k}$ of $10^k$. In don't think that is likely to happen often - any divisor of $10^{3k}$ that is that big $100$ as a factor.

For example, you'd need a power of $2$ that starts with mostly nines. You need a continued fraction convergent for $\log_2 10$ or $\log_5 10$ that is just unbelievably close. I'm not sure how to prove that isn't possible.

Basically, to find a solution in general, you need a $2^t$ or $5^t$ that starts with a lot of nines - at least the first half of them.I don't know how to prove that isn't possible, however.

1024 is close enough to $10^3$, but it doesn't divide $10^9$, and $10240$ is not close enough to $10^4$.

(If you plug in $s=1024,k=3$, you get $m=562.5$ and $n=23.4375$ and: $$\frac{562.5}{23.4375} = 24 = 23.4375 + 0.5625.$$

So at least the formulae work.)

It's pretty easy by exhaustion to prove for small values of $k$. We can use the continued fractions for $\log_2 10$ and $\log_5 10$ to give us examples - basically, we need an insanely large coefficient in the continued fraction.

Thomas Andrews
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