Solution

Answer: 59206

Let

$$n = 10a+b$$

in decimal notation, where $b$ is the last digit.

Right-rotating $n$ moves $b$ to the front:

$$R(n)=b\cdot 10^{k-1}+a$$

where $k$ is the number of digits of $n$.

We are told that

$$R(n)=m n$$

for some integer multiplier $m$.

We seek all $n$ with

$$10<n<10^{100}$$

satisfying this property, and we want the last $5$ digits of their sum.

Mathematical analysis

Suppose $n$ has $k$ digits.

Write

$$n=10a+b,$$

where:

$b\in{1,\dots,9}$ (the last digit cannot be $0$, otherwise the rotation loses a digit),
$0\le a<10^{k-1}$.

The rotated number is

$$R(n)=b\cdot 10^{k-1}+a.$$

The condition is

$$b\cdot 10^{k-1}+a = m(10a+b).$$

Rearrange:

$$b\cdot 10^{k-1}+a = 10ma+mb,$$

$$b(10^{k-1}-m)=a(10m-1).$$

Hence

$$a=\frac{b(10^{k-1}-m)}{10m-1}.$$

Then

$$n=10a+b$$

becomes

$$n =10\cdot \frac{b(10^{k-1}-m)}{10m-1}+b.$$

Combine terms:

$$n = \frac{10b(10^{k-1}-m)+b(10m-1)}{10m-1}.$$

The $-10bm$ and $+10bm$ cancel:

$$n = \frac{b(10^k-1)}{10m-1}.$$

So every solution must have the form

$$\boxed{ n=\frac{b(10^k-1)}{10m-1} }$$

with:

$1\le b\le 9$,
$1\le m\le 9$,
denominator dividing numerator.

Key divisibility condition

We need

$$10m-1 \mid b(10^k-1).$$

Since

$$10m-1 \in {9,19,29,39,49,59,69,79,89},$$

we can analyze the multiplicative order of $10$ modulo these numbers.

For each $m$, define

$$d=10m-1.$$

$$10^k\equiv 1 \pmod d,$$

then $d\mid (10^k-1)$, and every digit $b$ works.

Otherwise only some $b$ may compensate for the missing factors.

Because $k<100$, we can brute force all:

$k=2,\dots,99$,
$m=1,\dots,9$,
$b=1,\dots,9$,

and test divisibility.

The total search space is tiny:

$$98\times 9\times 9 = 7938.$$

Python implementation

MOD = 100000

total = 0

# k = number of digits
for k in range(2, 100):

    ten_k_minus_1 = pow(10, k) - 1

    # multiplier m
    for m in range(1, 10):

        d = 10 * m - 1

        # final digit b
        for b in range(1, 10):

            numerator = b * ten_k_minus_1

            # n must be integer
            if numerator % d == 0:

                n = numerator // d

                # verify the property directly
                s = str(n)

                rotated = int(s[-1] + s[:-1])

                if rotated == m * n:
                    total = (total + n) % MOD

print(total)

Code walkthrough

MOD = 100000
total = 0

We only need the last $5$ digits, so we maintain the sum modulo $100000$.

for k in range(2, 100):

The problem requires

$$10<n<10^{100},$$

so $n$ can have from $2$ to $99$ digits.

ten_k_minus_1 = pow(10, k) - 1

This computes

$$10^k-1,$$

the repeating-9 number.

for m in range(1, 10):

The rotation cannot increase the number by more than a factor of $9$, so $m\in{1,\dots,9}$.

d = 10 * m - 1

From the derived formula:

$$n=\frac{b(10^k-1)}{10m-1}.$$

for b in range(1, 10):

The last digit cannot be $0$.

numerator = b * ten_k_minus_1

if numerator % d == 0:

This checks whether

$$n = \frac{b(10^k-1)}{10m-1}$$

is an integer.

n = numerator // d

Construct the candidate.

s = str(n)
rotated = int(s[-1] + s[:-1])

Perform the right rotation explicitly.

if rotated == m * n:

Verify the defining property exactly.

total = (total + n) % MOD

Accumulate only the last five digits.

Small example verification

The problem example is:

$$142857$$

Right rotation:

$$714285.$$

Indeed:

$$714285 = 5 \times 142857.$$

Using the formula:

$k=6$,
$m=5$,
$b=7$,

we get

$$n=\frac{7(10^6-1)}{49} =\frac{7\cdot 999999}{49} =142857.$$

So the derivation matches the example perfectly.

Final verification

The search space is exhaustive:

every valid number must satisfy the derived formula,
all $k<100$, $m\in[1,9]$, and $b\in[1,9]$ are tested,
the defining property is verified directly.

Running the program gives:

$$59206$$

Thus the last five digits of the required sum are:

Answer: 59206

Project Euler Problem 168