Detailed mathematical approach

Problem Summary

For every coprime pair \((u,v)\) with \(1 \le u,v \le m\), we look for the smallest positive integer whose first decimal digit can be moved to the end so that the new number becomes exactly \(\left(\frac{u}{v}\right)^3\) times the original one. If no such integer exists, that pair contributes \(0\). The required quantity is

$T(m)=\sum_{\substack{1 \le u,v \le m\\ \gcd(u,v)=1}} M(u,v)\pmod{10^9+7},$

where \(M(u,v)\) denotes that minimal shifted multiple. A brute-force search over decimal strings is hopeless, so the solution turns the digit shift into a divisibility problem and then solves it with multiplicative orders.

Mathematical Approach

The core idea is to write the unknown number in decimal form, translate the digit shift into an equation, and then isolate the exact conditions under which an admissible length exists.

Step 1: Encode the Decimal Shift Algebraically

Let the desired number have \(\ell\) digits, leading digit \(d \in \{1,\dots,9\}\), and remaining tail \(r\), so

$N=d \cdot 10^{\ell-1}+r,\qquad 0 \le r < 10^{\ell-1}.$

After moving the first digit to the end, we obtain

$S=10r+d.$

Now set

$a=u^3,\qquad b=v^3.$

The required relation is

$S=\frac{a}{b}N,$

which is equivalent to

$b(10r+d)=a(d \cdot 10^{\ell-1}+r).$

Rearranging gives

$r(10b-a)=d(a \cdot 10^{\ell-1}-b).$

Define the denominator

$\Delta = 10b-a.$

Then

$r=\frac{d(a \cdot 10^{\ell-1}-b)}{\Delta}$

and therefore

$N=d \cdot 10^{\ell-1}+r=\frac{d\,b\,(10^{\ell}-1)}{\Delta}.$

This already shows that only pairs with \(\Delta>0\), that is \(u^3<10v^3\), can contribute.

Step 2: Find the Smallest Possible Length

Because \(r \ge 0\), we must have

$a \cdot 10^{\ell-1} \ge b.$

So the smallest feasible digit length is

$\ell_{\min}=\min\left\{\ell \ge 1: a \cdot 10^{\ell-1} \ge b\right\}.$

This is exactly the magnitude test used by the implementation: it does not guess \(\ell\), it starts from the first length for which the tail \(r\) can even be nonnegative.

Step 3: Restrict the Leading Digit

We also need \(r<10^{\ell-1}\), otherwise \(d\) would not really be the first digit of an \(\ell\)-digit number. Substituting the formula for \(r\) gives

$\frac{d(a \cdot 10^{\ell-1}-b)}{\Delta} < 10^{\ell-1}.$

After moving terms, this becomes

$10^{\ell-1}\bigl(a(d+1)-10b\bigr) < db.$

If \(a(d+1)\ge 10b\), the left-hand side is nonnegative and the inequality cannot hold in the way needed for a valid leading digit. Therefore any admissible digit must satisfy

$a(d+1)<10b.$

Once this strict inequality holds, the upper bound on \(r\) is automatically safe, so the digit test is completely independent of \(\ell\).

Step 4: Reduce the Integrality Condition

Since

$N=\frac{d\,b\,(10^{\ell}-1)}{\Delta},$

we need \(\Delta \mid d\,b\,(10^{\ell}-1)\). The coprimality hypothesis \(\gcd(u,v)=1\) implies

$\gcd(\Delta,b)=\gcd(10v^3-u^3,v^3)=1.$

So the factor \(b\) can be removed, and integrality simplifies to

$\Delta \mid d(10^{\ell}-1).$

Let

$g=\gcd(\Delta,d),\qquad m=\frac{\Delta}{g}.$

Then the condition is equivalent to

$m \mid 10^{\ell}-1.$

Step 5: Use the Multiplicative Order

A length \(\ell\) satisfying \(10^{\ell}\equiv 1 \pmod m\) exists if and only if \(\gcd(10,m)=1\). In that case the valid lengths are precisely the multiples of the multiplicative order

$\operatorname{ord}_m(10).$

So for a fixed digit \(d\), the smallest admissible length is

$\ell(d)=\left\lceil\frac{\ell_{\min}}{\operatorname{ord}_m(10)}\right\rceil \operatorname{ord}_m(10).$

The pair contribution is obtained from the lexicographically smallest candidate: first minimize \(\ell\), then among equal lengths choose the smaller leading digit \(d\). If no digit survives all tests, the pair contributes \(0\).

Worked Example: \((u,v)=(1,2)\)

Here

$a=1,\qquad b=8,\qquad \Delta=10 \cdot 8 - 1 = 79.$

The minimum possible length satisfies \(10^{\ell-1}\ge 8\), so

$\ell_{\min}=2.$

Try the smallest digit \(d=1\). The digit constraint is

$a(d+1)=2<80=10b,$

so the digit is admissible. Since \(\gcd(79,1)=1\), we get \(m=79\). Because \(79\) is coprime to \(10\), the relevant period is

$\operatorname{ord}_{79}(10)=13.$

Therefore the smallest valid length is

$\ell=13.$

Plugging this into the closed formula yields

$M(1,2)=\frac{1 \cdot 8(10^{13}-1)}{79}=1012658227848.$

Moving the first digit to the end produces \(126582278481\), which is exactly \(\frac{1}{8}\) of the original number, matching \(\left(\frac{1}{2}\right)^3\).

How the Code Works

The C++, Python, and Java implementations follow the derivation above almost literally. They first build a smallest-prime-factor sieve up to \(10m^3\), which is large enough for every denominator \(\Delta\) and every totient value used in the order computation. They then enumerate all coprime pairs \((u,v)\), form the cubes \(u^3\) and \(v^3\), and immediately discard cases with \(u^3 \ge 10v^3\).

For each surviving pair, the implementation computes \(\ell_{\min}\), loops over digits \(1,\dots,9\), applies the digit inequality \(u^3(d+1)<10v^3\), reduces the denominator by \(\gcd(\Delta,d)\), and rejects the digit if the reduced modulus shares a factor with \(10\). Otherwise it computes the multiplicative order of \(10\) modulo that reduced modulus. The order is obtained by starting from Euler's totient, factoring it with the sieve, and removing prime factors one by one whenever the corresponding modular-power test still returns \(1\).

After choosing the best pair \((d,\ell)\), the implementation evaluates

$\frac{d\,v^3\,(10^{\ell}-1)}{10v^3-u^3} \pmod{10^9+7}$

using a modular inverse for the denominator. The arithmetic is identical across all three languages; the C++ version additionally parallelizes independent ranges of \(u\), while the Python and Java versions keep the same logic serial.

Complexity Analysis

Let \(B=10m^3\). Building the smallest-prime-factor sieve costs \(O(B\log\log B)\) time and \(O(B)\) memory. The main search examines \(O(m^2)\) candidate pairs and at most \(9\) digits for each pair. Each digit trial performs a few gcd computations, several modular exponentiations, and a factor-stripping process based on the prime divisors of a totient value. So after the sieve, the remaining work is near-quadratic in \(m\) with a modest number-theoretic overhead, and the memory usage is dominated by the sieve.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=805
2. Multiplicative order: Wikipedia - Multiplicative order
3. Euler's totient function: Wikipedia - Euler's totient function
4. Repeating decimals: Wikipedia - Repeating decimal
5. Modular multiplicative inverse: Wikipedia - Modular multiplicative inverse