Problem 168: Number Rotations

Mathematical Foundation

Theorem 1 (Algebraic formulation). Let $n$ be a $d$-digit number ($d \geq 2$) with last digit $r$. Write $n = 10q + r$ where $q = \lfloor n/10 \rfloor$. The rotated number is $m = r \cdot 10^{d-1} + q$. The condition $m = kn$ for a positive integer $k$ is equivalent to:

Proof. From $m = kn$:

$\square$

Lemma 1 (Bounds on $k$). The multiplier $k$ satisfies $1 \leq k \leq 9$.

Proof. Since $m$ and $n$ are both $d$-digit numbers (or $m$ could have fewer digits if $r = 0$, but $r = 0$ leads to $m = q$ which has $d-1$ digits while $kn \geq n \geq 10^{d-1}$, a contradiction for $d \geq 2$), we need $r \geq 1$. Then $m < 10^d$ and $n \geq 10^{d-1}$, so $k = m/n < 10^d / 10^{d-1} = 10$. Also $k \geq 1$ since $m \geq 10^{d-1} \geq n/10$ and actually $m = kn \geq n$ requires $k \geq 1$. So $1 \leq k \leq 9$. $\square$

Theorem 2 (Digit-by-digit cyclic construction). For fixed $k \in \{1, \ldots, 9\}$ and last digit $r \in \{1, \ldots, 9\}$, the digits of $n$ satisfy a carry-propagation recurrence. Starting from the rightmost digit $a_d = r$, each preceding digit $a_{d-1}, a_{d-2}, \ldots$ is determined by:

where $c_i$ is the carry from position $i$. The sequence of digits is eventually periodic since the carry $c_i \in \{0, 1, \ldots, 8\}$ is bounded.

Proof. The condition $m = kn$ in terms of individual digits: multiplying $n = a_1 a_2 \cdots a_d$ by $k$ must produce $m = a_d a_1 a_2 \cdots a_{d-1}$. Column-by-column multiplication from right to left:

• Position $d$: $k \cdot a_d = 10c_{d-1} + a_{d-1}$
• Position $i$ ($2 \leq i \leq d-1$): $k \cdot a_i + c_i = 10c_{i-1} + a_{i-1}$
• Position $1$: $k \cdot a_1 + c_1 = a_d$ (with final carry 0 and result digit $a_d$)

Since each $a_i \in \{0, \ldots, 9\}$ and $c_i \in \{0, \ldots, 8\}$ (because $k \cdot 9 + 8 = 89 < 90$, so $c \leq 8$), the carry sequence determines the digit sequence. With 9 possible carry values, the sequence becomes periodic with period at most 9. For each valid cycle closing (where $c_1$ produces $a_d = r$), we get a valid $n$ of length equal to the cycle length. $\square$

Lemma 2. The last digit $r$ must satisfy $r \geq 1$, and for each $(k, r)$ pair, valid numbers $n$ exist for all cycle lengths $d$ that are multiples of the fundamental period of the carry recurrence, up to $d = 99$ (since $n < 10^{100}$).

Proof. As shown in Lemma 1, $r = 0$ is impossible. For each $(k, r)$, the carry recurrence starting from $c_d = 0$, $a_d = r$ generates digits $a_{d-1}, a_{d-2}, \ldots$ with carries $c_{d-1}, c_{d-2}, \ldots$. A valid $n$ of length $d$ requires returning to carry $c = 0$ with the final digit being $a_d = r$ after $d$ steps. This happens at all multiples of the fundamental cycle length. $\square$

Editorial

(The actual implementation tracks the carry recurrence carefully and checks the cycle-closing condition at each length.). We compute next digit and carry. We then check cycle closure: need carry = 0 and new_digit produces r. Finally, i.e., k * new_digit + carry_at_this_step should close.

Pseudocode

Start: digit = r, carry = 0
Compute next digit and carry
Check cycle closure: need carry = 0 and new_digit produces r
i.e., k * new_digit + carry_at_this_step should close
Actually: check if carry = 0 and the leading digit gives
k * a_1 + c_1 = r (the rotation condition)
Actually: need k*a_1 + c_1 = a_d = r with no further carry
This means the cycle has closed
but must verify the leading digit a_1 != 0
(no leading zeros in n)
Simpler: just check closing condition properly

Complexity Analysis

• Time: $O(9 \times 9 \times 100) = O(8{,}100)$ arithmetic operations. Effectively $O(1)$.
• Space: $O(100)$ for storing digits of the current candidate.

Answer

#include <bits/stdc++.h>
using namespace std;

/*
 * Problem 168: Number Rotations
 *
 * Find all n with 10 <= n < 10^100 such that rotating the last digit to
 * the front gives a multiple of n. Sum the last 5 digits of all such n.
 *
 * For each multiplier k (1..9) and last digit r (1..9), we generate digits
 * right-to-left: starting with digit r and carry 0, next digit = (k*prev_digit + carry) % 10,
 * new carry = (k*prev_digit + carry) / 10. After d steps, if we return to
 * (digit=r, carry=0), we have a valid d-digit number. Also need first digit != 0.
 */

int main(){
    const long long MOD = 100000;
    long long total = 0;

    for(int k = 1; k <= 9; k++){
        for(int r = 1; r <= 9; r++){
            // Generate digits right to left
            // digits[0] = rightmost digit = r (this is a_d)
            // digits[1] = a_{d-1}, etc.
            // digits[d-1] = a_1 (leftmost)
            // After d steps of generation, we need current digit = r and carry = 0
            // Also a_1 (= digits[d-1]) must be nonzero

            int digit = r;
            int carry = 0;
            // Store last 5 digits as we go (tracking powers of 10 mod 100000)
            // The digits are generated right to left: digit 0 is units, digit 1 is tens, etc.
            // last5 of n = sum of digit[i] * 10^i for i=0..4

            long long last5 = 0;
            long long pow10 = 1; // 10^i mod MOD

            for(int step = 0; step < 100; step++){
                // Current digit goes at position 'step' (from the right)
                if(step < 5){
                    last5 = (last5 + (long long)digit * pow10) % MOD;
                    pow10 = (pow10 * 10) % MOD;
                }

                // Check if we've completed a valid cycle (step+1 digits = d)
                // Need d >= 2, and after generating all d digits, the NEXT step
                // would produce (r, 0) again.
                // Actually: after placing digit at position 'step', compute next:
                int next_digit = (k * digit + carry) % 10;
                int next_carry = (k * digit + carry) / 10;

                int d = step + 1;
                if(d >= 2 && next_digit == r && next_carry == 0){
                    // Valid! The number has d digits.
                    // Check that the leftmost digit (digits[d-1] = current 'digit') is nonzero
                    if(digit != 0){
                        total = (total + last5) % MOD;
                    }
                }

                digit = next_digit;
                carry = next_carry;
            }
        }
    }

    cout << total << endl;
    return 0;
}

"""
Problem 168: Number Rotations

Find all n with 10 <= n < 10^100 such that rotating the last digit to
the front gives a multiple of n. Sum the last 5 digits of all such n.

For each multiplier k (1..9) and last digit r (1..9), generate digits
right-to-left using the recurrence from the multiplication condition.
"""

def solve():
    MOD = 100000
    total = 0

    for k in range(1, 10):
        for r in range(1, 10):
            digit = r
            carry = 0
            last5 = 0
            pow10 = 1  # 10^step mod MOD

            for step in range(100):
                # Place current digit at position 'step' (from right)
                if step < 5:
                    last5 = (last5 + digit * pow10) % MOD
                    pow10 = (pow10 * 10) % MOD

                # Compute next digit and carry
                val = k * digit + carry
                next_digit = val % 10
                next_carry = val // 10

                d = step + 1  # number of digits so far
                if d >= 2 and next_digit == r and next_carry == 0:
                    # Valid d-digit number; check leftmost digit nonzero
                    if digit != 0:
                        total = (total + last5) % MOD

                digit = next_digit
                carry = next_carry

    print(total)

solve()