1. **State the problem:** We need to show that $$-11100 \times 134 \equiv -1 \pmod{13}$$ without using a calculator. 2. **Reduce each number modulo 13:** Calculate $$-11100 \mod 13$$ and $$134 \mod 13$$. 3. **Calculate $$11100 \mod 13$$:** Divide 11100 by 13: $$11100 \div 13 = 853 \text{ remainder } 11$$ So, $$11100 \equiv 11 \pmod{13}$$ Therefore, $$-11100 \equiv -11 \equiv 13 - 11 = 2 \pmod{13}$$ 4. **Calculate $$134 \mod 13$$:** Divide 134 by 13: $$134 \div 13 = 10 \text{ remainder } 4$$ So, $$134 \equiv 4 \pmod{13}$$ 5. **Multiply the reduced values modulo 13:** $$(-11100) \times 134 \equiv 2 \times 4 = 8 \pmod{13}$$ 6. **Check if 8 is congruent to -1 modulo 13:** Since $$-1 \equiv 12 \pmod{13}$$ and $$8 \neq 12$$, the initial calculation seems off. 7. **Re-examine the calculation for $$-11100 \mod 13$$:** Calculate $$11100 \mod 13$$ more precisely: $$13 \times 853 = 11089$$ $$11100 - 11089 = 11$$ So, $$11100 \equiv 11 \pmod{13}$$ Therefore, $$-11100 \equiv -11 \equiv 13 - 11 = 2 \pmod{13}$$ (confirmed) 8. **Re-examine the calculation for $$134 \mod 13$$:** $$13 \times 10 = 130$$ $$134 - 130 = 4$$ So, $$134 \equiv 4 \pmod{13}$$ (confirmed) 9. **Multiply again:** $$2 \times 4 = 8 \pmod{13}$$ 10. **Check if the problem statement might have a typo or if the negative sign applies differently:** Try calculating $$11100 \times 134 \mod 13$$: $$11 \times 4 = 44 \equiv 44 - 39 = 5 \pmod{13}$$ 11. **Try $$-11100 \times 134 \equiv -1 \pmod{13}$$ by considering $$-1 \equiv 12 \pmod{13}$$:** We want: $$-11100 \times 134 \equiv 12 \pmod{13}$$ But we found $$8$$. 12. **Try reducing $$-11100$$ as $$-11100 \equiv -11100 + 13k$$ for some integer $$k$$ to get a different residue:** Since $$-11100 \equiv 2 \pmod{13}$$, this is correct. 13. **Try reducing $$134$$ differently:** No other residue possible. 14. **Try multiplying $$-11100 \times 134$$ directly modulo 13:** $$-11100 \times 134 \equiv 2 \times 4 = 8 \pmod{13}$$ 15. **Conclusion:** The product modulo 13 is 8, not -1. **Therefore, the statement $$-11100 \times 134 \equiv -1 \pmod{13}$$ is incorrect based on modular arithmetic calculations.**