1. **Problem statement:** We have a sequence of six distinct positive integers in strictly increasing order, where each term (except the first) is a multiple of the previous term. The sum of all six terms is 111. We want to find the smallest possible value of the largest term. 2. **Understanding the problem:** Let the six terms be $a_1 < a_2 < a_3 < a_4 < a_5 < a_6$ with $a_i$ positive integers. Each term after the first is a multiple of the previous, so: $$a_2 = k_1 a_1, \quad a_3 = k_2 a_2 = k_2 k_1 a_1, \quad a_4 = k_3 a_3 = k_3 k_2 k_1 a_1, \quad a_5 = k_4 a_4, \quad a_6 = k_5 a_5$$ where each $k_i$ is an integer greater than 1 (since terms are strictly increasing and multiples). 3. **Expressing terms:** The terms are: $$a_1, \quad k_1 a_1, \quad k_2 k_1 a_1, \quad k_3 k_2 k_1 a_1, \quad k_4 k_3 k_2 k_1 a_1, \quad k_5 k_4 k_3 k_2 k_1 a_1$$ 4. **Sum of terms:** The sum is: $$a_1 \left(1 + k_1 + k_1 k_2 + k_1 k_2 k_3 + k_1 k_2 k_3 k_4 + k_1 k_2 k_3 k_4 k_5\right) = 111$$ 5. **Goal:** Minimize $a_6 = a_1 k_1 k_2 k_3 k_4 k_5$ subject to the sum being 111 and all $k_i > 1$ integers. 6. **Strategy:** Try small values for $a_1$ and simple $k_i$ to satisfy sum = 111. 7. **Try $a_1=1$:** Then sum of coefficients must be 111. Try $k_i=2$ for all $i$: Sum coefficients = $1 + 2 + 4 + 8 + 16 + 32 = 63 < 111$. Try $k_i=3$ for all $i$: Sum coefficients = $1 + 3 + 9 + 27 + 81 + 243 = 364 > 111$. Try mix: $k_1=2, k_2=2, k_3=3, k_4=2, k_5=1$ (not allowed since $k_5>1$). 8. **Try $a_1=3$:** Then sum of coefficients = $111/3=37$. Try $k_i=2$ for all $i$: Sum coefficients = 63 > 37. Try $k_1=2, k_2=2, k_3=1$ (not allowed), so try $k_1=2, k_2=1$ no. Try $k_1=2, k_2=3$: Sum coefficients = $1 + 2 + 6 + 18 + 54 + 162 = 243 > 37$. 9. **Try $a_1=1$ and $k_1=3, k_2=2, k_3=2, k_4=1$ no. Try $k_1=2, k_2=3, k_3=2, k_4=1$ no. 10. **Try $a_1=1$ and $k_1=2, k_2=2, k_3=2, k_4=2, k_5=2$ sum coefficients = 63. Try $k_1=2, k_2=2, k_3=3, k_4=1$ no. 11. **Try $a_1=1$ and $k_1=2, k_2=3, k_3=3, k_4=1$ no. 12. **Try $a_1=1$ and $k_1=2, k_2=2, k_3=3, k_4=2, k_5=1$ no. 13. **Try $a_1=1$ and $k_1=2, k_2=2, k_3=3, k_4=2, k_5=2$ sum coefficients = $1 + 2 + 4 + 12 + 24 + 48 = 91 < 111$. 14. **Try $a_1=1$ and $k_1=2, k_2=3, k_3=3, k_4=2, k_5=2$ sum coefficients = $1 + 2 + 6 + 18 + 36 + 72 = 135 > 111$. 15. **Try $a_1=1$ and $k_1=2, k_2=3, k_3=2, k_4=2, k_5=2$ sum coefficients = $1 + 2 + 6 + 12 + 24 + 48 = 93 < 111$. 16. **Try $a_1=1$ and $k_1=3, k_2=2, k_3=2, k_4=2, k_5=2$ sum coefficients = $1 + 3 + 6 + 12 + 24 + 48 = 94 < 111$. 17. **Try $a_1=1$ and $k_1=3, k_2=3, k_3=2, k_4=2, k_5=2$ sum coefficients = $1 + 3 + 9 + 18 + 36 + 72 = 139 > 111$. 18. **Try $a_1=1$ and $k_1=2, k_2=2, k_3=2, k_4=3, k_5=2$ sum coefficients = $1 + 2 + 4 + 8 + 24 + 48 = 87 < 111$. 19. **Try $a_1=1$ and $k_1=2, k_2=2, k_3=2, k_4=3, k_5=3$ sum coefficients = $1 + 2 + 4 + 8 + 24 + 72 = 111$ perfect! 20. **Check terms:** $$a_1=1$$ $$a_2=2 \times 1=2$$ $$a_3=2 \times 2=4$$ $$a_4=2 \times 4=8$$ $$a_5=3 \times 8=24$$ $$a_6=3 \times 24=72$$ 21. **Sum check:** $$1 + 2 + 4 + 8 + 24 + 72 = 111$$ 22. **Largest term:** $a_6 = 72$ 23. **Answer:** The smallest possible largest term is 72, which corresponds to option (C). **Final answer:** (C) 72