1. The problem states we have a polynomial $$P(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n$$ with non-negative integer coefficients $a_i$. 2. Given: - $$P(1) = a_0 + a_1 + a_2 + \cdots + a_n = 25$$ - $$P(27) = a_0 + a_1 \cdot 27 + a_2 \cdot 27^2 + \cdots + a_n \cdot 27^n = 1771769$$ 3. Since all $a_i$ are non-negative integers and the sum $a_0 + a_1 + \cdots + a_n = 25$, and $27$ is the base used in $P(27)$, we can interpret $1771769$ in base 27 to find the coefficients $a_i$. 4. Convert $1771769$ to base $27$: - Divide $1771769$ by $27$: quotient $Q_1$, remainder $r_0$ - Continue dividing $Q_i$ by $27$ to get all coefficients: Calculations: $$1771769 \div 27 = 65658 \text{ remainder } 25$$ so $a_0 = 25$ $$65658 \div 27 = 2431 \text{ remainder } 21$$ so $a_1 = 21$ $$2431 \div 27 = 90 \text{ remainder } 1$$ so $a_2 = 1$ $$90 \div 27 = 3 \text{ remainder } 9$$ so $a_3 = 9$ $$3 \div 27 = 0 \text{ remainder } 3$$ so $a_4 = 3$ So polynomial coefficients are: $$a_0=25, a_1=21, a_2=1, a_3=9, a_4=3$$ Check sum: $$25 + 21 +1 +9 +3 = 59$$ This contradicts given $P(1)=25$. 5. This means the digits found are larger than the allowed sum $25$. So the hypothesis that coefficients correspond directly to base-27 digits fails here if $P(1) = 25$. 6. However, the problem states $a_i$ are non-negative integers and $P(1) = 25$ which is sum of coefficients. 7. Let's check if the coefficients extracted as digits in base 27 add to 25: Sum of digits found is 25 + 21 + 1 + 9 + 3 = 59, which is much larger. 8. It means $P(27) = 1771769$ and $P(1) = 25$ and coefficients are non-negative integer means the base-27 expansion is with digits equal to $a_i$. 9. This suggests that the coefficients $a_i$ are the digits in base 27 of $1771769$ but they sum to 25. 10. Let's add the digits obtained in step 4 in the order from highest power to lowest: $$3 + 9 + 1 + 21 + 25 = 59$$ which is not 25. 11. Another approach: sum of coefficients is 25. Since base is 27 (which is >=26), the coefficients also correspond to decimal digits in base 27 representation of $1771769$ with sum 25. 12. Possible only if some digits $a_i$ are zero, and sum to 25. Re-examine divisions carefully: - $1771769 \div 27 = 65658$ remainder $25$ ($a_0=25$) - $65658 \div 27 = 2431$ remainder $21$ ($a_1=21$) - $2431 \div 27 = 90$ remainder $1$ ($a_2=1$) - $90 \div 27 = 3$ remainder $9$ ($a_3=9$) - $3 \div 27 = 0$ remainder $3$ ($a_4=3$) Sum is $3 + 9 + 1 + 21 + 25 = 59$, contradicts sum $25$. 13. The key is coefficients must be less than or equal to 25 (since $P(1) = 25$). 14. Maybe the coefficients correspond to base 27 digits that sum to 25, but some coefficients are zero. 15. So try to express $1771769$ as sum of $a_i 27^i$ with sum of $a_i=25$. 16. Consider base 27 representation: $$1771769_{10} = a_0 + a_1(27) + a_2(27^2) + a_3(27^3) + a_4(27^4)$$ Calculate powers: $$27^0 =1$$ $$27^1 = 27$$ $$27^2 = 729$$ $$27^3 = 19683$$ $$27^4 = 531441$$ Try to find coefficients $a_i \\le 25$ such that: $$1771769 = a_4 \times 531441 + a_3 \times 19683 + a_2 \times 729 + a_1 \times 27 + a_0$$ Try for $a_4$: $$a_4 = \lfloor 1771769 / 531441 \rfloor = 3$$ Remaining: $$1771769 - 3 \times 531441 = 1771769 - 1594323 = 177446$$ $a_3$: $$\lfloor 177446 / 19683 \rfloor = 9$$ Remaining: $$177446 - 9 \times 19683 = 177446 - 177147 = 299$$ $a_2$: $$\lfloor 299 / 729 \rfloor = 0$$ Remaining: $$299$$ $a_1$: $$\lfloor 299 / 27 \rfloor = 11$$ Remaining: $$299 - 11 \times 27 = 299 - 297 = 2$$ $a_0 = 2$ Sum coefficients: $$a_0 + a_1 + a_2 + a_3 + a_4 = 2 + 11 + 0 + 9 + 3 = 25$$ matches given $P(1) = 25$. 17. Now compute the desired sum: $$a_0 + 2 a_1 + 3 a_2 + 4 a_3 + 5 a_4 = 2 + 2 \times 11 + 3 \times 0 + 4 \times 9 + 5 \times 3 = 2 + 22 + 0 + 36 + 15 = 75$$ **Final answer:** $$\boxed{75}$$