1. **State the problem:** We need to find consecutive positive integers whose sum is 2024 and determine the largest integer in such a sum. There must be at least 2 terms. 2. Let the first term be $a$ and the number of terms be $n$, where $n\geq 2$. 3. The sum of $n$ consecutive integers starting with $a$ is given by the formula: $$ S = n \times \frac{2a + (n-1)}{2} $$ Here, $S=2024$. 4. So, $$ 2024 = n \times \frac{2a + (n-1)}{2} \implies 4048 = n(2a + n -1) $$ 5. Rearranged: $$ 2a + n -1 = \frac{4048}{n} $$ 6. Since $a$ must be an integer, $\frac{4048}{n} - (n -1)$ must be even: $$ 2a = \frac{4048}{n} - n + 1 $$ 7. We test divisors $n$ of 4048 such that $a$ is positive integer. Also, minimum $n=2$. 8. Factor 4048: $$ 4048 = 2^4 \times 11 \times 23 $$ 9. Candidates for $n$ (divisors of 4048): 2,4,8,11,16,22,23,32,44,46,88,92,176,184,253,368,506,736,1012,2024,4048 10. For each $n$, check if $a$ is positive integer: - Calculate $a = \frac{\frac{4048}{n} - n + 1}{2}$ - Must be positive integer. 11. Check for each divisor: - $n=44$: $$ a = \frac{\frac{4048}{44} -44 +1}{2} = \frac{92 -44 +1}{2} = \frac{49}{2} = 24.5$$ Not integer. - $n=23$: $$ a= \frac{\frac{4048}{23} -23 +1}{2} = \frac{176 -23 +1}{2} = \frac{154}{2} = 77 $$. Since $a=77$ is positive integer, this works. 12. The largest integer in sum is: $$ a + n -1 = 77 + 23 -1 = 99 $$ 13. Check if others yield larger largest integer: Try $n=11$: $$ a= \frac{\frac{4048}{11} -11 +1}{2} = \frac{368 -11 +1}{2} = \frac{358}{2}=179 $$ Largest integer: $$ 179 + 11 -1 = 189 $$ (larger than 99) Try $n=8$: $$ a= \frac{506 -8 +1}{2} = \frac{499}{2} = 249.5 $$ no integer. Try $n=16$: $$ a= \frac{253 -16 +1}{2} = \frac{238}{2}=119 $$ Largest integer: $$ 119 + 16 -1 = 134 $$ less than 189 14. The largest largest integer found is for $n=11$ giving largest integer 189. 15. **Answer:** The largest integer in such a sum is $\boxed{189}$.