1. The problem asks to create a table of numbers that are multiples of 2 and whose total sum is 15000. 2. We know multiples of 2 are numbers like 2, 4, 6, 8, ... 3. Let's consider the first $n$ multiples of 2: $2, 4, 6, ..., 2n$. 4. The sum of the first $n$ multiples of 2 is given by the formula: $$\text{Sum} = 2 + 4 + 6 + ... + 2n = 2(1 + 2 + 3 + ... + n)$$ 5. The sum of the first $n$ natural numbers is: $$1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$$ 6. Substitute this into the sum formula: $$\text{Sum} = 2 \times \frac{n(n+1)}{2} = n(n+1)$$ 7. We want this sum to be 15000, so: $$n(n+1) = 15000$$ 8. This is a quadratic equation: $$n^2 + n - 15000 = 0$$ 9. Solve for $n$ using the quadratic formula: $$n = \frac{-1 \pm \sqrt{1 + 4 \times 15000}}{2} = \frac{-1 \pm \sqrt{60001}}{2}$$ 10. Calculate $\sqrt{60001} \approx 244.95$. 11. Taking the positive root: $$n = \frac{-1 + 244.95}{2} = \frac{243.95}{2} = 121.975$$ 12. Since $n$ must be an integer, try $n=121$ and $n=122$: - For $n=121$, sum = $121 \times 122 = 14762$ - For $n=122$, sum = $122 \times 123 = 15006$ 13. Neither equals exactly 15000, but $n=122$ is very close. 14. To get exactly 15000, we can adjust the last number or select a subset of multiples of 2. 15. For example, take the first 121 multiples of 2 and add an extra number to reach 15000: $$14762 + x = 15000 \Rightarrow x = 238$$ 16. Since 238 is a multiple of 2, the table can be: $$2, 4, 6, ..., 242, 238$$ 17. This set sums exactly to 15000. Final answer: A table of the first 121 multiples of 2 plus 238 sums to 15000.