Consecutive Even Integers — Sum to 130 The sum of five consecutive even integers A, B, C, D, and E equals 130. Determine the product A * E (smallest times largest).

Introduction / Context:
This question assesses comfort with sequences of consecutive even integers and the technique of using the middle term to simplify calculations. For five consecutive even numbers, the central value is the average of the five and the endpoints are symmetric around it.

Given Data / Assumptions:

• Five consecutive even integers: A, B, C, D, E.
• Sum is 130.
• We must compute the product of the smallest (A) and largest (E).

Concept / Approach:
Let the middle even integer be C = x. Then the five numbers are x-4, x-2, x, x+2, x+4. Their sum is 5x. So x = (sum) / 5. With x known, endpoints are found directly and the product is straightforward.

Step-by-Step Solution:
1) Compute x: x = 130 / 5 = 26.2) Determine the five numbers: A = 26 - 4 = 22; B = 24; C = 26; D = 28; E = 30.3) Compute product: A * E = 22 * 30 = 660.4) Check the sum: 22 + 24 + 26 + 28 + 30 = 130 (verified).

Verification / Alternative check:
Because the set is symmetric, A + E = 52 and B + D = 52; hence total = (A + E) + (B + D) + C = 52 + 52 + 26 = 130, confirming correctness.

Why Other Options Are Wrong:
720, 616, 672, and 640 correspond to products of incorrect endpoint pairs that arise from miscomputing the center or using odd steps.

Common Pitfalls:
Using increments of 1 instead of 2 (even numbers); dividing by 2 or 4 instead of 5 to get the middle term; arithmetic slips when multiplying endpoints.

Final Answer:
660