AMC 8 · 2016 · #19

Grade 6 arithmeticalgebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

sequences-arithmeticmean-median-mode-rangepattern-recognition easier-related-problempattern-recognitionidentify-subproblems ↑ Prerequisites: sequences-arithmeticmean-median-mode-range

📏 Short solution 💡 3 insights

📘 View easy version →

Problem

Take 25 even integers in a row, like ..., n-4, n-2, n, n+2, n+4, .... Their sum is 10,000. What is the largest one?

Pick an answer.

(A)

$mbox{ }360$

(B)

$mbox{ }388$

(C)

$mbox{ }412$

(D)

$mbox{ }416$

(E)

$mbox{ }424$

AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

Twenty-five consecutive even integers is too many to list, but the structure is highly regular. Tool #9 (Easier Problem) — try 3 or 5 consecutive evens first — exposes the pattern that the sum is always (count) × (middle term). Tool #5 (Pattern) lets us use that fact to jump straight to the middle term: 10,000 ÷ 25 = 400. Then Tool #7 (Subproblems) splits the rest into two clean pieces — find the middle term, then step out 12 places (by +2 each) to reach the largest. This dodges Tool #13 (Algebra) entirely.

1STEP 1

Test tiny cases: 4+6+8 = 3×6 and 2+4+6+8+10 = 5×6. Pattern: for an odd count of evenly-spaced terms, sum = count × middle term.

4 + 6 + 8 = 18 = 3 × 6, 2 + 4 + 6 + 8 + 10 = 30 = 5 × 6

2STEP 2

25 terms is an odd count, so sum = 25 × middle term. Divide: 10,000 ÷ 25 = 400 is the middle term.

middle term =

\frac{10,000}{25}

= 400

3STEP 3

With 25 terms the middle is the 13th term — 12 sit on each side — so term 13 equals 400.

\frac{25 + 1}{2}

= 13

4STEP 4

Term 13 to term 25 is 12 steps of +2, so the largest = 400 + 24 = 424 (E).

largest = 400 + 12 × 2 = 400 + 24 = 424 → (E)

Answer

mbox{ }424

Sanity check the smallest term too: 400 - 12 × 2 = 376. So the list runs 376, 378, …, 424. Quick pairing: (376 + 424) = 800, and pairing every term with its mirror around the middle gives 12 pairs of 800 plus the lone middle 400, total 12 × 800 + 400 = 9600 + 400 = 10,000. ✓ The answer 424 is the largest choice given and is consistent with the middle being a round 400.

💡Key takeaway

If you have an odd number of evenly-spaced numbers, the average IS the middle one — a Grade 6 idea that cracks this AMC 8 problem in one division!

More like this