2015 AIME I Problem 1

Attempt Problem 1 of the 2015 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AIME I solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

1.

The expressions A=1×2A = 1 \times 2 +3×4+ 3 \times 4 +5×6+ 5 \times 6 ++ \cdots +37×38+39+ 37 \times 38 + 39 and B=1B = 1 +2×3+ 2 \times 3 +4×5+ 4 \times 5 ++ \cdots +36×37+ 36 \times 37 +38×39+ 38 \times 39 are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers AA and B.B.

Answer: 722
Concepts:summationarithmetic sequencepairing and grouping
Difficulty rating: 1890
Solution:

Subtract term by term: BA=(139)+(2×31×2)+(4×53×4)++(38×3937×38). \begin{aligned} &B - A = (1 - 39) \\ &\quad {}+ (2 \times 3 - 1 \times 2) \\ &\quad {}+ (4 \times 5 - 3 \times 4) + \cdots \\ &\quad {}+ (38 \times 39 - 37 \times 38). \end{aligned} Each parenthesized difference has the form (2k+1)(2k)(2k+1)(2k) (2k1)(2k)=4k- (2k-1)(2k) = 4k for k=1,2,,19.k = 1, 2, \ldots, 19.

Therefore BA=38+4(1+2++19)=38+4190=722. \begin{aligned} B - A &= -38 \\ &\quad {}+ 4(1 + 2 + \cdots + 19) \\ &= -38 + 4 \cdot 190 = 722. \end{aligned}

Full Exam

Problem 1 in Other Years