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.

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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
Small Hint:

Compute B−AB - A by pairing each product in BB with the product in AA that shares its even factor

Big Hint:

Each pair contributes (2k+1)(2k)(2k+1)(2k) −(2k−1)(2k)=4k,- (2k-1)(2k) = 4k, and the leftover terms are 11 and −39-39

Solution:

Subtract term by term: B−A=(1−39)+(2×3−1×2)+(4×5−3×4)+⋯+(38×39−37×38). \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) −(2k−1)(2k)=4k- (2k-1)(2k) = 4k for k=1,2,…,19.k = 1, 2, \ldots, 19.

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

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