2015 AIME I Problems/Problem 1

Problem

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

Solution 1

We see that

$A=(1\times 2)+(3\times 4)+(5\times 6)+\cdots +(35\times 36)+(37\times 38)+39$

and

$B=1+(2\times 3)+(4\times 5)+(6\times 7)+\cdots +(36\times 37)+(38\times 39)$.

Therefore,

$B-A=-38+(2\times 2)+(2\times 4)+(2\times 6)+\cdots +(2\times 36)+(2\times 38)$

$=-38+4\times (1+2+3+\cdots+19)$

$=-38+4\times\frac{20\cdot 19}{2}=-38+760=\boxed{722}.$

Solution 2 (slower solution)

For those that aren't shrewd enough to recognize the above, we may use Newton's Little Formula to semi-bash the equations.

We write down the pairs of numbers after multiplication and solve each layer:


\[2, 12, 30, 56, 90...(39)\]

\[6, 18, 26, 34...\]

\[8, 8, 8...\]

and

\[(1) 6, 20, 42, 72...\] \[14, 22, 30...\]

\[8, 8, 8...\]


Then we use Newton's Little Formula for the sum of $n$ terms in a sequence.

Notice that there are $19$ terms in each sequence, plus the tails of $39$ and $1$ on the first and second equations, respectively.


So,


\[2\binom{19}{1}+10\binom{19}{2}+8\binom{19}{3}+1\]


\[6\binom{19}{1}+14\binom{19}{2}+8\binom{19}{3}+39\]


Subtracting $A$ from $B$ gives:


\[4\binom{19}{1}+4\binom{19}{2}-38\]


Which unsurprisingly gives us $\boxed{722}.$


-jackshi2006

See also

2015 AIME I (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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