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:

$(19choose1)/times 2$

2015 AIME I (Problems • Answer Key • Resources)
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2015 AIME I Problems/Problem 1

Contents

Problem

Solution 1

Solution 2 (slower solution)

See also