AIME Problem 1

Algebra Level 3

The expressions A = 1 × 2 + 3 × 4 + 5 × 6 + + 37 × 38 + 39 A=1\times2+3\times4+5\times6+\cdots+37\times38+39 and B = 1 + 2 × 3 + 4 × 5 + + 36 × 37 + 38 × 39 B=1+2\times3+4\times5+\cdots+36\times37+38\times39 are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers A A and B B .

This problem is part of this set .


The answer is 722.

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4 solutions

Bk Lim
Mar 26, 2015

Really easy pattern to follow, god knows why this is level 5 ! Btw, upvoted.

Venkata Karthik Bandaru - 6 years, 2 months ago

We note that:

{ A = k = 1 19 2 k ( 2 k 1 ) + 39 = 4 k = 1 19 k 2 2 k = 1 19 k + 39 B = 1 + k = 1 19 2 k ( 2 k + 1 ) = 1 + 4 k = 1 19 k 2 + 2 k = 1 19 k \begin{cases} \displaystyle A = \sum_{k=1}^{19} 2k(2k-1) + 39 = 4 \sum_{k=1}^{19} k^2 - 2 \sum_{k=1}^{19} k + 39 \\ \displaystyle B = 1 + \sum_{k=1}^{19} 2k(2k+1) = 1 + 4 \sum_{k=1}^{19} k^2 + 2 \sum_{k=1}^{19} k \end{cases}

Therefore,

A B = 38 4 k = 1 19 k = 38 4 ( 19 ) ( 20 ) 3 = 38 760 = 722 \begin{aligned} |A-B| & = \left| 38 - 4 \sum_{k=1}^{19} k \right| = \left| 38 - \frac{4(19)(20)}{3} \right| = \left| 38 - 760 \right| = \boxed{722} \end{aligned}

Sujoy Roy
Mar 21, 2015

Here, A = n = 1 19 [ ( 2 n 1 ) 2 n ] + 39 A=\sum_{n=1}^{19}[(2n-1)*2n]+39 and B = 1 + n = 1 19 [ 2 n ( 2 n + 1 ) ] B=1+\sum_{n=1}^{19}[2n*(2n+1)] .

Now, B A = 38 + n = 1 19 [ 2 n ( 2 n + 1 ( 2 n 1 ) ) ] B-A= -38+\sum_{n=1}^{19}[2n*(2n+1-(2n-1))]

= 38 + 4 n = 1 19 n = 38 + 4 19 20 2 = 722 =-38+4\sum_{n=1}^{19}n = -38+4*\frac{19*20}{2}=722 .

Prameela Reddy
Sep 22, 2017

B = 2.3 + 4.5 + 6.7 + 38 . 39 + 1

A = 1.2 + 3.4 +5.6 + 37 . 38 + 39

B - A = 2.2 + 4.2 + 6.2 + .........+ 38 . 2 - 38 = 4 ( 1 + 2 + 3 + ........+ 19) - 38 = 4 ( 19 * 20)/2 - 38 = 760 - 38 = 722

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