Difference between revisions of "2002 AMC 10B Problems/Problem 23"

Revision as of 10:24, 4 July 2013

Problem 23

Let $\{a_k\}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is

$\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89$

Solution

First of all, write $a_3$ and $a_4$ in terms of $a_2.$ $\begin{align*} a_3 &= a_{1+2} = 1 + a_2 + 2 = a_2 + 3\\ a_4 &= a_{2+2} = a_2 + a_2 + 4 = 2a_2 + 4 \end{align*}$

$a_6$ can be represented by $a_2$ in $2$ different ways.

$\begin{align*} a_{2+4} &= a_2 + a_4 + 8 = a_2 + 2a_2 + 4 + 8 = 3a_2 + 12\\ a_{3+3} &= 2a_3 + 9 = 6 + 2a_2 + 9 = 2a_2 + 15\\ \end{align*}$

Since both are equal to $a_6,$ you can set them equal to each other.

$\begin{align*} 3a_2 + 12 &= 2a_2 + 15\\ a_2 &= 3 \end{align*}$

Substitute the value of $a_2$ back into $a_6,$ and substitute that into $a_{12}.$

$a_6 = 2a_2+15 = 6+15 = 21$ $a_{12} = a_{6+6} = a_6+a_6+36 = 21+21+36 = \boxed{\mathrm{(D) \ } 78}$

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == See also ==
 {{AMC10 box|year=2002|ab=B|num-b=22|num-a=24}}
+{{MAA Notice}}

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Difference between revisions of "2002 AMC 10B Problems/Problem 23"

Revision as of 10:24, 4 July 2013

Problem 23

Solution

See also