Difference between revisions of "2002 AMC 10B Problems/Problem 23"
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− | == | + | == Solution 5 == |
− | + | Note that the sequence of triangular numbers <math>T_n=1+2+3+...+n</math> satisfies these conditions. It is immediately obvious that it satisfies <math>a_1=1</math>, and <math>a_{m+n}=a_m+a_n+mn</math> can be visually proven with the diagram below. | |
+ | |||
+ | <asy> | ||
+ | for(int i=5; i > 0; --i) { | ||
+ | for(int j=0; j < i; ++j) { | ||
+ | draw(circle((j+(5-i)/2,(5-i)*sqrt(3)/2),.2)); | ||
+ | }; | ||
+ | }; | ||
+ | path m1 = brace((2,-.3),(0,-.3),.2); | ||
+ | draw(m1); | ||
+ | label("$m$",m1,S); | ||
+ | |||
+ | path n1 = brace((4,-.3),(3,-.3),.2); | ||
+ | draw(n1); | ||
+ | label("$n$",n1,S); | ||
+ | |||
+ | |||
+ | draw((-.2*sqrt(3),-.2)--(2+.2*sqrt(3),-.2)--(1,.4+sqrt(3))--cycle); | ||
+ | label("$T_m$",(1,1/3*sqrt(3))); | ||
+ | |||
+ | draw((3-.2*sqrt(3),-.2)--(4+.2*sqrt(3),-.2)--(3.5,.4+.5*sqrt(3))--cycle); | ||
+ | label("$T_n$",(3.5,.5/3*sqrt(3))); | ||
+ | |||
+ | |||
+ | path m2 = brace((2+.15*sqrt(3),.15+2*sqrt(3)),(3+.15*sqrt(3),.15+sqrt(3)),.2); | ||
+ | draw(m2); | ||
+ | label("$m$",m2,(.5*sqrt(3),.5)); | ||
+ | |||
+ | path n2 = brace((1.5-.15*sqrt(3),.15+1.5*sqrt(3)),(2-.15*sqrt(3),.15+2*sqrt(3)),.2); | ||
+ | draw(n2); | ||
+ | label("$n$",n2,(-.5*sqrt(3),.5)); | ||
+ | |||
+ | |||
+ | draw((2.5,-.4+.5*sqrt(3))--(3+.4/3*sqrt(3),sqrt(3))--(2,.4+2*sqrt(3))--(1.5-.4/3*sqrt(3),1.5*sqrt(3))--cycle); | ||
+ | label("$mn$",(2.25,1.25*sqrt(3))); | ||
+ | </asy> | ||
+ | |||
+ | This means that we can use the triangular number formula <math>T_n = \frac{n(n+1)}{2}</math>, so the answer is <math>T_{12} = \frac{12(12+1)}{2} = \boxed{\mathrm{(D) \ } 78}</math>. ~[[User:emerald_block|emerald_block]] | ||
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=B|num-b=22|num-a=24}} | {{AMC10 box|year=2002|ab=B|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:38, 31 March 2020
Problem 23
Let be a sequence of integers such that and for all positive integers and Then is
Solution 1
When , . Hence, Adding these equations up, we have that
~AopsUser101
Solution 2
Substituting into : . Since , . Therefore, , and so on until . Adding the Left Hand Sides of all of these equations gives ; adding the Right Hand Sides of these equations gives . These two expressions must be equal; hence and . Substituting : . Thus we have a general formula for and substituting : .
Solution 3
We can literally just plug stuff in. No prerequisite is actually said in the sequence. Since , we know . After this, we can use to find . . Now, we can use and to find , or . Lastly, we can use to find .
Solution 4
We can set equal to , so we can say that
We set , we get .
We set m, we get .
Solving for is easy, just direct substitution.
Substituting, we get
Thus, the answer is .
~ euler123
Solution 5
Note that the sequence of triangular numbers satisfies these conditions. It is immediately obvious that it satisfies , and can be visually proven with the diagram below.
This means that we can use the triangular number formula , so the answer is . ~emerald_block
See also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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