Difference between revisions of "2002 AMC 10B Problems/Problem 23"
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<math> \mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89 </math> | <math> \mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89 </math> | ||
− | == Solution == | + | == Solution 1 == |
− | + | When <math>m=1</math>, <math>a_{n+1}=1+a_n+n</math>. Hence, | |
− | <cmath> | + | <cmath>a_{2}=1+a_1+1</cmath> |
− | + | <cmath>a_{3}=1+a_2+2</cmath> | |
+ | <cmath>a_{4}=1+a_3+3</cmath> | ||
+ | <cmath>\dots</cmath> | ||
+ | <cmath>a_{12}=1+a_{11}+11</cmath> | ||
+ | Adding these equations up, we have that <math>a_{12}=12+(1+2+3+...+11)=\boxed{\mathrm{(D) \ } 78}</math> | ||
− | + | ~AopsUser101 | |
− | + | == Solution 2 == | |
− | |||
− | |||
− | + | Substituting <math>n=1</math> into <math>a_{m+n}=a_m+a_n+mn</math>: <math>a_{m+1}=a_m+a_{1}+m</math>. | |
− | < | + | Since <math>a_1 = 1</math>, <math>a_{m+1}=a_m+m+1</math>. |
− | |||
− | + | Therefore, <math>a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)</math>, and so on until <math>a_2 = a_1 + 2</math>. | |
− | < | + | Adding the Left Hand Sides of all of these equations gives <math>a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2</math>. |
− | <cmath>a_{12} = a_{ | + | |
+ | Adding the Right Hand Sides of these equations gives | ||
+ | |||
+ | <math>(a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)</math>. | ||
+ | |||
+ | These two expressions must be equal; hence <math>a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2 = (a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)</math> and <math>a_m = a_1 + (m + (m-1) + (m-2) + \cdots + 2)</math>. | ||
+ | |||
+ | Substituting <math>a_1 = 1</math>: <math>a_m = 1 + (m + (m-1) + (m-2) + \cdots + 2) = 1+2+3+4+ \cdots +m = \frac{(m+1)(m)}{2}</math>. | ||
+ | |||
+ | Thus we have a general formula for <math>a_m</math> and substituting <math>m=12</math>: <math>a_{12} = \frac{(13)(12)}{2} = (13)(6) = \boxed{\mathrm{(D)} 78}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | We can literally just plug stuff in. No prerequisite is actually said in the sequence. Since <math>a_{m+n} = a_m+a_n +mn</math>, we know <math>a_2=a_1+a_1+1\cdot1=1+1+1=3</math>. After this, we can use <math>a_2</math> to find <math>a_4</math>. <math>a_4=a_2+a_2+2\cdot 2 = 3+3+4 = 10</math>. Now, we can use <math>a_2</math> and <math>a_4</math> to find <math>a_6</math>, or <math>a_6=a_4+a_2+4\cdot 2 = 10+3+8=21</math>. Lastly, we can use <math>a_6</math> to find <math>a_{12}</math>. <math>a_{12} = a_6+a_6+6\cdot 6 = 21+21+36= \boxed{\mathrm{(D) \ } 78}</math> | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | We can set <math>n</math> equal to <math>m</math>, so we can say that | ||
+ | <cmath>a_{m + m} = a_m + a_m + m*m</cmath> | ||
+ | <cmath>a_{2m} = 2a_m + m^2</cmath> | ||
+ | |||
+ | We set <math>2m = 12</math>, we get <math>m = 6</math>. | ||
+ | <cmath>a_{12} = 2a_6 + 36</cmath> | ||
+ | |||
+ | We set <math>2m = 6</math>m, we get <math>m = 3</math>. | ||
+ | <cmath>a_6 = 2a_3 + 9</cmath> | ||
+ | |||
+ | Solving for <math>a_3</math> is easy, just direct substitution. | ||
+ | <cmath>a_2 = 1 + 1 + 1 = 3</cmath> | ||
+ | <cmath>a_3 = a_{2 + 1} = 3 + 1 + 2 = 6</cmath> | ||
+ | |||
+ | Substituting, we get | ||
+ | <cmath>a_6 = 2(6) + 9 = 21</cmath> | ||
+ | <cmath>a_{12} = 2(21) + 36 = 78</cmath> | ||
+ | |||
+ | Thus, the answer is <math>\boxed{D}</math>. | ||
+ | |||
+ | ~ euler123 | ||
+ | |||
+ | == 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]] | ||
+ | |||
+ | |||
+ | == Solution 6 == | ||
+ | Follow solution 4 with plugging in m=n, but for the last part since since the subscript is 2n, and we aren't given anything that has odd numbers ( without substitution into the original equation), we need to find two numbers that add up to 12, which result from the multiplication of two even numbers. We realize that 4+8 works, since 4=2x2 and 8=4x2. We then proceed in substituting n=2 and n=4 into the equation, then add to obtain 78. | ||
+ | ~Charmainema07292010 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=zraGzYAh0uM ~David | ||
+ | |||
+ | == See also == | ||
+ | {{AMC10 box|year=2002|ab=B|num-b=22|num-a=24}} | ||
+ | {{MAA Notice}} |
Latest revision as of 05:38, 23 October 2023
Contents
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 .
Solution 6
Follow solution 4 with plugging in m=n, but for the last part since since the subscript is 2n, and we aren't given anything that has odd numbers ( without substitution into the original equation), we need to find two numbers that add up to 12, which result from the multiplication of two even numbers. We realize that 4+8 works, since 4=2x2 and 8=4x2. We then proceed in substituting n=2 and n=4 into the equation, then add to obtain 78. ~Charmainema07292010
Video Solution
https://www.youtube.com/watch?v=zraGzYAh0uM ~David
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.