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

(Solution 4)
(new solution built off of additional comment)
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~ euler123
 
~ euler123
  
== Additional Comment==
+
== Solution 5 ==
  
This is also the formula for the triangular numbers <math>T_n=1+2+3+...+n</math>, as seen in Solution 2
+
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);
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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 $\{a_k\}$ 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 1

When $m=1$, $a_{n+1}=1+a_n+n$. Hence, \[a_{2}=1+a_1+2\] \[a_{3}=1+a_2+3\] \[a_{4}=1+a_3+4\] \[\dots\] \[a_{12}=1+a_{11}+11\] Adding these equations up, we have that $a_{12}=12+(1+2+3+...+11)=\boxed{78}$

~AopsUser101

Solution 2

Substituting $n=1$ into $a_{m+n}=a_m+a_n+mn$: $a_{m+1}=a_m+a_{1}+m$. Since $a_1 = 1$, $a_{m+1}=a_m+m+1$. Therefore, $a_m = a_{m-1} + m, a_{m-1}=a_{m-2}+(m-1), a_{m-2} = a_{m-3} + (m-2)$, and so on until $a_2 = a_1 + 2$. Adding the Left Hand Sides of all of these equations gives $a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2$; adding the Right Hand Sides of these equations gives $(a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)$. These two expressions must be equal; hence $a_m + a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_2 = (a_{m-1} + a_{m-2} + a_{m-3} + \cdots + a_1) + (m + (m-1) + (m-2) + \cdots + 2)$ and $a_m = a_1 + (m + (m-1) + (m-2) + \cdots + 2)$. Substituting $a_1 = 1$: $a_m = 1 + (m + (m-1) + (m-2) + \cdots + 2) = 1+2+3+4+ \cdots +m = \frac{(m+1)(m)}{2}$. Thus we have a general formula for $a_m$ and substituting $m=12$: $a_{12} = \frac{(13)(12)}{2} = (13)(6) = \boxed{\mathrm{(D) \ } 78}$.

Solution 3

We can literally just plug stuff in. No prerequisite is actually said in the sequence. Since $a_{m+n} = a_m+a_n +mn$, we know $a_2=a_1+a_1+1\cdot1=1+1+1=3$. After this, we can use $a_2$ to find $a_4$. $a_4=a_2+a_2+2\cdot 2 = 3+3+4 = 10$. Now, we can use $a_2$ and $a_4$ to find $a_6$, or $a_6=a_4+a_2+4\cdot 2 = 10+3+8=21$. Lastly, we can use $a_6$ to find $a_{12}$. $a_{12} = a_6+a_6+6\cdot 6 = 21+21+36= \boxed{(D) 78}$

Solution 4

We can set $n$ equal to $m$, so we can say that \[a_{m + m} = a_m + a_m + m*m\] \[a_{2m} = 2a_m + m^2\]

We set $2m = 12$, we get $m = 6$. \[a_{12} = 2a_6 + 36\]

We set $2m = 6$m, we get $m = 3$. \[a_6 = 2a_3 + 9\]

Solving for $a_3$ is easy, just direct substitution. \[a_2 = 1 + 1 + 1 = 3\] \[a_3 = a_{2 + 1} = 3 + 1 + 2 = 6\]

Substituting, we get \[a_6 = 2(6) + 9 = 21\] \[a_{12} = 2(21) + 36 = 78\]

Thus, the answer is $\boxed{D}$.

~ euler123

Solution 5

Note that the sequence of triangular numbers $T_n=1+2+3+...+n$ satisfies these conditions. It is immediately obvious that it satisfies $a_1=1$, and $a_{m+n}=a_m+a_n+mn$ 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 $T_n = \frac{n(n+1)}{2}$, so the answer is $T_{12} = \frac{12(12+1)}{2} = \boxed{\mathrm{(D) \ } 78}$. ~emerald_block

See also

2002 AMC 10B (ProblemsAnswer KeyResources)
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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