Difference between revisions of "2018 AMC 12B Problems/Problem 9"

(Reformatted answer choices and writing in this page.)
(Improved Sol 2 significantly.)
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== Solution 2 ==
 
== Solution 2 ==
We have
+
Recall that the sum of the first <math>100</math> positive integers is <math>\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.</math> It follows that
<cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) = \sum^{100}_{i=1} (100i+5050) = 100 \cdot 5050 + 5050 \cdot 100 = \boxed{\textbf{(E) }1{,}010{,}000}.</cmath>
+
<cmath>\begin{align*}
 +
\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \left(\sum^{100}_{j=1} i + \sum^{100}_{j=1} j\right) \\
 +
&= \sum^{100}_{i=1} (100i+5050) \\
 +
&= 100\sum^{100}_{i=1} i + \sum^{100}_{i=1} 5050 \\
 +
&= 100\cdot5050+5050\cdot100 \\
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&= \boxed{\textbf{(E) }1{,}010{,}000}.
 +
\end{align*}</cmath>
 +
~Vfire ~MRENTHUSIASM
  
 
== Solution 3 ==
 
== Solution 3 ==
We have
+
Recall that the sum of the first <math>100</math> positive integers is <math>\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.</math> It follows that
 
<cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j)  = \sum^{100}_{i=1} \sum^{100}_{i=1} 2i =  100\cdot(5050\cdot2) =  \boxed{\textbf{(E) }1{,}010{,}000}.</cmath>
 
<cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j)  = \sum^{100}_{i=1} \sum^{100}_{i=1} 2i =  100\cdot(5050\cdot2) =  \boxed{\textbf{(E) }1{,}010{,}000}.</cmath>
  

Revision as of 12:55, 20 September 2021

Problem

What is \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ?\]

$\textbf{(A) }100{,}100 \qquad \textbf{(B) }500{,}500\qquad \textbf{(C) }505{,}000 \qquad \textbf{(D) }1{,}001{,}000 \qquad \textbf{(E) }1{,}010{,}000 \qquad$

Solution 1

We can start by writing out the first couple of terms: \[\begin{array}{ccccccccc} (1+1) &+ &(1+2) &+ &(1+3) &+ &\dots &+ &(1+100) \\ (2+1) &+ &(2+2) &+ &(2+3) &+ &\dots &+ &(2+100) \\ (3+1) &+ &(3+2) &+ &(3+3) &+ &\dots &+ &(3+100) \\ [-1ex] &&&&\vdots&&&& \\ (100+1) &+ &(100+2) &+ &(100+3) &+ &\dots &+ &(100+100) \end{array}\] Looking at the first terms in the parentheses, we can see that $1+2+3+\dots+100$ occurs $100$ times. It goes vertically and exists $100$ times horizontally. Looking at the second terms in the parentheses, we can see that $1+2+3+\dots+100$ occurs $100$ times. It goes horizontally and exists $100$ times vertically.

Thus, we have \[2\left(\dfrac{100\cdot101}{2}\cdot 100\right)=\boxed{\textbf{(E) }1{,}010{,}000}.\]

Solution 2

Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ It follows that \begin{align*} \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \left(\sum^{100}_{j=1} i + \sum^{100}_{j=1} j\right) \\ &= \sum^{100}_{i=1} (100i+5050) \\ &= 100\sum^{100}_{i=1} i + \sum^{100}_{i=1} 5050 \\ &= 100\cdot5050+5050\cdot100 \\ &= \boxed{\textbf{(E) }1{,}010{,}000}. \end{align*} ~Vfire ~MRENTHUSIASM

Solution 3

Recall that the sum of the first $100$ positive integers is $\sum^{100}_{k=1} k = \frac{101\cdot100}{2}=5050.$ It follows that \[\sum^{100}_{i=1} \sum^{100}_{j=1} (i+j)  = \sum^{100}_{i=1} \sum^{100}_{i=1} 2i =  100\cdot(5050\cdot2) =  \boxed{\textbf{(E) }1{,}010{,}000}.\]

Solution 4

The minimum term is $1 + 1 = 2$, and the maximum term is $100 + 100 = 200$. The average of the $100 \cdot 100 = 10{,}000$ terms is the average of the minimum and maximum terms, which is $\frac{2+200}{2}=101$. The sum is therefore $101 \cdot 10{,}000 = \boxed{\textbf{(E) }1{,}010{,}000}$.

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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 12 Problems and Solutions

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