Difference between revisions of "2018 AMC 12B Problems/Problem 9"
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<cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ? </cmath> | <cmath> \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) ? </cmath> | ||
− | <math> \textbf{(A) }100,100 \qquad | + | <math> \textbf{(A) }100{,}100 \qquad |
− | \textbf{(B) }500,500\qquad | + | \textbf{(B) }500{,}500\qquad |
− | \textbf{(C) }505,000 \qquad | + | \textbf{(C) }505{,}000 \qquad |
− | \textbf{(D) }1,001,000 \qquad | + | \textbf{(D) }1{,}001{,}000 \qquad |
− | \textbf{(E) }1,010,000 \qquad </math> | + | \textbf{(E) }1{,}010{,}000 \qquad </math> |
== Solution 1 == | == Solution 1 == | ||
+ | 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>\begin{align*} | ||
+ | \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{j=1}i + \sum^{100}_{i=1} \sum^{100}_{j=1}j \\ | ||
+ | &= \sum^{100}_{i=1} (100i) + 100 \sum^{100}_{j=1}j \\ | ||
+ | &= 100 \sum^{100}_{i=1}i + 100 \sum^{100}_{j=1}j \\ | ||
+ | &= 100\cdot5050 + 100\cdot5050 \\ | ||
+ | &= \boxed{\textbf{(E) }1{,}010{,}000}. | ||
+ | \end{align*}</cmath> | ||
+ | ~MRENTHUSIASM | ||
− | + | == Solution 2 == | |
+ | 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>\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*}</cmath> | ||
+ | ~Vfire ~MRENTHUSIASM | ||
− | < | + | == Solution 3 == |
− | < | + | 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> Since the nested summation is symmetric with respect to <math>i</math> and <math>j,</math> it follows that |
− | < | + | <cmath>\begin{align*} |
− | < | + | \sum^{100}_{i=1} \sum^{100}_{j=1} (i+j) &= \sum^{100}_{i=1} \sum^{100}_{i=1} (2i) \\ |
− | <cmath>(100 | + | &= 2\sum^{100}_{i=1} \sum^{100}_{i=1} i \\ |
+ | &= 2\sum^{100}_{i=1} 5050 \\ | ||
+ | &= 2\cdot(5050\cdot100) \\ | ||
+ | &= \boxed{\textbf{(E) }1{,}010{,}000}. | ||
+ | \end{align*}</cmath> | ||
+ | ~Vfire ~MRENTHUSIASM | ||
− | + | == Solution 4 == | |
− | + | The sum contains <math>100\cdot100=10000</math> terms, and the average value of both <math>i</math> and <math>j</math> is <math>\frac{101}{2}.</math> Therefore, the sum becomes <cmath>10000\cdot\left(\frac{101}{2}+\frac{101}{2}\right)=\boxed{\textbf{(E) }1{,}010{,}000}.</cmath> | |
+ | ~Rejas ~MRENTHUSIASM | ||
− | + | == Solution 5 == | |
− | + | We start by writing out the first few terms: | |
+ | <cmath>\begin{array}{ccccccccc} | ||
+ | (1+1) &+ &(1+2) &+ &(1+3) &+ &\cdots &+ &(1+100) \\ | ||
+ | (2+1) &+ &(2+2) &+ &(2+3) &+ &\cdots &+ &(2+100) \\ | ||
+ | (3+1) &+ &(3+2) &+ &(3+3) &+ &\cdots &+ &(3+100) \\ [-1ex] | ||
+ | &&&&\vdots&&&& \\ | ||
+ | (100+1) &+ &(100+2) &+ &(100+3) &+ &\cdots &+ &(100+100). | ||
+ | \end{array}</cmath> | ||
+ | From the first terms in the parentheses, the sum <math>1+2+3+\cdots+100</math> occurs <math>100</math> times vertically. | ||
− | + | From the second terms in the parentheses, the sum <math>1+2+3+\cdots+100</math> occurs <math>100</math> times horizontally. | |
− | < | + | Recall that the sum of the first <math>100</math> positive integers is <math>1+2+3+\cdots+100=\frac{101\cdot100}{2}=5050.</math> Therefore, the answer is <cmath>2\cdot\left(5050\cdot100\right)=\boxed{\textbf{(E) }1{,}010{,}000}.</cmath> |
+ | ~RandomPieKevin ~MRENTHUSIASM | ||
+ | == Solution 6 == | ||
+ | When we expand the nested summation as shown in Solution 5, note that: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>The term <math>2</math> occurs <math>1</math> time. <p> | ||
+ | The term <math>3</math> occurs <math>2</math> times. <p> | ||
+ | The term <math>4</math> occurs <math>3</math> times. <p> | ||
+ | <math>\cdots</math> <p> | ||
+ | The term <math>101</math> occurs <math>100</math> times. <p> | ||
+ | More generally, the term <math>k+1</math> occurs <math>k</math> times for <math>k\in\{1,2,3,\ldots,100\}.</math><p></li> | ||
+ | <li>The term <math>102</math> occurs <math>99</math> times. <p> | ||
+ | The term <math>103</math> occurs <math>98</math> times. <p> | ||
+ | The term <math>104</math> occurs <math>97</math> times. <p> | ||
+ | <math>\cdots</math> <p> | ||
+ | The term <math>200</math> occurs <math>1</math> time. <p> | ||
+ | More generally, the term <math>k+101</math> occurs <math>100-k</math> times for <math>k\in\{1,2,3,\ldots,99\}.</math><p></li> | ||
+ | </ol> | ||
+ | Together, the nested summation becomes | ||
+ | <cmath>\begin{align*} | ||
+ | \sum^{100}_{k=1}\left[(k+1)k\right] + \sum^{99}_{k=1}\left[(k+101)(100-k)\right] &= \sum^{100}_{k=1}\left[k^2+k\right] + \sum^{99}_{k=1}\left[-k^2-k+10100\right] \\ | ||
+ | &= \sum^{100}_{k=1}k^2 + \sum^{100}_{k=1}k - \sum^{99}_{k=1}k^2 - \sum^{99}_{k=1}k + \sum^{99}_{k=1}10100 \\ | ||
+ | &= \left(\sum^{100}_{k=1}k^2 - \sum^{99}_{k=1}k^2\right) + \left(\sum^{100}_{k=1}k - \sum^{99}_{k=1}k\right) + \sum^{99}_{k=1}10100 \\ | ||
+ | &= 100^2+100+10100\cdot99 \\ | ||
+ | &= \boxed{\textbf{(E) }1{,}010{,}000}. | ||
+ | \end{align*}</cmath> | ||
+ | ~MRENTHUSIASM | ||
− | == Solution | + | ==Video Solution (HOW TO THINK CRITICALLY!!!)== |
− | + | https://youtu.be/flvmGmDmwZ8 | |
− | |||
+ | ~Education, the Study of Everything | ||
==See Also== | ==See Also== |
Latest revision as of 22:03, 27 May 2023
Contents
Problem
What is
Solution 1
Recall that the sum of the first positive integers is
It follows that
~MRENTHUSIASM
Solution 2
Recall that the sum of the first positive integers is
It follows that
~Vfire ~MRENTHUSIASM
Solution 3
Recall that the sum of the first positive integers is
Since the nested summation is symmetric with respect to
and
it follows that
~Vfire ~MRENTHUSIASM
Solution 4
The sum contains terms, and the average value of both
and
is
Therefore, the sum becomes
~Rejas ~MRENTHUSIASM
Solution 5
We start by writing out the first few terms:
From the first terms in the parentheses, the sum
occurs
times vertically.
From the second terms in the parentheses, the sum occurs
times horizontally.
Recall that the sum of the first positive integers is
Therefore, the answer is
~RandomPieKevin ~MRENTHUSIASM
Solution 6
When we expand the nested summation as shown in Solution 5, note that:
- The term
occurs
time.
The term
occurs
times.
The term
occurs
times.
The term
occurs
times.
More generally, the term
occurs
times for
- The term
occurs
times.
The term
occurs
times.
The term
occurs
times.
The term
occurs
time.
More generally, the term
occurs
times for
Together, the nested summation becomes
~MRENTHUSIASM
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.