2018 AMC 12B Problem 9

Attempt Problem 9 of the 2018 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12B solutions, or check the answer key.

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9.

What is ∑i=1100∑j=1100(i+j)? \sum_{i=1}^{100}\sum_{j=1}^{100}(i+j)?
100,100100{,}100
500,500500{,}500
505,000505{,}000
1,001,0001{,}001{,}000
1,010,0001{,}010{,}000
Answer: E
Concepts:summationarithmetic sequence
Small hint:
Split the sum into ∑∑i+∑∑j\sum\sum i+\sum\sum j
Big hint:
The sum of the first 100100 positive integers is 50505050
Solution:
Splitting the sum, ∑i=1100∑j=1100(i+j)=∑i=1100∑j=1100i+∑i=1100∑j=1100j=100∑i=1100i+100∑j=1100j. \begin{gathered} \sum_{i=1}^{100}\sum_{j=1}^{100}(i+j) \\ =\sum_{i=1}^{100}\sum_{j=1}^{100}i \\ {}+\sum_{i=1}^{100}\sum_{j=1}^{100}j \\ =100\sum_{i=1}^{100}i \\ {}+100\sum_{j=1}^{100}j. \end{gathered}
Since ∑k=1100k=5050,\sum_{k=1}^{100}k=5050, this equals 100⋅5050100\cdot5050 +100⋅5050+100\cdot5050 =1,010,000.=1{,}010{,}000.
Thus, the correct answer is E.
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