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=1100j=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
Difficulty rating: 1620
Solution:

Splitting the sum, i=1100j=1100(i+j)=i=1100j=1100i+i=1100j=1100j=100i=1100i+100j=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 1005050100\cdot5050 +1005050+100\cdot5050 =1,010,000.=1{,}010{,}000.

Thus, the correct answer is E.

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