2018 AMC 12B Problem 9

Below is the professionally curated solution for Problem 9 of the 2018 AMC 12B, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:summationarithmetic sequence

Difficulty rating: 1620

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

Solution:

Splitting the sum, i=1100j=1100(i+j)=i=1100j=1100i+i=1100j=1100j=100i=1100i+100j=1100j. \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.

Since k=1100k=5050,\sum_{k=1}^{100}k=5050, this equals 1005050+1005050=1,010,000.100\cdot5050+100\cdot5050=1{,}010{,}000.

Thus, the correct answer is E.

Problem 9 in Other Years