AMC 10 · 2018 · #9

Grade 6 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation
double-countingtriangular-numberssequences-arithmeticfundamental-counting-principle easier-related-problemdouble-countingpattern-recognition ↑ Prerequisites: sequences-arithmetictriangular-numbers
📏 Medium solution 💡 3 insights
Problem
Two summation signs are stacked. The outer counter runs through 1 up to 100, and for each outer value the inner counter runs through 1 up to 100 as well. For every pair you write down the sum of the two counters. Find the total of all those numbers added together.

Pick an answer.

(A)
$100{,}100$
(B)
$500{,}500$
(C)
$505{,}000$
(D)
$1{,}001{,}000$
(E)
$1{,}010{,}000$
How to solve
Strategy Organize Information in More Ways

The 10,000 terms are already organized one way — pair by pair — and that way is useless, because nobody adds 10,000 things by hand. Tool #15 (Organize Information in More Ways) re-sorts the same terms into a 100 × 100 grid and asks a different question: how many times does each number get counted? Repeated counting turns into multiplication, and the whole sum collapses. Tool #9 (Solve an Easier Related Problem) makes the structure visible first by shrinking 100 down to 3, where all nine terms fit on one line. Tool #7 (Identify Subproblems) splits the awkward i+j into two clean sums that can be handled separately, and Tool #5 (Look for a Pattern) supplies the pairing trick for 1+2+…+100.

1STEP 1

Shrink 100 down to 3

Shrink a hundred down to three.

Σ³_i=1 Σ³_j=1 (i+j) = (2+3+4)+(3+4+5)+(4+5+6) = 36
2STEP 2

Split i+j into two sums

Split the sum into two.

Σ¹⁰⁰_i=1 Σ¹⁰⁰_j=1 (i+j) = Σ¹⁰⁰_i=1 Σ¹⁰⁰_j=1 i + Σ¹⁰⁰_i=1 Σ¹⁰⁰_j=1 j
3STEP 3

Count how often each number appears

Each number appears a hundred times.

Σ¹⁰⁰_i=1 Σ¹⁰⁰_j=1 i = 100Σ¹⁰⁰_i=1 i, Σ¹⁰⁰_i=1 Σ¹⁰⁰_j=1 j = 100Σ¹⁰⁰_j=1 j
4STEP 4

Add 1 through 100 by pairing

Pairing gives the sum one through a hundred.

1+2+…+100 = 50 × 101 = 5050
5STEP 5

Multiply and finish

Multiplying and adding gives one million ten thousand.

100 · 5050 + 100 · 5050 = 505,000 + 505,000 = 1,010,000 → (E)
Answer
1,010,000
Check with averages. There are 100 × 100 = 10,000 terms. The value of i averages (1+100)/2 = 50.5 and so does j, so a typical term i+j is about 101. That predicts 10,000 × 101 = 1,010,000, matching (E) exactly. The small case also checks out: the same reasoning with 3 gives 2 · 3 · (1+2+3) = 36, which is what was counted by hand in the first step. Choice (C) 505,000 is the trap for anyone who computes one half and forgets that the j parts contribute just as much, and (B) 500,500 is close to 100 × 5050 but not equal to it.
💡Key takeaway

When a sum has far too many terms to add one at a time, stop adding and start counting how many times each number shows up — then multiply.

  • Shrink 100 down to 3
  • Split i+j into two sums
  • Count how often each number appears
  • Add 1 through 100 by pairing
  • Multiply and finish

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