AMC 10 · 2021 · #5

Grade 7 number-theorycounting

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

fraction-arithmeticgcdsystematic-enumerationpair-counting systematic-enumerationidentify-subproblems ↑ Prerequisites: fraction-arithmeticgcd

📏 Medium solution 💡 3 insights

Problem

A fraction is called special when its numerator and denominator are positive integers adding to 15. It need not be in lowest terms. Add two special fractions, possibly the same one twice, and ask when the result is an integer. Count how many different integers can be produced.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Make a Systematic List

There are only 14 special fractions, so the whole universe of the problem fits on one line. Listing them all is cheap and removes any guessing. Pairing all 14 with all 14 would be 105 additions, which is too many by hand, so after the list we re-sort the fractions by their reduced denominator. That one re-sort kills almost every pair at once and leaves three tiny groups to add.

1STEP 1

List all special fractions

List all the special fractions.

a + b = 15, a ≥ 1, b ≥ 1 → a/b ∈ { 1/14, 2/13, 3/12, 4/11, 5/10, 6/9, 7/8, 8/7, 9/6, 10/5, 11/4, 12/3, 13/2, 14/1 }

2STEP 2

Reduce each to lowest terms

Reduce each to lowest terms.

1/14, 2/13, 1/4, 4/11, 1/2, 2/3, 7/8, 8/7, 3/2, 2, 11/4, 4, 13/2, 14

3STEP 3

Denominators must match

The denominators must match for an integer sum.

p/q + r/s = n → r/s = (nq - p)/q, gcd(nq - p, q) = gcd(p, q) = 1 → s = q

4STEP 4

Sort into denominator groups

Sort into denominator groups.

q = 1: {2, 4, 14} q = 2: { 1/2, 3/2, 13/2 } q = 4: { 1/4, 11/4 } 2p/q ∈ Z, gcd(p, q) = 1 → q ∣ 2

5STEP 5

Add inside each group

Add inside each group.

{2, 4, 14} → {4, 6, 8, 16, 18, 28}, { 1/2, 3/2, 13/2 } → {1, 2, 3, 7, 8, 13}, { 1/4, 11/4 } → {3}

6STEP 6

Merge the lists and count

Merging the lists gives 11.

{4, 6, 8, 16, 18, 28} ∪ {1, 2, 3, 7, 8, 13} ∪ {3} = {1, 2, 3, 4, 6, 7, 8, 13, 16, 18, 28}, 13 - 2 = 11

Answer

The final set can be checked one entry at a time against actual special fractions: 1 = 5/10 + 5/10, 3 = 3/12 + 11/4, 7 = 5/10 + 13/2, 28 = 14/1 + 14/1, and so on, so all 11 values are genuinely reachable. The raw count of group sums is 13, which is exactly choice (E) and is the trap for anyone who forgets that 8 and 3 each appear twice; catching only one of the two repeats gives 12, choice (D). Since 11 is what survives after both repeats are removed, it is consistent.

💡Key takeaway

Two fractions in lowest terms can only add to a whole number when their denominators are already the same, so sorting by denominator turns a huge pile of pairs into three short lists.

List all special fractions
Reduce each to lowest terms
Denominators must match
Sort into denominator groups
Add inside each group
Merge the lists and count