AMC 10 · 2015 · #10

Grade 7 arithmetic
Archetype Units-Digit Tracking └ Sub-archetype Tail Collapses to Zero
units-digit-trackingparitymultiples identify-subproblems ↑ Prerequisites: units-digit-tracking
📏 Medium solution 💡 2 insights
Problem
Multiply together every odd negative integer that is strictly greater than -2015. Two things are asked about that single product: whether it is positive or negative, and what its units (ones) digit is.

Pick an answer.

(A)
It is a negative number ending with a 1.
(B)
It is a positive number ending with a 1.
(C)
It is a negative number ending with a 5.
(D)
It is a positive number ending with a 5.
(E)
It is a negative number ending with a 0.

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The answer choices ask two separate things at once — a sign and a units digit — so the cleanest move is tool #7 (Identify Subproblems): solve "is it positive or negative?" and "what is the ones digit?" independently, then combine. The sign comes from counting how many negative factors there are. For the units digit, tool #5 (Look for a Pattern) shows that once a product of odd numbers includes the factor -5, the ones digit locks onto 5 and never changes. Finally tool #3 (Eliminate Possibilities) matches the sign-plus-digit result to exactly one of the five prose choices.

1STEP 1

List the factors and split the question

The odd negatives above -2015 run -1, -3, -5, …, -2013; we need only the sign and ones digit, so split it into two subproblems.

P = (-1)·(-3)·(-5)…(-2013)
2STEP 2

Subproblem 1: find the sign

The factors match the odd numbers 1, 3, 5, …, 2013, so there are 1007 of them — an odd count of negatives makes the product negative.

(2013-1)/2+1 = 1007 (odd) → P ＜ 0
3STEP 3

Subproblem 2: find the units digit

Because a factor is -5 the product is a multiple of 5, and every factor is odd, so the ones digit locks at 5, never 0.

5 ∣ P and P is odd → units digit = 5
4STEP 4

Combine and pick the choice

Combine the two: the product is negative with ones digit 5, which matches exactly one prose choice — (C).

P ＜ 0, units digit = 5 → (C)
Answer
It is a negative number ending with a 5.
Sanity-check both claims on a small version with the same shape, the odd negatives down to -9: (-1)(-3)(-5)(-7)(-9) = -945. There are 5 factors (odd count), so it is negative, and it ends in 5 — a negative number ending in 5, exactly the structure we claimed. The real product just has more factors, but the parity of the count stays odd and the trapped factor of 5 keeps the ones digit at 5, so (C) holds.
💡Key takeaway

When a question asks for only the sign and the last digit, solve each one separately — count the negatives for the sign, and watch the ones digit lock onto 5 once a 5 joins a product of odd numbers.

  • List the factors and split the question
  • Subproblem 1: find the sign
  • Subproblem 2: find the units digit
  • Combine and pick the choice

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