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2015 AMC 10B Problem 10

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2015 AMC 10B Problem 10AMC 10 Step by Step
Name:                             Date: September 13, 2026
Modular Arithmetic · difficulty 2 of 5 · about 2 min · position P1-10

What is the sign and units digit of the product of all the odd negative integers strictly greater than −2015-2015 ?

  1. A)

    It is a negative number ending with a 1.\text{It is a negative number ending with a 1.}

  2. B)

    It is a positive number ending with a 1.\text{It is a positive number ending with a 1.}

  3. C)

    It is a negative number ending with a 5.\text{It is a negative number ending with a 5.}

  4. D)

    It is a positive number ending with a 5.\text{It is a positive number ending with a 5.}

  5. E)

    It is a negative number ending with a 0.\text{It is a negative number ending with a 0.}

Answer

C

Key insight

There are 1007 factors, an odd count, so the product is negative; it contains the factor -5 and only odd factors, so it ends in 5.

Solution

The integers in question are −1,−3,−5,…,−2013-1, -3, -5, \dots, -2013 (not −2015-2015, since "strictly greater"). Writing them as −(2m−1)-(2m-1) for m=1,…,1007m = 1, \dots, 1007, there are 10071007 of them.

Sign: each factor is negative and there are 10071007 factors, an odd number, so the product is negative.

Units digit: the absolute value is 1⋅3⋅5⋯20131 \cdot 3 \cdot 5 \cdots 2013. This product is odd (all factors odd) and divisible by 55 (it contains the factor 55). An odd multiple of 55 ends in 55.

So the product is negative and ends in 55: (C) It is a negative number ending with a 5.\boxed{\textbf{(C)}\ \text{It is a negative number ending with a 5.}}

Why this works

Two independent invariants settle the two questions. The sign of a product is determined by the parity of the number of negative factors, and a units digit is a mod-1010 fact: divisible by 55 pins the last digit to 00 or 55, and oddness picks 55. No multiplication is needed at all.

The trap

Miscounting the factors as 1008 (including -2015 or an off-by-one) and calling the product positive, choice (D).

Common mistakes

  • Miscounting the factors as 1008 (including −2015-2015 or an off-by-one) and calling the product positive, choice (D).
  • Thinking a product of many numbers "must" end in 00 (choice (E)); that needs a factor of 22, and every factor here is odd.

Techniques

Set up the equation/formula and compute; no special trick needed · Use an invariant, parity, or coloring argument