Difference between revisions of "2015 AMC 10B Problems/Problem 10"

(Solution)
m (Problem)
Line 1: Line 1:
 
==Problem==
 
==Problem==
 
What are the sign and units digit of the product of all the odd negative integers strictly greater than <math>-2015</math>?
 
What are the sign and units digit of the product of all the odd negative integers strictly greater than <math>-2015</math>?
 +
<br>
 +
<math>\textbf{(A) }</math> It is a negative number ending with a 1. <br>
 +
<math>\textbf{(B) }</math> It is a positive number ending with a 1. <br>
 +
<math>\textbf{(C) }</math> It is a negative number ending with a 5. <br>
 +
<math>\textbf{(D) }</math> It is a positive number ending with a 5. <br>
 +
<math>\textbf{(E) }</math> It is a negative number ending with a 0. <br>
 +
 
==Solution==
 
==Solution==
 
Since <math>-5>-2015</math>, the product must end with a <math>5</math>.
 
Since <math>-5>-2015</math>, the product must end with a <math>5</math>.

Revision as of 16:36, 2 May 2020

Problem

What are the sign and units digit of the product of all the odd negative integers strictly greater than $-2015$?
$\textbf{(A) }$ It is a negative number ending with a 1.
$\textbf{(B) }$ It is a positive number ending with a 1.
$\textbf{(C) }$ It is a negative number ending with a 5.
$\textbf{(D) }$ It is a positive number ending with a 5.
$\textbf{(E) }$ It is a negative number ending with a 0.

Solution

Since $-5>-2015$, the product must end with a $5$.

The multiplicands are the odd negative integers from $-1$ to $-2013$. There are $\frac{|-2013+1|}2+1=1006+1$ of these numbers. Since $(-1)^{1007}=-1$, the product is negative.

Therefore, the answer must be $\boxed{\text{(\textbf C) It is a negative number ending with a 5.}}$

Alternatively, we can calculate the units digit of each group of 5 odd numbers using modular arithmetic

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png