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

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==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>?
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<br>
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<math>\textbf{(A) }</math> It is a negative number ending with a 1. <br>
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<math>\textbf{(B) }</math> It is a positive number ending with a 1. <br>
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<math>\textbf{(C) }</math> It is a negative number ending with a 5. <br>
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<math>\textbf{(D) }</math> It is a positive number ending with a 5. <br>
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<math>\textbf{(E) }</math> It is a negative number ending with a 0. <br>
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==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>.
  
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The multiplicands are the odd negative integers from <math>-1</math> to <math>-2013</math>. There are <math>\frac{2013+1}2=1007</math> of these numbers. Since <math>(-1)^{1007}=-1</math>, the product is negative.
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The multiplicands are the odd negative integers from <math>-1</math> to <math>-2013</math>. There are <math>\frac{|-2013+1|}2+1=1006+1</math> of these numbers. Since <math>(-1)^{1007}=-1</math>, the product is negative.
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 +
Therefore, the answer must be <math>\boxed{\text{(\textbf C) It is a negative number ending with a 5.}}</math>
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Alternatively, we can calculate the units digit of each group of 5 odd numbers using modular arithmetic.
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==Video Solution 1==
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https://youtu.be/r5QASm5hnII
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/UoU-cvet1pg
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~savannahsolver
  
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Therefore, the answer must be <math>\boxed{\text{(\textbf B) It is a negative number ending with a 5.}}</math>
 
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=B|num-b=9|num-a=11}}
 
{{AMC10 box|year=2015|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:13, 2 August 2022

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.

Video Solution 1

https://youtu.be/r5QASm5hnII

~Education, the Study of Everything

Video Solution

https://youtu.be/UoU-cvet1pg

~savannahsolver

See Also

2015 AMC 10B (Problems • Answer Key • Resources)
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

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