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>? | ||
| − | ==Solution== | + | <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 1== | ||
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>. | ||
| Line 7: | Line 14: | ||
Therefore, the answer must be <math>\boxed{\text{(\textbf C) It is a negative number ending with a 5.}}</math> | Therefore, the answer must be <math>\boxed{\text{(\textbf C) It is a negative number ending with a 5.}}</math> | ||
| + | |||
| + | Alternatively, we can calculate the units digit of each group of 5 odd numbers using modular arithmetic. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | To find the sign of the product, we need to count the number of integers in this set. We can do this by counting the positive odd integers from <math>1</math> to <math>2013</math>. This is equal to <math>\frac{2013-1}{2}+1=1007</math>. Since there are <math>1007</math> negative numbers being multiplied, and <math>1007</math> is an odd number, the product will be negative. To find the units digit of the product, we only need to multiply the units digits of the numbers. This is equal to <math>1\times3\times5\times7\times9... 1\times3</math>. Notice how it repeats in groups of <math>1\times3\times5\times7\times9...</math> until the end where we just do an extra <math>1\times3</math>. Let's go by cycles of <math>1\times3\times5\times7\times9</math>. The first cycle we have <math>1\times3\times5\times7\times9=945</math>, and our second cycle we have <math>1\times3\times5\times7\times9\times1\times3\times5\times7\times9=945\times945</math>. Notice how the units digits is always 5, so whatever we multiply it by always has a units digits of 5, so our answer is <math>\boxed{\text{(\textbf C) It is a negative number ending with a 5.}}</math> | ||
| + | |||
| + | ~Baihly2024 | ||
| + | |||
| + | ==Video Solution 1== | ||
| + | https://youtu.be/r5QASm5hnII | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/UoU-cvet1pg | ||
| + | |||
| + | ~savannahsolver | ||
==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 18:47, 12 October 2025
Problem
What are the sign and units digit of the product of all the odd negative integers strictly greater than 
?
It is a negative number ending with a 1.
It is a positive number ending with a 1.
It is a negative number ending with a 5.
It is a positive number ending with a 5.
It is a negative number ending with a 0.
Solution 1
Since 
, the product must end with a 
.
The multiplicands are the odd negative integers from 
to 
. There are 
of these numbers. Since 
, the product is negative.
Therefore, the answer must be 
Alternatively, we can calculate the units digit of each group of 5 odd numbers using modular arithmetic.
Solution 2
To find the sign of the product, we need to count the number of integers in this set. We can do this by counting the positive odd integers from 
to 
. This is equal to 
. Since there are 
negative numbers being multiplied, and is an odd number, the product will be negative. To find the units digit of the product, we only need to multiply the units digits of the numbers. This is equal to 
. Notice how it repeats in groups of 
until the end where we just do an extra 
. Let's go by cycles of 
. The first cycle we have 
, and our second cycle we have 
. Notice how the units digits is always 5, so whatever we multiply it by always has a units digits of 5, so our answer is
~Baihly2024
Video Solution 1
~Education, the Study of Everything
Video Solution
~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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
