Difference between revisions of "2011 AMC 8 Problems/Problem 8"
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==Solution 2== | ==Solution 2== | ||
| − | The sum of an even number added to an odd number is always odd. The smallest possible sum is <math>3</math>, and the largest possible sum is <math>11</math>. The odd numbers in between can be achieved by replacing chips with <math>{ | + | The sum of an even number added to an odd number is always odd. The smallest possible sum is <math>3</math>, and the largest possible sum is <math>11</math>. The odd numbers in between can be achieved by replacing chips with <math>{\pm }2</math> within the same bag. Therefore, we can conclude that there are <math>(11-3)/2(*)+1=\boxed{\textbf{(B)}\ 5}</math> possible sums. |
*Note that we are only accounting for odd numbers, thus the <math>/2</math>. | *Note that we are only accounting for odd numbers, thus the <math>/2</math>. | ||
| − | + | ~megaboy6679 | |
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/qPLOGcNPdTQ | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=7|num-a=9}} | {{AMC8 box|year=2011|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:04, 18 November 2024
Problem
Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips?
Solution 1
By adding a number from Bag A and a number from Bag B together, the values we can get are 
Therefore the number of different values is 
.
Solution 2
The sum of an even number added to an odd number is always odd. The smallest possible sum is 
, and the largest possible sum is 
. The odd numbers in between can be achieved by replacing chips with 
within the same bag. Therefore, we can conclude that there are 
possible sums.
- Note that we are only accounting for odd numbers, thus the
.
.~megaboy6679
Video Solution by WhyMath
See Also
| 2011 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
