Difference between revisions of "2005 AMC 8 Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
| − | There are <math>5</math> days from May 1 to May 5. If we set the first day as <math>n</math>, then the second day can be expressed as <math>n+6</math> | + | There are <math>5</math> days from May 1 to May 5. If we set the first day as <math>n</math>, then the second day can be expressed as <math>n+6</math>, the third as <math>n+12</math>, and so on, for five days. |
| + | The sum <math>n+(n+6)+(n+12)+(n+18)+(n+24)</math> is obviously equal | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2005|num-b=11|num-a=13}} | {{AMC8 box|year=2005|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 04:14, 23 July 2018
Contents
Problem
Big Al, the ape, ate 100 bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many bananas did Big Al eat on May 5?
Solution
There are 
days from May 1 to May 5. The number of bananas he eats each day is an arithmetic sequence. He eats 
bananas on May 5, and 
bananas on May 1. The sum of this arithmetic sequence is equal to 
.
Solution 2
There are days from May 1 to May 5. If we set the first day as , then the second day can be expressed as 
, the third as 
, and so on, for five days.
The sum 
is obviously equal
See Also
| 2005 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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. 
