Difference between revisions of "2015 AIME I Problems/Problem 1"
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==Problem== | ==Problem== | ||
| − | The expressions <math>A</math> = <math> 1 \times 2 + 3 \times 4 + 5 \times 6 + | + | The expressions <math>A</math> = <math> 1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + 37 \times 38 + 39 </math> and <math>B</math> = <math> 1 + 2 \times 3 + 4 \times 5 + \cdots + 36 \times 37 + 38 \times 39 </math> are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers <math>A</math> and <math>B</math>. |
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| + | ==Solution== | ||
| + | |||
| + | We see that | ||
| + | |||
| + | <math>A=(1\times 2)+(3\times 4)+(5\times 6)+\cdots +(35\times 36)+(37\times 38)+39</math> | ||
| + | |||
| + | and | ||
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| + | <math>B=1+(2\times 3)+(4\times 5)+(6\times 7)+\cdots +(36\times 37)+(38\times 39)</math>. | ||
| + | |||
| + | Therefore, | ||
| + | |||
| + | <math>B-A=-38+(2\times 2)+(2\times 4)+(2\times 6)+\cdots +(2\times 36)+(2\times 38)</math> | ||
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| + | <math>=-38+4\times (1+2+3+\cdots+19)</math> | ||
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| + | <math>=-38+4\times\frac{20\cdot 19}{2}=-38+760=\boxed{722}.</math> | ||
== See also == | == See also == | ||
Revision as of 15:58, 20 March 2015
Problem
The expressions 
= 
and 
= 
are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers and .
Solution
We see that
and
.
Therefore,
See also
| 2015 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
