Difference between revisions of "2015 AIME I Problems/Problem 1"
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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>. | 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>. | ||
+ | |||
+ | ==Video Solution For Problems 1-3== | ||
+ | https://www.youtube.com/watch?v=5HAk-6qlOH0 | ||
+ | |||
==Solution 1== | ==Solution 1== | ||
We have <cmath>|A-B|=|1+3(4-2)+5(6-4)+ \cdots + 37(38-36)-39(1-38)|</cmath><cmath>\implies |2(1+3+5+7+ \cdots +37)-1-39(37)|</cmath><cmath>\implies |361(2)-1-39(37)|=|722-1-1443|=|-722|\implies \boxed{722}</cmath> | We have <cmath>|A-B|=|1+3(4-2)+5(6-4)+ \cdots + 37(38-36)-39(1-38)|</cmath><cmath>\implies |2(1+3+5+7+ \cdots +37)-1-39(37)|</cmath><cmath>\implies |361(2)-1-39(37)|=|722-1-1443|=|-722|\implies \boxed{722}</cmath> |
Latest revision as of 00:26, 9 January 2023
Contents
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
.
Video Solution For Problems 1-3
https://www.youtube.com/watch?v=5HAk-6qlOH0
Solution 1
We have
Solution 2
We see that
and
.
Therefore,
Solution 3 (slower solution)
For those that aren't shrewd enough to recognize the above, we may use Newton's Little Formula to semi-bash the equations.
We write down the pairs of numbers after multiplication and solve each layer:
and
Then we use Newton's Little Formula for the sum of terms in a sequence.
Notice that there are terms in each sequence, plus the tails of
and
on the first and second equations, respectively.
So,
Subtracting from
gives:
Which unsurprisingly gives us
-jackshi2006
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.