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>. | ||
+ | ==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> | ||
− | ==Solution== | + | ==Solution 2== |
We see that | We see that | ||
Line 20: | Line 22: | ||
<math>=-38+4\times\frac{20\cdot 19}{2}=-38+760=\boxed{722}.</math> | <math>=-38+4\times\frac{20\cdot 19}{2}=-38+760=\boxed{722}.</math> | ||
+ | |||
+ | ==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: | ||
+ | |||
+ | |||
+ | <cmath>2, 12, 30, 56, 90...(39)</cmath> | ||
+ | |||
+ | <cmath>6, 18, 26, 34...</cmath> | ||
+ | |||
+ | <cmath>8, 8, 8...</cmath> | ||
+ | |||
+ | and | ||
+ | |||
+ | <cmath>(1) 6, 20, 42, 72...</cmath> | ||
+ | <cmath>14, 22, 30...</cmath> | ||
+ | |||
+ | <cmath>8, 8, 8...</cmath> | ||
+ | |||
+ | |||
+ | Then we use Newton's Little Formula for the sum of <math>n</math> terms in a sequence. | ||
+ | |||
+ | Notice that there are <math>19</math> terms in each sequence, plus the tails of <math>39</math> and <math>1</math> on the first and second equations, respectively. | ||
+ | |||
+ | |||
+ | So, | ||
+ | |||
+ | |||
+ | <cmath>2\binom{19}{1}+10\binom{19}{2}+8\binom{19}{3}+1</cmath> | ||
+ | |||
+ | |||
+ | <cmath>6\binom{19}{1}+14\binom{19}{2}+8\binom{19}{3}+39</cmath> | ||
+ | |||
+ | |||
+ | Subtracting <math>A</math> from <math>B</math> gives: | ||
+ | |||
+ | |||
+ | <cmath>4\binom{19}{1}+4\binom{19}{2}-38</cmath> | ||
+ | |||
+ | |||
+ | Which unsurprisingly gives us <math>\boxed{722}.</math> | ||
+ | |||
+ | |||
+ | -jackshi2006 | ||
== See also == | == See also == |
Latest revision as of 06:04, 2 January 2022
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 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.