Difference between revisions of "2015 AIME I Problems/Problem 1"
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Jackshi2006 (talk | contribs) (→Solution 2 (slower solution)) |
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Notice that there are 19 terms in each sequence, plus the tails of 39 and 1 on the first and second equations, respectively. | Notice that there are 19 terms in each sequence, plus the tails of 39 and 1 on the first and second equations, respectively. | ||
− | So: | + | So: (this is as far as my latex knowledge goes) |
− | <math> | + | ((19 choose 1) x 2) + ((19 choose 2) x 10) + ((19 choose 3) x 8) + 1 |
+ | |||
+ | ((19 choose 1) x 6) + ((19 choose 2) x 14) + ((19 choose 3) x 8) + 39 | ||
+ | |||
+ | Subtracting A from B gives: | ||
+ | |||
+ | ((19 choose 1) x 4) + ((19 choose 2) x 4) - 38 | ||
+ | |||
+ | Which unsurprisingly gives us <math>/boxed{722}.</math> | ||
== See also == | == See also == |
Revision as of 14:08, 25 March 2020
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 see that
and
.
Therefore,
Solution 2 (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 n terms in a sequence.
Notice that there are 19 terms in each sequence, plus the tails of 39 and 1 on the first and second equations, respectively.
So: (this is as far as my latex knowledge goes)
((19 choose 1) x 2) + ((19 choose 2) x 10) + ((19 choose 3) x 8) + 1
((19 choose 1) x 6) + ((19 choose 2) x 14) + ((19 choose 3) x 8) + 39
Subtracting A from B gives:
((19 choose 1) x 4) + ((19 choose 2) x 4) - 38
Which unsurprisingly gives us
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.