1985 AIME Problems/Problem 1
Contents
[hide]Problem
Let , and for
, let
. Calculate the product
.
Solution
Since ,
. Setting
and
in this equation gives us respectively
,
,
and
so
Notice that the value of
was completely unneeded!
Another Way to think about the solution
Every time you multiply in the next term of the sequence, all the numbers before are flipped from the numerator to the denominator or the denominator to the numerator because they are divided. So, 97, the first term, will appear in the multiplied out form in this pattern: where
is the numerator and
is the denominator. The second term (2) will appear in the pattern
where
means that the number is skipped the first term. All the pairs of
cross out and you find that only the even terms have an
left over while all the values of the odd terms are crossed out.
Solution 2
Another way to do this is to realize that most of our numbers will be canceled out in the multiplication in the end, and to just list out the terms of our product and cancel:
See also
1985 AIME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |