Difference between revisions of "2023 AMC 10B Problems/Problem 6"
E is 2.71828 (talk | contribs) m (→Problem) |
E is 2.71828 (talk | contribs) m (→Solution 2) |
||
| Line 25: | Line 25: | ||
Therefore, there are <math>\boxed{\textbf{(B) }674}</math> evens. | Therefore, there are <math>\boxed{\textbf{(B) }674}</math> evens. | ||
| + | |||
| + | ~e_is_2.71828 | ||
Revision as of 16:59, 15 November 2023
Problem
Let 
, and 
for 
. How many terms in the sequence 
are even?
Solution 1
We calculate more terms:
We find a pattern: if 
is a multiple of 
, then the term is even, or else it is odd.
There are 
multiples of from 
to 
.
~Mintylemon66
Solution 2
Like in the other solution, we find a pattern, except in a more rigorous way.
Since we start with and , the next term is 
.
We start with odd, then odd, then (the sum of odd and odd) even, (the sum of odd and even) odd, and so on. Basically the pattern goes: odd, odd, even, odd odd, even, odd, odd even…
When we take 
we get 
with a remainder of one. So we have full cycles, and an extra odd at the end.
Therefore, there are 
evens.
~e_is_2.71828
