Difference between revisions of "2023 AMC 10B Problems/Problem 6"
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| − | On my copy of the AMC 10B, the order of the answers is different, so the correct answer is (E) 674 | + | On my copy of the AMC 10B, the order of the answers is different, so the correct answer is (E) 674 |
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| + | ==Solution 1== | ||
| + | |||
| + | We calculate more terms: | ||
| + | |||
| + | <math>1,3,4,5,9,14,...</math> | ||
| + | |||
| + | We find a pattern: if <math>n</math> is a multiple of <math>3</math>, then the term is even, or else it is odd. | ||
| + | There are <math>\lfloor \frac{2023}{3} \rfloor =\boxed{\textbf{(B) }674}</math> multiples of <math>3</math> from <math>1</math> to <math>2023</math>. | ||
| + | |||
| + | ~Mintylemon66 | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 18:50, 15 November 2023
Problem
Let 
, and 
for 
. How many terms in the sequence 
are even?
On my copy of the AMC 10B, the order of the answers is different, so the correct answer is (E) 674
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
