Question 1210224: Find the number of sequences containing three terms, such that
* The second term is equal to the sum of the first term plus one.
* The third term is equal to twice the second term.
* Each term is an integer in \{0, 1, 2, \dots, 100\}.
Found 2 solutions by CPhill, greenestamps:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let the three terms of the sequence be $a_1, a_2, a_3$.
According to the given conditions:
\begin{enumerate}
\item The second term is equal to the sum of the first term plus one: $a_2 = a_1 + 1$.
\item The third term is equal to twice the second term: $a_3 = 2a_2$.
\item Each term is an integer in $\{0, 1, 2, \dots, 100\}$.
\end{enumerate}
From condition 1, since $a_2 = a_1 + 1$, and $a_2$ is an integer, $a_1$ must also be an integer.
From condition 2, since $a_3 = 2a_2$, and $a_2$ is an integer, $a_3$ must also be an integer.
We have the following relationships between the terms:
$a_2 = a_1 + 1$
$a_3 = 2(a_1 + 1) = 2a_1 + 2$
Since each term is in the set $\{0, 1, 2, \dots, 100\}$, we have the following inequalities:
$0 \le a_1 \le 100$
$0 \le a_2 \le 100 \implies 0 \le a_1 + 1 \le 100$
$0 \le a_3 \le 100 \implies 0 \le 2a_1 + 2 \le 100$
Let's analyze the second inequality:
$0 \le a_1 + 1 \implies a_1 \ge -1$
$a_1 + 1 \le 100 \implies a_1 \le 99$
Combining these, we get $-1 \le a_1 \le 99$.
Let's analyze the third inequality:
$0 \le 2a_1 + 2 \implies 2a_1 \ge -2 \implies a_1 \ge -1$
$2a_1 + 2 \le 100 \implies 2a_1 \le 98 \implies a_1 \le 49$
Combining these, we get $-1 \le a_1 \le 49$.
For all three conditions to be satisfied, $a_1$ must satisfy all three inequalities:
$0 \le a_1 \le 100$
$-1 \le a_1 \le 99$
$-1 \le a_1 \le 49$
The intersection of these three ranges for $a_1$ is $0 \le a_1 \le 49$.
Since $a_1$ must be an integer, the possible values for $a_1$ are $0, 1, 2, \dots, 49$.
The number of possible integer values for $a_1$ is $49 - 0 + 1 = 50$.
Each valid value of $a_1$ uniquely determines the sequence $(a_1, a_1+1, 2(a_1+1))$.
For example:
If $a_1 = 0$, the sequence is $(0, 1, 2)$.
If $a_1 = 49$, the sequence is $(49, 50, 100)$.
The number of such sequences is equal to the number of possible values for $a_1$, which is 50.
Final Answer: The final answer is $\boxed{50}$
Answer by greenestamps(13198)  (Show Source):
You can put this solution on YOUR website!
According to the definition of the sequence, each sequence is of the form
n, n+1, 2n+2
The numbers that can be used are the integers from 0 to 100, inclusive. So the "last" sequence is the one in which 2n+2 is equal to 100.
2n+2 = 100
2n = 98
n = 49
The "first" sequence is clearly the one in which n is 0.
So the sequences that satisfy the conditions of the problem are those where n is from 0 to 49, inclusive, which means 50 such sequences.
ANSWER: 50
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