SOLUTION: Consider the sequence 1, 3, 4, 9, 10, 12, 13, ... which consists of every positive integer that can be expressed as a sum of distinct powers of 3. What is the first term that

Question 1210225: Consider the sequence
1, 3, 4, 9, 10, 12, 13, ...
which consists of every positive integer that can be expressed as a sum of distinct powers of 3.
What is the first term that is greater than 20?
Found 2 solutions by CPhill, greenestamps:
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's analyze how the numbers in the sequence are formed by sums of distinct powers of 3:
Powers of 3 are:
$3^0 = 1$
$3^1 = 3$
$3^2 = 9$
$3^3 = 27$
$3^4 = 81$
and so on.
The sequence consists of numbers that can be written in the form $c_0 \cdot 3^0 + c_1 \cdot 3^1 + c_2 \cdot 3^2 + c_3 \cdot 3^3 + \dots$, where each $c_i$ is either 0 or 1 (because the powers of 3 must be distinct).
Let's generate the terms of the sequence in increasing order:
Using only $3^0 = 1$:
$1 \cdot 3^0 = 1$
Using $3^0 = 1$ and $3^1 = 3$:
$1 \cdot 3^0 = 1$ (already listed)
$1 \cdot 3^1 = 3$
$1 \cdot 3^0 + 1 \cdot 3^1 = 1 + 3 = 4$
Using $3^0 = 1$, $3^1 = 3$, and $3^2 = 9$:
$1 \cdot 3^0 = 1$
$1 \cdot 3^1 = 3$
$1 \cdot 3^2 = 9$
$1 \cdot 3^0 + 1 \cdot 3^1 = 1 + 3 = 4$
$1 \cdot 3^0 + 1 \cdot 3^2 = 1 + 9 = 10$
$1 \cdot 3^1 + 1 \cdot 3^2 = 3 + 9 = 12$
$1 \cdot 3^0 + 1 \cdot 3^1 + 1 \cdot 3^2 = 1 + 3 + 9 = 13$
The sequence generated so far in increasing order is: 1, 3, 4, 9, 10, 12, 13, ...
We are looking for the first term that is greater than 20. Let's continue generating terms by including the next power of 3, which is $3^3 = 27$.
The terms formed using $3^0, 3^1, 3^2$ are all less than $1 + 3 + 9 = 13$. Now, let's consider sums that include $3^3 = 27$:
$1 \cdot 3^3 = 27$
$1 \cdot 3^0 + 1 \cdot 3^3 = 1 + 27 = 28$
$1 \cdot 3^1 + 1 \cdot 3^3 = 3 + 27 = 30$
$1 \cdot 3^2 + 1 \cdot 3^3 = 9 + 27 = 36$
$1 \cdot 3^0 + 1 \cdot 3^1 + 1 \cdot 3^3 = 1 + 3 + 27 = 31$
$1 \cdot 3^0 + 1 \cdot 3^2 + 1 \cdot 3^3 = 1 + 9 + 27 = 37$
$1 \cdot 3^1 + 1 \cdot 3^2 + 1 \cdot 3^3 = 3 + 9 + 27 = 39$
$1 \cdot 3^0 + 1 \cdot 3^1 + 1 \cdot 3^2 + 1 \cdot 3^3 = 1 + 3 + 9 + 27 = 40$
Looking at the sequence 1, 3, 4, 9, 10, 12, 13, ..., the next terms will be formed by sums of distinct powers of 3 greater than 13. The next power of 3 is 27. The smallest positive integer that can be expressed as a sum of distinct powers of 3 and is greater than 13 will involve the smallest power of 3 that is greater than 13, which is 27.
The smallest term in the sequence that includes $3^3 = 27$ is $1 \cdot 3^3 = 27$.
Since 27 is greater than 20, and all previous terms in the sequence (formed by sums of 1, 3, and 9) are less than or equal to $1+3+9=13$, the first term in the sequence that is greater than 20 is 27.
Final Answer: The final answer is $\boxed{27}$
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

The last given number in the sequence is .

Since that number is the sum of all the distinct powers of 3 up to the second power, the next number in the sequence must involve the third power of 3.

3 to the third power is 27, which is greater than 20, so it is the next number in the sequence.

ANSWER: 27

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