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Question 1210226: Consider the sequence
1, 5, 6, 25, 26, 30, 31, ...
which consists of every positive integer that can be expressed as a sum of distinct powers of 5.
What is the first term that is greater than 50?
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 5:
Powers of 5 are:
$5^0 = 1$
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
and so on.
The sequence consists of numbers that can be written in the form $c_0 \cdot 5^0 + c_1 \cdot 5^1 + c_2 \cdot 5^2 + c_3 \cdot 5^3 + \dots$, where each $c_i$ is either 0 or 1 (because the powers of 5 must be distinct).
Let's generate the terms of the sequence in increasing order:
Using only $5^0 = 1$:
$1 \cdot 5^0 = 1$
Using $5^0 = 1$ and $5^1 = 5$:
$1 \cdot 5^0 = 1$ (already listed)
$1 \cdot 5^1 = 5$
$1 \cdot 5^0 + 1 \cdot 5^1 = 1 + 5 = 6$
Using $5^0 = 1$, $5^1 = 5$, and $5^2 = 25$:
$1 \cdot 5^0 = 1$
$1 \cdot 5^1 = 5$
$1 \cdot 5^2 = 25$
$1 \cdot 5^0 + 1 \cdot 5^1 = 1 + 5 = 6$
$1 \cdot 5^0 + 1 \cdot 5^2 = 1 + 25 = 26$
$1 \cdot 5^1 + 1 \cdot 5^2 = 5 + 25 = 30$
$1 \cdot 5^0 + 1 \cdot 5^1 + 1 \cdot 5^2 = 1 + 5 + 25 = 31$
The sequence generated so far in increasing order is: 1, 5, 6, 25, 26, 30, 31, ...
We are looking for the first term that is greater than 50. Let's continue generating terms by including the next power of 5, which is $5^3 = 125$.
The terms formed using $5^0, 5^1, 5^2$ are all less than or equal to $1 + 5 + 25 = 31$. Now, let's consider sums that include $5^3 = 125$:
$1 \cdot 5^3 = 125$
$1 \cdot 5^0 + 1 \cdot 5^3 = 1 + 125 = 126$
$1 \cdot 5^1 + 1 \cdot 5^3 = 5 + 125 = 130$
$1 \cdot 5^2 + 1 \cdot 5^3 = 25 + 125 = 150$
$1 \cdot 5^0 + 1 \cdot 5^1 + 1 \cdot 5^3 = 1 + 5 + 125 = 131$
$1 \cdot 5^0 + 1 \cdot 5^2 + 1 \cdot 5^3 = 1 + 25 + 125 = 151$
$1 \cdot 5^1 + 1 \cdot 5^2 + 1 \cdot 5^3 = 5 + 25 + 125 = 155$
$1 \cdot 5^0 + 1 \cdot 5^1 + 1 \cdot 5^2 + 1 \cdot 5^3 = 1 + 5 + 25 + 125 = 156$
Looking at the sequence 1, 5, 6, 25, 26, 30, 31, ..., the next terms will be formed by sums of distinct powers of 5 greater than 31. The next power of 5 is 125. The smallest positive integer that can be expressed as a sum of distinct powers of 5 and is greater than 31 will involve the smallest power of 5 that is greater than 31, which is 125.
The smallest term in the sequence that includes $5^3 = 125$ is $1 \cdot 5^3 = 125$.
Since 125 is greater than 50, and all previous terms in the sequence (formed by sums of 1, 5, and 25) are less than or equal to $1+5+25=31$, the first term in the sequence that is greater than 50 is 125.
Final Answer: The final answer is $\boxed{125}$
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The last number shown in the sequence is 
That is the sum of all the distinct powers of 5 up to the second power, so the next number in the sequence must involve the third power of 5, which is 125.
And since 125 is greater than 50, it is the next number in the sequence.
ANSWER: 125
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