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Problem 22
When a positive integer \(N\) is fed into a machine, the output is a number calculated according to the rule shown below.
For example, starting with an input of \(N=7\), the machine will output \(3 \cdot 7+1=22\). Then if the output is repeatedly inserted into the machine five more times, the final output is 26 .
\[
7 \rightarrow 22 \rightarrow 11 \rightarrow 34 \rightarrow 17 \rightarrow 52 \rightarrow 26
\]When the same 6 -step process is applied to a different starting value of \(N\), the final output is 1 . What is the sum of all such integers \(N\) ?
\[
N \rightarrow \ldots \rightarrow \_\rightarrow-\rightarrow \_\rightarrow \rightarrow 1
\]Solution:
Here the final output is 1 and we do the calculation in backward.
Here we will consider all possible inputs from which we can get the particular output.
\[
\{1\} \rightarrow\{2\} \rightarrow\{4\} \rightarrow\{1,8\} \rightarrow\{2,16\} \rightarrow\{4,5,32\} \rightarrow\{1,8,10,64\}
\]
Here 2 must come from 4 (as there is no integer \(n\) satisfying \(3 n+1=2\) )But 16 could come from 32 or 5.
If we consider input is 5, output will be \(3 \ times 5 +1=16\).
If we consider input is 16, output will be \(\frac{32}{2}=16 \).
After 6 steps in the last set there are 4 numbers.
Sum of the numbers is \(1+8+10+64=83\).
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