Math, asked by s2208debaijuni273, 6 months ago

If a positive integer is fed into the number crunching machine, it produces one of two
results:
If the integer fed in is EVEN, the machine divides the number by 2.
If the integer fed in is ODD, the machine subtracts one from the number.
If a negative integer is fed into the number crunching machine, it produces its additive
inverse.
Question1. Find the result when the following numbers are fed into the machine:
123
(ii) -72
Question2. What can be the input to the machine if the output is 18?
(a) 36
(b) 19 (c)-18 (d) All of the above
Question3. It is possible to feed the output from the machine back into the input, and
continue to do so until the output reaches zero. For example, will an initial input of 11, the
following would occur:
11
10 - 5 4 2
We see that 6 steps are required to reach zero.
Give the number of steps required to reach zero if you start with -24. Show the steps.

Answers

Answered by rp3993668

Answer:

When a positive integer $N$ is fed into a machine, the output is a number calculated according to the rule shown below.

[asy] size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label("$N$",(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label("if $N$ is even",(9.25,1.25),N); label("if $N$ is odd",(9.25,-1.25),N); label("$\frac N2$",(12,1.25)); label("$3N+1$",(12.6,-1.25)); [/asy] 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 $26.$\[7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26\]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 $N?$\[N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1\]$\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83$

Solution 1

We start with the final output of $1$ and work backwards, taking care to consider all possible inputs that could have resulted in any particular output. This produces the following set of possibilities at each stage: \[\{1\}\rightarrow\{2\}\rightarrow\{4\}\rightarrow\{1,8\}\rightarrow\{2,16\}\rightarrow\{4,5,32\}\rightarrow\{1,8,10,64\}\] where, for example, $2$ must come from $4$ (as there is no integer $n$ satisfying $3n+1=2$), but $16$ could come from $32$ or $5$ (as $\frac{32}{2} = 3 \cdot 5 + 1 = 16$, and $32$ is even while $5$ is odd). By construction, the last set in this sequence contains all the numbers which will lead to the number $1$ at the end of the $6$-step process, and their sum is $1+8+10+64=\boxed{\textbf{(E) }83}$.

Solution 2 (variant of Solution 1)

As in Solution 1, we work backwards from $1$, this time showing the possible cases in a tree diagram:

Prob22-diagram.png

The possible numbers are those at the "leaves" of the tree (the ends of the various branches), which are $1$, $8$, $64$, and $10$. Thus the answer is $1+8+64+10=\boxed{\textbf{(E) }83}$.

Solution 3 (algebraic)

We begin by finding the inverse of the function that the machine uses. Call the input $I$ and the output $O$. If $I$ is even, $O=\frac{I}{2}$, and if $I$ is odd, $O=3I+1$. We can therefore see that $I=2O$ when $I$ is even and $I=\frac{O-1}{3}$ when $I$ is odd. Therefore, starting with $1$, if $I$ is even, $I=2$, and if $I$ is odd, $I=0$, but the latter is not valid since $0$ is not actually odd. This means that the 2nd-to-last number in the sequence has to be $2$. Now, substituting $2$ into the inverse formulae, if $I$ is even, $I=4$ (which is indeed even), and if $I$ is odd, $I=\frac{1}{3}$, which is not an integer. This means the 3rd-to-last number in the sequence has to be $4$. Substituting in $4$, if $I$ is even, $I=8$, but if $I$ is odd, $I=1$. Both of these are valid solutions, so the 4th-to-last number can be either $1$ or $8$. If it is $1$, then by the argument we have just made, the 5th-to-last number has to be $2$, the 6th-to-last number has to be $4$, and the 7th-to-last number, which is the first number, must be either $1$ or $8$. In this way, we have ultimately found two solutions: $N=1$ and $N=8$.

On the other hand, if the 4th-to-last number is $8$, substituting $8$ into the inverse formulae shows that the 5th-to-last number is either $16$ or $\frac{7}{3}$, but the latter is not an integer. Substituting $16$ shows that if $I$ is even, $I=32$, but if I is odd, $I=5$, and both of these are valid. If the 6th-to-last number is $32$, then the first number must be $64$, since