Number Factory

Logic Level 1

The calculator above takes in an integer input at the lower left. Then, a path is taken to the top right - at each step moving up or right - and the operations given are performed on the input sequentially.

If a path gives an output of 43, which of the following choices could have been the initial input?

1 2 3 4 5 6 7 8

2 solutions

Worranat Pakornrat
Mar 30, 2017

It's easier to reverse the process from the end product to its source, as shown below:

That is, starting from $43$ , its nearest operator is $\boxed{+1}$ , so we can reverse the operation to be $43-1 = 42$ .

Then from $42$ , there are two legs to go: to divide by $2$ or $3$ , yielding $21$ or $14$ respectively. In the question, it is stated that the input is an integer, and with all the processes, the input will continue to be an integer. However, for the first leg, when $21+1 = 22$ , it's not a multiple of $5$ or $3$ while $21+2 = 23$ , it's prime and not a multiple of $3$ . Thus, this leg is not applicable.

Continuing on the latter leg, $14+3=17$ on the upper leg is prime, it's thus not a multiple of $7$ . On the other hand, $14+1 = 15$ which can be divided by both $3$ and $5$ in the next operations, yielding $5$ and $3$ respectively.

Then $5-2 = 3$ and $3-2 =1$ are not even numbers, so that leaves only one path left $3-1 = 2$ and $\dfrac{2}{2}$ = 1).

Thus, $\boxed{1}$ is the only plausible input digit, and there is only one path to reach the desired result, as demonstrated by the pink trail below:

Moderator note:

To clarify, as mentioned in the comments, only the initial input is required to be an integer; intermediate steps are allowed to be non-integers.

It is easy to demonstrate that every intermediate step must be an integer.

We have an integer [1;8] in step 0. In every next step we have either a multiplication by an integer or an addition to an integer. Since a multiplication of integers or addition of integers always results in an integer, every intermediate step is always an integer.

We further proceed just as Worranat suggested. For such a small problem this is probably the optimal solution.

Gsm Offln - 4 years, 1 month ago

Fun problem! The range of achievable outputs for digit inputs appears to be $21$ to $505$ , but I haven't calculated the number of distinct possible outputs. That would be a bit tedious and best left to a computer program.

Brian Charlesworth - 4 years, 2 months ago

Thank you. ;)

Worranat Pakornrat - 4 years, 2 months ago

Note: We cannot directly claim that "Since 22 is not a multiple of 3 or 5, this leg is not applicable", since it might have be possible that some other operation would have allowed us to divide by 3.
We just need a small fix to this problem. Do you see what to do?

Calvin Lin Staff - 4 years, 2 months ago

I don't quite see. Do you mean there is other plausible operation route?

Worranat Pakornrat - 4 years, 2 months ago

Not quite. In order to conclude that "the number has to be a multiple of 5 (or in the other case, 3)", what we require is a certain condition on the input. Otherwise, without any restriction, the input could have been $\frac{22}{5}$ , which when multiplied by 5 gives us 22.

Calvin Lin Staff - 4 years, 2 months ago

@Calvin Lin – But in the question, we already stated that the input is an integer, and with all the processes, the input will continue to be an integer, not a fraction. Isn't that enough?

Worranat Pakornrat - 4 years, 2 months ago

@Worranat Pakornrat – Right. My point is that you didn't state that in the solution, and left that to the reader to "mind read". That's the small fix that I'm referring to.

IE Notice that every input and output will be an integer. Hence, an output from " $\times n$ " will be a multiple of $n$ . So, when backtracking, if the output is not a multiple of $n$ , then there isn't a valid input we could use.

Ah, I see I said "small fix to problem" instead of "small fix to solution". Sorry for the confusion.

Calvin Lin Staff - 4 years, 2 months ago

@Calvin Lin – That's alright. I already added in some clarification. Hope you enjoyed the problem then. ;)

Worranat Pakornrat - 4 years, 2 months ago

Misleading problem statement.

Héctor Corte León - 4 years, 1 month ago

In what way is it misleading? That would allows us to clarify it for the rest of the community.

Calvin Lin Staff - 4 years, 1 month ago

The way I read it, you could take any route and achieve the same result, which is obviously not the case. That is misleading

Gabriel Souza - 4 years, 1 month ago

@Gabriel Souza – I've added a note to help clarify that. It should be clear that different paths would lead to different outputs.

Calvin Lin Staff - 4 years, 1 month ago

The input is an integer; the integers are closed under addition, subtraction, and multiplication; thus, every intermediate number is an integer.

Richard Desper - 4 years, 1 month ago

Betty BellaItalia
Apr 15, 2017

Number Factory

2 solutions

Moderator note:

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