How to get 32 by using +1 , +1 , ×3 , ×3 , ÷2 , ÷2, ^2, ^2?

Question

How to get 32 by using +1,+1,×3,×3,÷2,÷2,^2,^2?
(The last two operators are 'squaring'.)

Answer 1

I like Glorfindel's answer as it takes you through an example solution, and I recommend his answer. Just to show some underlying beauty and complexity of this puzzle, I created a diagram (directed multigraph) that shows all possible results (that yield integers) of permuting these operations. You'll have to open the image in another tab and magnify to follow it in detail. I hope you enjoy.

Note that the green nodes are those where at least one sequence of the operations terminates there.

Starting at zero:

Starting at one:

Since there might be some interest in this case, here's a similar map.

It is not actually possible to get from 1 to 32 using exactly these 8 operations.

Bonus Information

I checked every graph that starts at 0 up through 32, and confirmed that none of the other graphs terminate at 32. What's more, the answer given by Glorfindel is the only solution. This puzzle makes me happy.

Answer 2

Assuming you start with 0 (like on a calculator), this should work:

$$0 \times 3 = 0$$ $$0 + 1 = 1$$ $$1 \times 3 = 3$$ $$3 + 1 = 4$$ $$4^2 = 16$$ $$16 \div 2 = 8$$ $$8^2 = 64$$ $$64 \div 2 = 32$$

Answer 3

I don't know how I got to this question from Stack Overflow... But here is the solution and the Python code for checking every combination:

((((((((0*3)+1)*3)+1)**2)/2.0)**2)/2.0)

lis = ['+1', '+1', '*3', '*3', '/2.0', '/2.0', '**2', '**2']
result = set()
def sol(lis, current, res, result):

    if not lis and res == 32:
        result.add(current)
        return

    for i in range(len(lis)):
        c = f"({current}{lis[i]})"
        sol(lis[:i] + lis[i+1:], c, eval("res"+lis[i]), result)
sol(lis, 0, 0, result)
print(result)
```

Answer 4

I noticed that there was a lot of emphasis around powers of 2, and that gave me an idea:

I started from the solution and brute-forced my way backwards. This puzzle can be rephrased as something like: Starting at 32, use these operators sequentially to get to 0: $$-1, -1, \div 3, \div 3, \times 2, \times 2, \sqrt(n), \sqrt(n)$$

Notably:

This set of inverse operations has a smaller search space because dividing by 3 and taking the square root are much more restrictive in their use.

My work:

$$32 \times 2 = 64$$ $$\sqrt(64) = 8$$ $$8 \times 2 = 16$$ $$\sqrt(16)= 4$$ $$4-1 = 3$$ $$3 \div 3 = 1$$ $$1-1 = 0$$ $$0 \div 3 = 0$$ Then all you have to do is re-invert the operations (and reverse their order) to get the answer of $$\times 3, +1, \times 3, +1, (n)^2, \div 2, (n)^2, \div 2$$

Answer 5

This solution reduces the problem to a simpler one using each operation's effect on prime factorization.

We begin with $0+1=1=2^03^0$. Each step is represented by the form: $2^n3^m$.
We have the following operations:

$(*3)$ yields $2^n3^{m+1}$

$(/2)$ yields $2^{n-1}3^m$

$(\hat{}2)$ yields $2^{2n}3^{2m}$

and $(+1)$ "randomly" shuffles/introduces new exponents. Our desired number is $2^5=32$.
The only reasonable first step is to increase the exponents from zero, so we can apply: $$2^03^0*3=2^03^1$$ We then use our only chance to get rid of the $3$ factor entirely: $$2^03^1+1=2^23^0$$ We must get from $2^2\rightarrow2^5$, so let's reduce the problem to solving $2\rightarrow5$ using the remaining four operations (on the exponent): $(-1), (-1), (*2), (*2)$. Let's start backwards from $5$:

The last operation must have been $6-1=5$ (because $2.5*2$ is unreachable),

Next must have been $3*2=6$, (cannot be $7-1=6$, because $2*2*2\neq7$),

$3$ is odd, so we came from $4-1=3$, and finally $2*2=4$.

Converting everything back, we have performed the following:
$$(((0+1)*3+1)^2/2)^2/2=32$$

Pretty contrived but it kind of forces the solution without exhausting all the possibilities!

Answer 6

I had a lot of fun with this one! First, as others did, I worked out it could "only" be done starting from one particular number, then saw Glorfindel's answer and realised that if

4 is a waypoint, then so is -4, because -4^2 is also 16.

So I worked backwards from that and found

a new starting point of -8/9.

Then it occurred to me that

all the operations are reversible on the complex plane

so I modified the operations accordingly and wrote a solver, also as an exercise in the C++ STL, tuples, lambdas and all that fancy stuff, wrote the output to a file and fed it into gnuplot.

This is the resulting plot, showing

all 161,280 possible starting points
NB: there are duplicates I haven't removed;
there are more results than 8! (40,320) because a complex number has two square roots

Edit: After posting I found a way to remove duplicates, which if I got the logic right means there are

4950 unique answers, using a difference of 0.0001 in both parts.

Edit: Can I post an example?

32 -1 /3 -sqrt -sqrt -1 *2 *2 /3 = -1.3333 - i2.3906

Answer 7

In response to the assertion that it cannot be done starting with 1, here is a method that may work, depending on exactly how the rules are applied.

$$(1) \times 3 = 3$$ $$3 + 1 = 4$$ $${4 \div 2} = 2$$ $$2^2 = 4$$ $${4 \div 2} = 2$$ $$2^{2+1 \times 3} = 2^5$$ $$2^5 = 32$$

Answer 8

I am assuming that we are looking for an integer solution. With this assumption, here is a top down approach that nobody else has used. The first question that needs to be asked is is that If a solution exists then what will be the last operator that would be used .

The last operator cannot be ×3 or ^2 as no integer × 3 = 32 nor (any integer)^2 = 32 . Therefore, the last operator to be used has to be either +1 or ÷2 . Therefore the numbers that are possible right before the last operator was used was either 31 (since 31+1= 32) or 64 ( since 64/2= 32). The following picture shows that there are only 5 ways to get to 32 using a total of 2 operations.

The image below shows a complete expansion of one of the 5 paths shown above. As has been mentioned in the previous answers, we can see in the image below that there is indeed a path starting from 0 such that using all the 8 operators exactly once, we can get to 32.