Fill in the boxes to get the right equation

Question

Here is a math puzzle I had a little bit of hard time with

No computers please

There is a solution without inverting 6 to 9

Answer 1

The trick is that

Two of the letters are actually roman numerals. D = 500 and C = 100.
$25 - 12 + D / C = 3 * 6$
$13 + 5 = 18$
This uses all "numbers from below" once.

Answer 2

Partial Answer:

This answer follows by BODMAS or BEDMAS or PEDMAS.

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Umm...

THERE IS NO SOLUTION! (without lateral thinking; without inverting the $6$, for example)

Let's call the numbers we can choose from, the Option Numbers.

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25 cannot be in the third and fourth box.

Proof:

This is our equation: $$\Box-\Box+\Box\:/\:\Box=\Box\times\Box.\tag{$\small \rm given$}$$ $12$, $6$ and $3$ do not divide $25$, so the third box can only be $25$ if the fourth box is $25$. Suppose that involves a solution. Then we have $$\begin{align}\Box - \Box + \boxed{25}\:/\:\boxed{25} &= \Box - \Box + 1 \\ &= \Box\times \Box.\end{align}$$

The largest number for the left hand side is $25-3+1=23$ so the right hand side cannot be greater than $23$. But $23$ is prime and both $22$ and $21$ have two distinct prime factors (although none of option numbers are prime), so the RHS cannot be greater than $20$.

Also, $20=5\times 4 = 10\times 2$ which uses none of the option numbers as well, and since $19$ is prime, that means the RHS cannot be greater than $18$ which is $3\times 6$ or $6\times 3$. But also, every other product strictly involving the option numbers is greater than $18$, so the RHS cannot be lower than $18$ either.

If the RHS cannot be greater or lower than $18$, then it is equal to $18$. $$\Box-\Box+\Box\:/\:\Box=18.\tag*{$(3\times 6$ or $6\times 3)$}$$

Now $18=6\times 3$ which uses two of the option numbers. So now we must find option numbers such that $$\Box-\Box+1=\boxed6\times \boxed3 =18$$ Therefore $\Box-\Box=18-1=17$. Of course the first box has to have a bigger value than $17$, because $17$ is positive and all the option numbers are positive. The only option number bigger than $17$ is $25$. So $\boxed{25}-\Box=17$. Therefore the second box has a value of $25-17=8$ but $8$ is not an option number.

This is a contradiction, so $25$ cannot be in the third box, and thus fourth one, too.

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$\Box\:/\:\Box=2$ or $4$.

Proof:

Now $\Box\:/\: \Box$ has to be an integer since $18$ is an integer, therefore the numerator box (third one) has an option number greater than the denominator box (fourth one). Since $3$ is the lowest option number, then $3$ cannot be in the third box. That leaves $12$ or $6$, so that leaves the fourth box to be $6$ or $3$. Therefore, this fraction must be equal to $12/6$, $6/3$ or $12/3$ which is $2$, $2$ or $4$. And since $2=2$, then the fraction is either $2$ or $4$.

We thus have the equations: $$\begin{align}\Box-\Box+2&=18 \\ \small{\rm or} \quad \Box-\Box+4&=18.\end{align}$$ Therefore, $$\begin{align}\Box-\Box&=18-2=16 \\ \small{\rm or} \quad \Box-\Box&=18-4=12.\end{align}$$

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And finally,

From the previous proof, THERE EXISTS NO SOLUTION!

Proof:

Now considering the first equation, the first box has to have an option number greater than $16$. The only option number like that is $25$. We thus have $$\boxed{25}-\Box=16$$ therefore $\Box=25-16=9$. But $9$ is not an option number. That is a contradiction, so the first equation cannot exist. $$\require{cancel}{\xcancel {\Box-\Box=16}}$$

Considering the second equation, the first box needs to be greater than $12$. It can't be $12$, it has to be greater than $12$. Again, the only option number greater than $12$ is $25$. We thus have $$\boxed{25}-\Box=12$$ therefore $\Box=25-12=13$. But $13$ is not an option number. That is a contradiction so the second equation cannot exist. $$\require{cancel}{\xcancel {\Box-\Box=12}}$$ But if both equations cannot exist, then...

...THERE IS NO SOLUTION!

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Therefore,

Some lateral-thinking must be required, unless you do not follow by BODMAS or BEDMAS or PEDMAS.

Answer 3

There doesn't seem to be anything that says that only one number can be placed into each box. Thus

$$12 - 25 + 66 \div 3 = 3 \times 3$$

would be a valid solution.

It just requires putting

two $6$s in the same box.

Answer 4

The puzzle explicitly states: Each number from below must be used once at least once.

Our numbers are $12, 6, 25, 3$. Without changing any of the numbers, using integer math instead of decimals, and following the rule above:

$12 - 3 + 6 / 25 = 3 * 3$

Following Order of Operations:

$3 * 3 = 9$
$6 / 25 = 0$
$3 + 0 = 3$
$12 - 3 = 9$
$9 = 9$

Answer 5

how about

$25-9+12/6=3\times6$

to do that

I rotated 6 into 9 as you suspected which is valid for the tag provided.

Answer 6

My solution is

$25 - 12 + 25 / 3 = 3 \times 6$

because

the numbers are octal base, and converting to decimal base

gives

$21 - 10 + 21 / 3 = 3 \times 6$

Answer 7

Using the tag:

Each number must be used. It seems like there are 4 numbers: 12, 6, 25, 3. However, I'm guessing there are 6 numbers (lateral thinking): 1, 2, 6, 2, 5, 3. So one of the answers (there may be more with this logic): is
6 - 5 + 3 / 1 = 2 * 2
3 - 5 + 6 / 1 = 2 * 2 is another order