# Fill in the boxes to get the right equation

Here is a math puzzle I had a little bit of hard time with There is a solution without inverting 6 to 9

• With regard to operator order on the left hand side, is the division performed first, followed by the subtraction and then the addition? – hexomino Sep 21 '18 at 13:20
• Yes division before addition or subtraction – DrD Sep 21 '18 at 13:21
• Glad you included the "no computers please" line :P – Mr Pie Sep 21 '18 at 13:24
• This is my own puzzle @Gareth McCaughan. My Grandapa told me!! – DrD Sep 21 '18 at 14:04
• @user477343 there is: I have just found one. – Weather Vane Sep 21 '18 at 15:25

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.

• What a way to start as a new contributor!! Kudos @Usermomome. Great Lateral Thinking – DrD Sep 21 '18 at 17:02
• Agreed with @DEEM. This is a beautiful answer; it's clear, does not break any of the given rules and makes perfect sense overall! $(+1)$, and welcome to the Puzzling Stack Exchange (Puzzling.SE)! :D – Mr Pie Sep 22 '18 at 0:20

This answer follows by BODMAS or BEDMAS or PEDMAS.

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.

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.

$$\Box\:/\:\Box=2$$ 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}

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!

Therefore,

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

• check the tags in the question :) – Oray Sep 21 '18 at 14:19
• @Oray I did, but DEEM wrote that he/she found a solution without inverting a $6$ to a $9$, and I cannot think of anything else more lateral :P – Mr Pie Sep 21 '18 at 14:32
• @user477343 This is a great answer, and though I hate to do so, I can't help it because it's driving me crazy lol; your OOP is incorrect. PEMDAS is what you're looking for. Multiplication always comes prior to division. – Taco Sep 21 '18 at 15:44
• @PerpetualJ That is not true, I think. MD and AS can swap either way. Say I have: $a+b-c$. What do you do first? Add or subtract? It is either way. Multiplication is literally adding a certain number of times (pun not intended) and division is subtracting a certain number of times, so it is either way for them too. See here for example :P – Mr Pie Sep 21 '18 at 15:57
• This is such an impressive analysis @user477343. You must be an engineer :) – DrD Sep 22 '18 at 12:38

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.

• @Gareth I just saw your comment on the question above, after posting this solution. I'm surprised you didn't post an answer yourself! – GentlePurpleRain Sep 21 '18 at 15:42
• OP responded "No more than one number in the square please" – eye_am_groot Sep 21 '18 at 15:51
• @Greg: I'm only putting one number in each; I'm just putting one number twice in one of them... :P (This is a valid answer to the question as posed. That criterion was not in the question.) – GentlePurpleRain Sep 21 '18 at 15:56
• lol... I guess... – eye_am_groot Sep 21 '18 at 15:59
• I didn't post an answer because I hadn't found (or indeed looked for) one :-). – Gareth McCaughan Sep 22 '18 at 11:16

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:

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

• ... Since when does 6/25 = 0. As a mathematician, I find this a ground breaking result XD I except a paper on ArXiv will follow shortly? – Brevan Ellefsen Sep 23 '18 at 2:28
• @BrevanEllefsen I stated that I was using integer only math. Integers are whole numbers and thus any decimal values are dropped. Hence 0.24 becomes 0. – Taco Sep 23 '18 at 3:10

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

to do that

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

• I didn't copy this - didn't notice - UV. – Weather Vane Sep 21 '18 at 14:08
• @WeatherVane np :) – Oray Sep 21 '18 at 14:09
• Happy that you reached the same conclusion. – Weather Vane Sep 21 '18 at 14:09

My solution is

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

because

the numbers are octal base, and converting to decimal base

gives

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

• I have already submitted this answer -.- – Oray Sep 21 '18 at 14:07
• @Oray this is a new, different answer. – Weather Vane Sep 21 '18 at 15:13

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