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

Blockquote

No computers please

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 at 13:20
  • Yes division before addition or subtraction – DEEM Sep 21 at 13:21
  • Glad you included the "no computers please" line :P – user477343 Sep 21 at 13:24
  • 1
    This is my own puzzle @Gareth McCaughan. My Grandapa told me!! – DEEM Sep 21 at 14:04
  • 1
    @user477343 there is: I have just found one. – Weather Vane Sep 21 at 15:25
up vote 18 down vote accepted

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.

  • 1
    What a way to start as a new contributor!! Kudos @Usermomome. Great Lateral Thinking – DEEM Sep 21 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 – user477343 Sep 22 at 0:20

Partial Answer:

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 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 – user477343 Sep 21 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. – PerpetualJ Sep 21 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 – user477343 Sep 21 at 15:57
  • 1
    This is such an impressive analysis @user477343. You must be an engineer :) – DEEM Sep 22 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 at 15:42
  • OP responded "No more than one number in the square please" – Greg Sep 21 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 at 15:56
  • lol... I guess... – Greg Sep 21 at 15:59
  • I didn't post an answer because I hadn't found (or indeed looked for) one :-). – Gareth McCaughan Sep 22 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 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. – PerpetualJ Sep 23 at 3:10

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.

  • 1
    I didn't copy this - didn't notice - UV. – Weather Vane Sep 21 at 14:08
  • @WeatherVane np :) – Oray Sep 21 at 14:09
  • Happy that you reached the same conclusion. – Weather Vane Sep 21 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 at 14:07
  • @Oray this is a new, different answer. – Weather Vane Sep 21 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.