Mixing
Fill in the boxes to get the right equation
Clash Royale CLAN TAG#URR8PPP
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favorite
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
mathematics lateral-thinking no-computers formation-of-numbers
edited 48 mins ago
GentlePurpleRain♦
16k565132
asked 2 hours ago
DEEM
3,8651074
 |Â
show 11 more comments
up vote
2
down vote
favorite
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
mathematics lateral-thinking no-computers formation-of-numbers
edited 48 mins ago
GentlePurpleRain♦
16k565132
asked 2 hours ago
DEEM
3,8651074
With regard to operator order on the left hand side, is the division performed first, followed by the subtraction and then the addition?
– hexomino
2 hours ago
Yes division before addition or subtraction
– DEEM
2 hours ago
Glad you included the "no computers please" line :P
– user477343
2 hours ago
1
This is my own puzzle @Gareth McCaughan. My Grandapa told me!!
– DEEM
1 hour ago
1
@user477343 there is: I have just found one.
– Weather Vane
32 mins ago
 |Â
show 11 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
mathematics lateral-thinking no-computers formation-of-numbers
edited 48 mins ago
GentlePurpleRain♦
16k565132
asked 2 hours ago
DEEM
3,8651074
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
mathematics lateral-thinking no-computers formation-of-numbers
mathematics lateral-thinking no-computers formation-of-numbers
edited 48 mins ago
GentlePurpleRain♦
16k565132
asked 2 hours ago
DEEM
3,8651074
edited 48 mins ago
GentlePurpleRain♦
16k565132
asked 2 hours ago
DEEM
3,8651074
edited 48 mins ago
GentlePurpleRain♦
16k565132
edited 48 mins ago
GentlePurpleRain♦
16k565132
edited 48 mins ago
GentlePurpleRain♦
16k565132
16k565132
asked 2 hours ago
DEEM
3,8651074
asked 2 hours ago
DEEM
3,8651074
asked 2 hours ago
DEEM
3,8651074
3,8651074
With regard to operator order on the left hand side, is the division performed first, followed by the subtraction and then the addition?
– hexomino
2 hours ago
Yes division before addition or subtraction
– DEEM
2 hours ago
Glad you included the "no computers please" line :P
– user477343
2 hours ago
1
This is my own puzzle @Gareth McCaughan. My Grandapa told me!!
– DEEM
1 hour ago
1
@user477343 there is: I have just found one.
– Weather Vane
32 mins ago
 |Â
show 11 more comments
With regard to operator order on the left hand side, is the division performed first, followed by the subtraction and then the addition?
– hexomino
2 hours ago
Yes division before addition or subtraction
– DEEM
2 hours ago
Glad you included the "no computers please" line :P
– user477343
2 hours ago
1
This is my own puzzle @Gareth McCaughan. My Grandapa told me!!
– DEEM
1 hour ago
1
@user477343 there is: I have just found one.
– Weather Vane
32 mins ago
With regard to operator order on the left hand side, is the division performed first, followed by the subtraction and then the addition?
– hexomino
2 hours ago
With regard to operator order on the left hand side, is the division performed first, followed by the subtraction and then the addition?
– hexomino
2 hours ago
Yes division before addition or subtraction
– DEEM
2 hours ago
Yes division before addition or subtraction
– DEEM
2 hours ago
Glad you included the "no computers please" line :P
– user477343
2 hours ago
Glad you included the "no computers please" line :P
– user477343
2 hours ago
1
1
This is my own puzzle @Gareth McCaughan. My Grandapa told me!!
– DEEM
1 hour ago
This is my own puzzle @Gareth McCaughan. My Grandapa told me!!
– DEEM
1 hour ago
1
1
@user477343 there is: I have just found one.
– Weather Vane
32 mins ago
@user477343 there is: I have just found one.
– Weather Vane
32 mins ago
 |Â
show 11 more comments
5 Answers
5
active
oldest
votes
up vote
1
down vote
how about
$25-9+12/6=3times6$
to do that
I rotated 6 into 9 as you suspected which is valid for the tag provided.
answered 2 hours ago
Oray
14.4k435141
1
I didn't copy this - didn't notice - UV.
– Weather Vane
1 hour ago
@WeatherVane np :)
– Oray
1 hour ago
Happy that you reached the same conclusion.
– Weather Vane
1 hour ago
add a comment |Â
up vote
1
down vote
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: $$Box-Box+Box:/:Box=BoxtimesBox.tag$small rm given$$$ $12$, $6$ 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 $$Box - Box + 1 = Boxtimes Box.$$
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, $20=5times 4 = 10times 2$ which uses none of the option numbers as well, and since $19$ is prime, that means the RHS cannot be greater than $18$. 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 $18$. $$Box-Box+Box:/:Box=18.tag*$(3times 6$ or $6times 3)$$$
Now $18=6times 3$ which uses two of the option numbers. So now we must find option numbers such that $$Box-Box+1=boxed6times boxed3 =18$$ 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 $25$. 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: $$beginalignBox-Box+2&=18 \ smallrm or quad Box-Box+4&=18.endalign$$ Therefore, $$beginalignBox-Box&=18-2=16 \ smallrm or quad Box-Box&=18-4=12.endalign$$
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 $$25-Box=16$$ therefore $Box=25-16=9$. But $9$ is not an option number. That is a contradiction, so the first equation cannot exist. $$requirecancelxcancel Box-Box=16$$
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 $$25-Box=12$$ therefore $Box=25-12=13$. But $13$ is not an option number. That is a contradiction so the second equation cannot exist. $$requirecancelxcancel Box-Box=12$$ 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.
answered 2 hours ago
user477343
3,3911745
check the tags in the question :)
– Oray
1 hour ago
@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
1 hour ago
@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
13 mins ago
add a comment |Â
up vote
1
down vote
My solution is
$25 - 12 + 25 / 3 = 3 times 6$
because
the numbers are octal base, and converting to decimal base
gives
$21 - 10 + 21 / 3 = 3 times 6$
answered 1 hour ago
Weather Vane
2687
I have already submitted this answer -.-
– Oray
1 hour ago
@Oray this is a new, different answer.
– Weather Vane
44 mins ago
add a comment |Â
up vote
0
down vote
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
answered 1 hour ago
Greg
61118
add a comment |Â
up vote
0
down vote
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.
answered 17 mins ago
GentlePurpleRain♦
16k565132
@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♦
14 mins ago
OP responded "No more than one number in the square please"
– Greg
5 mins ago
@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♦
12 secs ago
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
how about
$25-9+12/6=3times6$
to do that
I rotated 6 into 9 as you suspected which is valid for the tag provided.
answered 2 hours ago
Oray
14.4k435141
1
I didn't copy this - didn't notice - UV.
– Weather Vane
1 hour ago
@WeatherVane np :)
– Oray
1 hour ago
Happy that you reached the same conclusion.
– Weather Vane
1 hour ago
add a comment |Â
up vote
1
down vote
how about
$25-9+12/6=3times6$
to do that
I rotated 6 into 9 as you suspected which is valid for the tag provided.
answered 2 hours ago
Oray
14.4k435141
1
I didn't copy this - didn't notice - UV.
– Weather Vane
1 hour ago
@WeatherVane np :)
– Oray
1 hour ago
Happy that you reached the same conclusion.
– Weather Vane
1 hour ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
how about
$25-9+12/6=3times6$
to do that
I rotated 6 into 9 as you suspected which is valid for the tag provided.
answered 2 hours ago
Oray
14.4k435141
how about
$25-9+12/6=3times6$
to do that
I rotated 6 into 9 as you suspected which is valid for the tag provided.
answered 2 hours ago
Oray
14.4k435141
edited 2 hours ago
answered 2 hours ago
Oray
14.4k435141
answered 2 hours ago
Oray
14.4k435141
answered 2 hours ago
Oray
14.4k435141
14.4k435141
1
I didn't copy this - didn't notice - UV.
– Weather Vane
1 hour ago
@WeatherVane np :)
– Oray
1 hour ago
Happy that you reached the same conclusion.
– Weather Vane
1 hour ago
add a comment |Â
1
I didn't copy this - didn't notice - UV.
– Weather Vane
1 hour ago
@WeatherVane np :)
– Oray
1 hour ago
Happy that you reached the same conclusion.
– Weather Vane
1 hour ago
1
1
I didn't copy this - didn't notice - UV.
– Weather Vane
1 hour ago
I didn't copy this - didn't notice - UV.
– Weather Vane
1 hour ago
@WeatherVane np :)
– Oray
1 hour ago
@WeatherVane np :)
– Oray
1 hour ago
Happy that you reached the same conclusion.
– Weather Vane
1 hour ago
Happy that you reached the same conclusion.
– Weather Vane
1 hour ago
add a comment |Â
up vote
1
down vote
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: $$Box-Box+Box:/:Box=BoxtimesBox.tag$small rm given$$$ $12$, $6$ 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 $$Box - Box + 1 = Boxtimes Box.$$
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, $20=5times 4 = 10times 2$ which uses none of the option numbers as well, and since $19$ is prime, that means the RHS cannot be greater than $18$. 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 $18$. $$Box-Box+Box:/:Box=18.tag*$(3times 6$ or $6times 3)$$$
Now $18=6times 3$ which uses two of the option numbers. So now we must find option numbers such that $$Box-Box+1=boxed6times boxed3 =18$$ 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 $25$. 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: $$beginalignBox-Box+2&=18 \ smallrm or quad Box-Box+4&=18.endalign$$ Therefore, $$beginalignBox-Box&=18-2=16 \ smallrm or quad Box-Box&=18-4=12.endalign$$
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 $$25-Box=16$$ therefore $Box=25-16=9$. But $9$ is not an option number. That is a contradiction, so the first equation cannot exist. $$requirecancelxcancel Box-Box=16$$
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 $$25-Box=12$$ therefore $Box=25-12=13$. But $13$ is not an option number. That is a contradiction so the second equation cannot exist. $$requirecancelxcancel Box-Box=12$$ 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.
answered 2 hours ago
user477343
3,3911745
check the tags in the question :)
– Oray
1 hour ago
@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
1 hour ago
@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
13 mins ago
add a comment |Â
up vote
1
down vote
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: $$Box-Box+Box:/:Box=BoxtimesBox.tag$small rm given$$$ $12$, $6$ 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 $$Box - Box + 1 = Boxtimes Box.$$
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, $20=5times 4 = 10times 2$ which uses none of the option numbers as well, and since $19$ is prime, that means the RHS cannot be greater than $18$. 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 $18$. $$Box-Box+Box:/:Box=18.tag*$(3times 6$ or $6times 3)$$$
Now $18=6times 3$ which uses two of the option numbers. So now we must find option numbers such that $$Box-Box+1=boxed6times boxed3 =18$$ 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 $25$. 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: $$beginalignBox-Box+2&=18 \ smallrm or quad Box-Box+4&=18.endalign$$ Therefore, $$beginalignBox-Box&=18-2=16 \ smallrm or quad Box-Box&=18-4=12.endalign$$
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 $$25-Box=16$$ therefore $Box=25-16=9$. But $9$ is not an option number. That is a contradiction, so the first equation cannot exist. $$requirecancelxcancel Box-Box=16$$
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 $$25-Box=12$$ therefore $Box=25-12=13$. But $13$ is not an option number. That is a contradiction so the second equation cannot exist. $$requirecancelxcancel Box-Box=12$$ 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.
answered 2 hours ago
user477343
3,3911745
check the tags in the question :)
– Oray
1 hour ago
@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
1 hour ago
@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
13 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
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: $$Box-Box+Box:/:Box=BoxtimesBox.tag$small rm given$$$ $12$, $6$ 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 $$Box - Box + 1 = Boxtimes Box.$$
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, $20=5times 4 = 10times 2$ which uses none of the option numbers as well, and since $19$ is prime, that means the RHS cannot be greater than $18$. 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 $18$. $$Box-Box+Box:/:Box=18.tag*$(3times 6$ or $6times 3)$$$
Now $18=6times 3$ which uses two of the option numbers. So now we must find option numbers such that $$Box-Box+1=boxed6times boxed3 =18$$ 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 $25$. 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: $$beginalignBox-Box+2&=18 \ smallrm or quad Box-Box+4&=18.endalign$$ Therefore, $$beginalignBox-Box&=18-2=16 \ smallrm or quad Box-Box&=18-4=12.endalign$$
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 $$25-Box=16$$ therefore $Box=25-16=9$. But $9$ is not an option number. That is a contradiction, so the first equation cannot exist. $$requirecancelxcancel Box-Box=16$$
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 $$25-Box=12$$ therefore $Box=25-12=13$. But $13$ is not an option number. That is a contradiction so the second equation cannot exist. $$requirecancelxcancel Box-Box=12$$ 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.
answered 2 hours ago
user477343
3,3911745
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: $$Box-Box+Box:/:Box=BoxtimesBox.tag$small rm given$$$ $12$, $6$ 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 $$Box - Box + 1 = Boxtimes Box.$$
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, $20=5times 4 = 10times 2$ which uses none of the option numbers as well, and since $19$ is prime, that means the RHS cannot be greater than $18$. 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 $18$. $$Box-Box+Box:/:Box=18.tag*$(3times 6$ or $6times 3)$$$
Now $18=6times 3$ which uses two of the option numbers. So now we must find option numbers such that $$Box-Box+1=boxed6times boxed3 =18$$ 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 $25$. 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: $$beginalignBox-Box+2&=18 \ smallrm or quad Box-Box+4&=18.endalign$$ Therefore, $$beginalignBox-Box&=18-2=16 \ smallrm or quad Box-Box&=18-4=12.endalign$$
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 $$25-Box=16$$ therefore $Box=25-16=9$. But $9$ is not an option number. That is a contradiction, so the first equation cannot exist. $$requirecancelxcancel Box-Box=16$$
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 $$25-Box=12$$ therefore $Box=25-12=13$. But $13$ is not an option number. That is a contradiction so the second equation cannot exist. $$requirecancelxcancel Box-Box=12$$ 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.
answered 2 hours ago
user477343
3,3911745
edited 1 hour ago
answered 2 hours ago
user477343
3,3911745
answered 2 hours ago
user477343
3,3911745
answered 2 hours ago
user477343
3,3911745
3,3911745
check the tags in the question :)
– Oray
1 hour ago
@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
1 hour ago
@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
13 mins ago
add a comment |Â
check the tags in the question :)
– Oray
1 hour ago
@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
1 hour ago
@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
13 mins ago
check the tags in the question :)
– Oray
1 hour ago
check the tags in the question :)
– Oray
1 hour ago
@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
1 hour ago
@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
1 hour ago
@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
13 mins ago
@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
13 mins ago
add a comment |Â
up vote
1
down vote
My solution is
$25 - 12 + 25 / 3 = 3 times 6$
because
the numbers are octal base, and converting to decimal base
gives
$21 - 10 + 21 / 3 = 3 times 6$
answered 1 hour ago
Weather Vane
2687
I have already submitted this answer -.-
– Oray
1 hour ago
@Oray this is a new, different answer.
– Weather Vane
44 mins ago
add a comment |Â
up vote
1
down vote
My solution is
$25 - 12 + 25 / 3 = 3 times 6$
because
the numbers are octal base, and converting to decimal base
gives
$21 - 10 + 21 / 3 = 3 times 6$
answered 1 hour ago
Weather Vane
2687
I have already submitted this answer -.-
– Oray
1 hour ago
@Oray this is a new, different answer.
– Weather Vane
44 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
My solution is
$25 - 12 + 25 / 3 = 3 times 6$
because
the numbers are octal base, and converting to decimal base
gives
$21 - 10 + 21 / 3 = 3 times 6$
answered 1 hour ago
Weather Vane
2687
My solution is
$25 - 12 + 25 / 3 = 3 times 6$
because
the numbers are octal base, and converting to decimal base
gives
$21 - 10 + 21 / 3 = 3 times 6$
answered 1 hour ago
Weather Vane
2687
edited 34 mins ago
answered 1 hour ago
Weather Vane
2687
answered 1 hour ago
Weather Vane
2687
answered 1 hour ago
Weather Vane
2687
2687
I have already submitted this answer -.-
– Oray
1 hour ago
@Oray this is a new, different answer.
– Weather Vane
44 mins ago
add a comment |Â
I have already submitted this answer -.-
– Oray
1 hour ago
@Oray this is a new, different answer.
– Weather Vane
44 mins ago
I have already submitted this answer -.-
– Oray
1 hour ago
I have already submitted this answer -.-
– Oray
1 hour ago
@Oray this is a new, different answer.
– Weather Vane
44 mins ago
@Oray this is a new, different answer.
– Weather Vane
44 mins ago
add a comment |Â
up vote
0
down vote
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
answered 1 hour ago
Greg
61118
add a comment |Â
up vote
0
down vote
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
answered 1 hour ago
Greg
61118
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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
answered 1 hour ago
Greg
61118
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
answered 1 hour ago
Greg
61118
edited 39 mins ago
answered 1 hour ago
Greg
61118
answered 1 hour ago
Greg
61118
answered 1 hour ago
Greg
61118
61118
add a comment |Â
add a comment |Â
up vote
0
down vote
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.
answered 17 mins ago
GentlePurpleRain♦
16k565132
@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♦
14 mins ago
OP responded "No more than one number in the square please"
– Greg
5 mins ago
@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♦
12 secs ago
add a comment |Â
up vote
0
down vote
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.
answered 17 mins ago
GentlePurpleRain♦
16k565132
@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♦
14 mins ago
OP responded "No more than one number in the square please"
– Greg
5 mins ago
@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♦
12 secs ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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.
answered 17 mins ago
GentlePurpleRain♦
16k565132
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.
answered 17 mins ago
GentlePurpleRain♦
16k565132
answered 17 mins ago
GentlePurpleRain♦
16k565132
answered 17 mins ago
GentlePurpleRain♦
16k565132
answered 17 mins ago
GentlePurpleRain♦
16k565132
16k565132
@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♦
14 mins ago
OP responded "No more than one number in the square please"
– Greg
5 mins ago
@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♦
12 secs ago
add a comment |Â
@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♦
14 mins ago
OP responded "No more than one number in the square please"
– Greg
5 mins ago
@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♦
12 secs ago
@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♦
14 mins ago
@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♦
14 mins ago
OP responded "No more than one number in the square please"
– Greg
5 mins ago
OP responded "No more than one number in the square please"
– Greg
5 mins ago
@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♦
12 secs ago
@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♦
12 secs ago
add a comment |Â
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With regard to operator order on the left hand side, is the division performed first, followed by the subtraction and then the addition?
– hexomino
2 hours ago
Yes division before addition or subtraction
– DEEM
2 hours ago
Glad you included the "no computers please" line :P
– user477343
2 hours ago
1
This is my own puzzle @Gareth McCaughan. My Grandapa told me!!
– DEEM
1 hour ago
1
@user477343 there is: I have just found one.
– Weather Vane
32 mins ago