Mixing
Fill in the boxes to get the right equation v2
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
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My take on a recent puzzle. Place numbers into boxes to create a correct equation, using each number at least once:
$$Box-Box+BoxtimesBox=Box:/:Box$$
This time the four decimal numbers are 3, 6, 7, 12.
mathematics lateral-thinking no-computers
asked 5 hours ago
Roman Odaisky
1183
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Roman Odaisky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
3
down vote
favorite
My take on a recent puzzle. Place numbers into boxes to create a correct equation, using each number at least once:
$$Box-Box+BoxtimesBox=Box:/:Box$$
This time the four decimal numbers are 3, 6, 7, 12.
mathematics lateral-thinking no-computers
asked 5 hours ago
Roman Odaisky
1183
New contributor
Roman Odaisky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Does this follow operator precedence rules? I mean, is the multiplication made before the addition, or after it?
– Racso
5 hours ago
@Rasco Of course it does, how else?
– Roman Odaisky
4 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
My take on a recent puzzle. Place numbers into boxes to create a correct equation, using each number at least once:
$$Box-Box+BoxtimesBox=Box:/:Box$$
This time the four decimal numbers are 3, 6, 7, 12.
mathematics lateral-thinking no-computers
asked 5 hours ago
Roman Odaisky
1183
New contributor
Roman Odaisky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
My take on a recent puzzle. Place numbers into boxes to create a correct equation, using each number at least once:
$$Box-Box+BoxtimesBox=Box:/:Box$$
This time the four decimal numbers are 3, 6, 7, 12.
mathematics lateral-thinking no-computers
mathematics lateral-thinking no-computers
asked 5 hours ago
Roman Odaisky
1183
New contributor
Roman Odaisky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Roman Odaisky
1183
New contributor
Roman Odaisky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Roman Odaisky
1183
New contributor
Roman Odaisky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
Roman Odaisky
1183
asked 5 hours ago
Roman Odaisky
1183
1183
New contributor
Roman Odaisky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Roman Odaisky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Roman Odaisky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Does this follow operator precedence rules? I mean, is the multiplication made before the addition, or after it?
– Racso
5 hours ago
@Rasco Of course it does, how else?
– Roman Odaisky
4 hours ago
add a comment |Â
1
Does this follow operator precedence rules? I mean, is the multiplication made before the addition, or after it?
– Racso
5 hours ago
@Rasco Of course it does, how else?
– Roman Odaisky
4 hours ago
1
1
Does this follow operator precedence rules? I mean, is the multiplication made before the addition, or after it?
– Racso
5 hours ago
Does this follow operator precedence rules? I mean, is the multiplication made before the addition, or after it?
– Racso
5 hours ago
@Rasco Of course it does, how else?
– Roman Odaisky
4 hours ago
@Rasco Of course it does, how else?
– Roman Odaisky
4 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
1
down vote
accepted
Edit: Got it.
-$7 + 3 * 3 = 12/6$ (Still don’t have to fill in every box.)
Great puzzle!
answered 3 hours ago
Excited Raichu
1487
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What are you going to do with the genvyvat fynfu?
– Roman Odaisky
3 hours ago
It’s not possible for me to vtaber vg?
– Excited Raichu
3 hours ago
Would be qvivqr ol mreb!
– Weather Vane
3 hours ago
I suppose so. Svkrq vg!
– Excited Raichu
3 hours ago
This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
– Roman Odaisky
3 hours ago
add a comment |Â
up vote
1
down vote
My answer is
$boxed12 - boxed12 + boxed: 3 : times boxed: 3 : = boxed63 : / : boxed: 7 : $
because
the question says "place numbers into boxes" so I placed two numbers into the 5th box.
Note: There is one solution to the obvious question:
$boxed: 7 : - boxed12 + boxed: 3 : times boxed: 3 : = boxed12 : / : boxed: 3 : $
But it is illegal because it does not use all the numbers.
Solved with pencil and paper only.
answered 2 hours ago
Weather Vane
2887
Also very nice. I was pretty sure repeated application of this idea can lead to a solution, but to solve the puzzle doing it only once looks remarkable to me.
– Roman Odaisky
1 hour ago
add a comment |Â
up vote
0
down vote
Partial Answer:
Well, well, well.
THERE EXISTS NO SOLUTION! (without lateral thinking, although inverting the $6$ to a $9$ has no solution either, in particular).
Let's call the numbers we can choose from, the Option Numbers.
The right hand side (RHS) will either equal $1$, $2$ or $4$ (without lateral thinking).
Proof:
This is our equation: $$Box-Box+BoxtimesBox=Box:/:Boxtag$smallrm given$$$ $7$ can only be in the fifth box if the sixth box is also $7$, as that is the only number out of the option numbers that divide $7$. Otherwise, the left hand side (LHS) would not be an integer. So, a possibility is that the RHS is equal to $7/7=1$.
Excluding $7$ now, $3$ cannot be in the fifth box because it is the lowest option number, and thus the fraction will not be an integer otherwise. That leaves $6$ and $12$.
So the fraction is either $12/6$, $12/3$ or $6/3$ which is $2$, $2$ or $4$ respectively. Since $2=2$, then the RHS is either equal to $2$ or $4$. But we mentioned that it can also be equal to $1$, so the possible values the RHS can equal are $$1,2,4.$$
Therefore, $$Box-Box+BoxtimesBox=1,2;textor;4.$$
$Box-Box$ equals a number in between $-14$ and $-83$ inclusive.
Proof:
From the previous proof, we know that the LHS is either $1$, $2$ or $4$. All the option numbers are positive, and the minimum product that can be made from them is $3times 6$ or $6times 3$ which is $18$; also, the maximum value that the RHS can equal is $4$. So the maximum value of $Box-Box$ is... $$Box-Box=4-18=-14.tag$smallrm as ; the ; maximum ; value$$$ Now, doing the opposite to find the minimum value, we find maximum product of option numbers and minimum value RHS can equal. That makes $84$ and $1$ respectively (since $84=12times 7$ or $7times 12$). So the minimum value of $Box-Box$ is... $$Box-Box=1-84=-83.tag$smallrm as ; the ; minimum ; value$$$
Therefore, the expression, $Box-Box$, must equal something in between $-14$ and $-83$ inclusive. Since it is negative, the first box must have an option number smaller than the second box. Now we know where to place numbers, we can make this range between $-14$ and $-83$ smaller!
$Box-Box$ must equal something in between $-1$ and $-9$ inclusive.
Proof:
Let the first box be the lowest option number (namely, $3$) and the second box be the highest option number (namely, $12$). We obtain that $$boxed3-boxed12=-9.tag$smallrm as ; the ; minimum ; value$$$ This is the minimum value because the pair $(3,12)$ is the furthest away from each other out of all the option numbers; and since we are dealing with negatives, the maximum (furthest away) turns to minimum (because positive becomes negative). Now look at the option numbers: $3$, $6$, $7$, $12$. Which pair is the closest to each other? That pair is $(6,7)$ which differs by only $1$. Therefore, $$boxed6-boxed7=-1.tag$smallrm as ; the ; maximum ; value$$$
And therefore, $Box-Box$ must equal something in between $-1$ and $-9$ inclusive.
And finally,
The maximum value of $Boxtimes Box$ must be $12$... uh oh. That doesn't seem right... it's a contradiction; THERE EXISTS NO SOLUTION (without lateral thinking)!
Proof:
If we get the maximum value the LHS can equal (namely, $4$) and subtract the minimum value of $Box-Box$ (namely, $-9$), we get $$4-(-9)=4+9=13.$$ Therefore, the maximum value of $Boxtimes Box$ is $13$. But $13$ is prime, so the maximum value reduces to $12$.
But we now have a problem. The minimum product that can be made from the option numbers is $3times 6$ or $6times 3$ which is $18$. Therefore, the minimum value of $Boxtimes Box$ is $18$. $18>12$. This is a contradiction. We do not need to consider the minimum value of LHS anymore, now (can you guess why?).
Therefore,
THERE IS NO SOLUTION! (without lateral thinking).
This was tougher than the puzzle that inspired the OP... but I did it. In fact,
Invert $6$ to make $9$ and substitute in all the proofs (with minimum value of product being $7times 3$ or $3times 7$ which is $21$). You will still obtain a contradiction, so there is no solution with an inverted $6$ as well.
answered 4 hours ago
user477343
3,3181745
Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
– PotatoLatte
4 hours ago
add a comment |Â
up vote
0
down vote
OK, so:
$$ 12-7+6*3 = 2E_16/10_2 $$
You restrict the decimal numbers we can use to 3, 6, 7 and 12, but you don't restrict other numeric bases. Here, I'm using a base 16 number (2E, equivalent to 46 in base 10) and a binary number (10, equivalent to 2).
answered 4 hours ago
Racso
2886
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Racso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Edit: Got it.
-$7 + 3 * 3 = 12/6$ (Still don’t have to fill in every box.)
Great puzzle!
answered 3 hours ago
Excited Raichu
1487
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What are you going to do with the genvyvat fynfu?
– Roman Odaisky
3 hours ago
It’s not possible for me to vtaber vg?
– Excited Raichu
3 hours ago
Would be qvivqr ol mreb!
– Weather Vane
3 hours ago
I suppose so. Svkrq vg!
– Excited Raichu
3 hours ago
This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
– Roman Odaisky
3 hours ago
add a comment |Â
up vote
1
down vote
accepted
Edit: Got it.
-$7 + 3 * 3 = 12/6$ (Still don’t have to fill in every box.)
Great puzzle!
answered 3 hours ago
Excited Raichu
1487
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What are you going to do with the genvyvat fynfu?
– Roman Odaisky
3 hours ago
It’s not possible for me to vtaber vg?
– Excited Raichu
3 hours ago
Would be qvivqr ol mreb!
– Weather Vane
3 hours ago
I suppose so. Svkrq vg!
– Excited Raichu
3 hours ago
This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
– Roman Odaisky
3 hours ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Edit: Got it.
-$7 + 3 * 3 = 12/6$ (Still don’t have to fill in every box.)
Great puzzle!
answered 3 hours ago
Excited Raichu
1487
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Edit: Got it.
-$7 + 3 * 3 = 12/6$ (Still don’t have to fill in every box.)
Great puzzle!
answered 3 hours ago
Excited Raichu
1487
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
answered 3 hours ago
Excited Raichu
1487
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
Excited Raichu
1487
answered 3 hours ago
Excited Raichu
1487
1487
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What are you going to do with the genvyvat fynfu?
– Roman Odaisky
3 hours ago
It’s not possible for me to vtaber vg?
– Excited Raichu
3 hours ago
Would be qvivqr ol mreb!
– Weather Vane
3 hours ago
I suppose so. Svkrq vg!
– Excited Raichu
3 hours ago
This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
– Roman Odaisky
3 hours ago
add a comment |Â
What are you going to do with the genvyvat fynfu?
– Roman Odaisky
3 hours ago
It’s not possible for me to vtaber vg?
– Excited Raichu
3 hours ago
Would be qvivqr ol mreb!
– Weather Vane
3 hours ago
I suppose so. Svkrq vg!
– Excited Raichu
3 hours ago
This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
– Roman Odaisky
3 hours ago
What are you going to do with the genvyvat fynfu?
– Roman Odaisky
3 hours ago
What are you going to do with the genvyvat fynfu?
– Roman Odaisky
3 hours ago
It’s not possible for me to vtaber vg?
– Excited Raichu
3 hours ago
It’s not possible for me to vtaber vg?
– Excited Raichu
3 hours ago
Would be qvivqr ol mreb!
– Weather Vane
3 hours ago
Would be qvivqr ol mreb!
– Weather Vane
3 hours ago
I suppose so. Svkrq vg!
– Excited Raichu
3 hours ago
I suppose so. Svkrq vg!
– Excited Raichu
3 hours ago
This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
– Roman Odaisky
3 hours ago
This is indeed the intended solution. I wonder whether someone can out-lateral me though :−)
– Roman Odaisky
3 hours ago
add a comment |Â
up vote
1
down vote
My answer is
$boxed12 - boxed12 + boxed: 3 : times boxed: 3 : = boxed63 : / : boxed: 7 : $
because
the question says "place numbers into boxes" so I placed two numbers into the 5th box.
Note: There is one solution to the obvious question:
$boxed: 7 : - boxed12 + boxed: 3 : times boxed: 3 : = boxed12 : / : boxed: 3 : $
But it is illegal because it does not use all the numbers.
Solved with pencil and paper only.
answered 2 hours ago
Weather Vane
2887
Also very nice. I was pretty sure repeated application of this idea can lead to a solution, but to solve the puzzle doing it only once looks remarkable to me.
– Roman Odaisky
1 hour ago
add a comment |Â
up vote
1
down vote
My answer is
$boxed12 - boxed12 + boxed: 3 : times boxed: 3 : = boxed63 : / : boxed: 7 : $
because
the question says "place numbers into boxes" so I placed two numbers into the 5th box.
Note: There is one solution to the obvious question:
$boxed: 7 : - boxed12 + boxed: 3 : times boxed: 3 : = boxed12 : / : boxed: 3 : $
But it is illegal because it does not use all the numbers.
Solved with pencil and paper only.
answered 2 hours ago
Weather Vane
2887
Also very nice. I was pretty sure repeated application of this idea can lead to a solution, but to solve the puzzle doing it only once looks remarkable to me.
– Roman Odaisky
1 hour ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
My answer is
$boxed12 - boxed12 + boxed: 3 : times boxed: 3 : = boxed63 : / : boxed: 7 : $
because
the question says "place numbers into boxes" so I placed two numbers into the 5th box.
Note: There is one solution to the obvious question:
$boxed: 7 : - boxed12 + boxed: 3 : times boxed: 3 : = boxed12 : / : boxed: 3 : $
But it is illegal because it does not use all the numbers.
Solved with pencil and paper only.
answered 2 hours ago
Weather Vane
2887
My answer is
$boxed12 - boxed12 + boxed: 3 : times boxed: 3 : = boxed63 : / : boxed: 7 : $
because
the question says "place numbers into boxes" so I placed two numbers into the 5th box.
Note: There is one solution to the obvious question:
$boxed: 7 : - boxed12 + boxed: 3 : times boxed: 3 : = boxed12 : / : boxed: 3 : $
But it is illegal because it does not use all the numbers.
Solved with pencil and paper only.
answered 2 hours ago
Weather Vane
2887
edited 56 mins ago
answered 2 hours ago
Weather Vane
2887
answered 2 hours ago
Weather Vane
2887
answered 2 hours ago
Weather Vane
2887
2887
Also very nice. I was pretty sure repeated application of this idea can lead to a solution, but to solve the puzzle doing it only once looks remarkable to me.
– Roman Odaisky
1 hour ago
add a comment |Â
Also very nice. I was pretty sure repeated application of this idea can lead to a solution, but to solve the puzzle doing it only once looks remarkable to me.
– Roman Odaisky
1 hour ago
Also very nice. I was pretty sure repeated application of this idea can lead to a solution, but to solve the puzzle doing it only once looks remarkable to me.
– Roman Odaisky
1 hour ago
Also very nice. I was pretty sure repeated application of this idea can lead to a solution, but to solve the puzzle doing it only once looks remarkable to me.
– Roman Odaisky
1 hour ago
add a comment |Â
up vote
0
down vote
Partial Answer:
Well, well, well.
THERE EXISTS NO SOLUTION! (without lateral thinking, although inverting the $6$ to a $9$ has no solution either, in particular).
Let's call the numbers we can choose from, the Option Numbers.
The right hand side (RHS) will either equal $1$, $2$ or $4$ (without lateral thinking).
Proof:
This is our equation: $$Box-Box+BoxtimesBox=Box:/:Boxtag$smallrm given$$$ $7$ can only be in the fifth box if the sixth box is also $7$, as that is the only number out of the option numbers that divide $7$. Otherwise, the left hand side (LHS) would not be an integer. So, a possibility is that the RHS is equal to $7/7=1$.
Excluding $7$ now, $3$ cannot be in the fifth box because it is the lowest option number, and thus the fraction will not be an integer otherwise. That leaves $6$ and $12$.
So the fraction is either $12/6$, $12/3$ or $6/3$ which is $2$, $2$ or $4$ respectively. Since $2=2$, then the RHS is either equal to $2$ or $4$. But we mentioned that it can also be equal to $1$, so the possible values the RHS can equal are $$1,2,4.$$
Therefore, $$Box-Box+BoxtimesBox=1,2;textor;4.$$
$Box-Box$ equals a number in between $-14$ and $-83$ inclusive.
Proof:
From the previous proof, we know that the LHS is either $1$, $2$ or $4$. All the option numbers are positive, and the minimum product that can be made from them is $3times 6$ or $6times 3$ which is $18$; also, the maximum value that the RHS can equal is $4$. So the maximum value of $Box-Box$ is... $$Box-Box=4-18=-14.tag$smallrm as ; the ; maximum ; value$$$ Now, doing the opposite to find the minimum value, we find maximum product of option numbers and minimum value RHS can equal. That makes $84$ and $1$ respectively (since $84=12times 7$ or $7times 12$). So the minimum value of $Box-Box$ is... $$Box-Box=1-84=-83.tag$smallrm as ; the ; minimum ; value$$$
Therefore, the expression, $Box-Box$, must equal something in between $-14$ and $-83$ inclusive. Since it is negative, the first box must have an option number smaller than the second box. Now we know where to place numbers, we can make this range between $-14$ and $-83$ smaller!
$Box-Box$ must equal something in between $-1$ and $-9$ inclusive.
Proof:
Let the first box be the lowest option number (namely, $3$) and the second box be the highest option number (namely, $12$). We obtain that $$boxed3-boxed12=-9.tag$smallrm as ; the ; minimum ; value$$$ This is the minimum value because the pair $(3,12)$ is the furthest away from each other out of all the option numbers; and since we are dealing with negatives, the maximum (furthest away) turns to minimum (because positive becomes negative). Now look at the option numbers: $3$, $6$, $7$, $12$. Which pair is the closest to each other? That pair is $(6,7)$ which differs by only $1$. Therefore, $$boxed6-boxed7=-1.tag$smallrm as ; the ; maximum ; value$$$
And therefore, $Box-Box$ must equal something in between $-1$ and $-9$ inclusive.
And finally,
The maximum value of $Boxtimes Box$ must be $12$... uh oh. That doesn't seem right... it's a contradiction; THERE EXISTS NO SOLUTION (without lateral thinking)!
Proof:
If we get the maximum value the LHS can equal (namely, $4$) and subtract the minimum value of $Box-Box$ (namely, $-9$), we get $$4-(-9)=4+9=13.$$ Therefore, the maximum value of $Boxtimes Box$ is $13$. But $13$ is prime, so the maximum value reduces to $12$.
But we now have a problem. The minimum product that can be made from the option numbers is $3times 6$ or $6times 3$ which is $18$. Therefore, the minimum value of $Boxtimes Box$ is $18$. $18>12$. This is a contradiction. We do not need to consider the minimum value of LHS anymore, now (can you guess why?).
Therefore,
THERE IS NO SOLUTION! (without lateral thinking).
This was tougher than the puzzle that inspired the OP... but I did it. In fact,
Invert $6$ to make $9$ and substitute in all the proofs (with minimum value of product being $7times 3$ or $3times 7$ which is $21$). You will still obtain a contradiction, so there is no solution with an inverted $6$ as well.
answered 4 hours ago
user477343
3,3181745
Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
– PotatoLatte
4 hours ago
add a comment |Â
up vote
0
down vote
Partial Answer:
Well, well, well.
THERE EXISTS NO SOLUTION! (without lateral thinking, although inverting the $6$ to a $9$ has no solution either, in particular).
Let's call the numbers we can choose from, the Option Numbers.
The right hand side (RHS) will either equal $1$, $2$ or $4$ (without lateral thinking).
Proof:
This is our equation: $$Box-Box+BoxtimesBox=Box:/:Boxtag$smallrm given$$$ $7$ can only be in the fifth box if the sixth box is also $7$, as that is the only number out of the option numbers that divide $7$. Otherwise, the left hand side (LHS) would not be an integer. So, a possibility is that the RHS is equal to $7/7=1$.
Excluding $7$ now, $3$ cannot be in the fifth box because it is the lowest option number, and thus the fraction will not be an integer otherwise. That leaves $6$ and $12$.
So the fraction is either $12/6$, $12/3$ or $6/3$ which is $2$, $2$ or $4$ respectively. Since $2=2$, then the RHS is either equal to $2$ or $4$. But we mentioned that it can also be equal to $1$, so the possible values the RHS can equal are $$1,2,4.$$
Therefore, $$Box-Box+BoxtimesBox=1,2;textor;4.$$
$Box-Box$ equals a number in between $-14$ and $-83$ inclusive.
Proof:
From the previous proof, we know that the LHS is either $1$, $2$ or $4$. All the option numbers are positive, and the minimum product that can be made from them is $3times 6$ or $6times 3$ which is $18$; also, the maximum value that the RHS can equal is $4$. So the maximum value of $Box-Box$ is... $$Box-Box=4-18=-14.tag$smallrm as ; the ; maximum ; value$$$ Now, doing the opposite to find the minimum value, we find maximum product of option numbers and minimum value RHS can equal. That makes $84$ and $1$ respectively (since $84=12times 7$ or $7times 12$). So the minimum value of $Box-Box$ is... $$Box-Box=1-84=-83.tag$smallrm as ; the ; minimum ; value$$$
Therefore, the expression, $Box-Box$, must equal something in between $-14$ and $-83$ inclusive. Since it is negative, the first box must have an option number smaller than the second box. Now we know where to place numbers, we can make this range between $-14$ and $-83$ smaller!
$Box-Box$ must equal something in between $-1$ and $-9$ inclusive.
Proof:
Let the first box be the lowest option number (namely, $3$) and the second box be the highest option number (namely, $12$). We obtain that $$boxed3-boxed12=-9.tag$smallrm as ; the ; minimum ; value$$$ This is the minimum value because the pair $(3,12)$ is the furthest away from each other out of all the option numbers; and since we are dealing with negatives, the maximum (furthest away) turns to minimum (because positive becomes negative). Now look at the option numbers: $3$, $6$, $7$, $12$. Which pair is the closest to each other? That pair is $(6,7)$ which differs by only $1$. Therefore, $$boxed6-boxed7=-1.tag$smallrm as ; the ; maximum ; value$$$
And therefore, $Box-Box$ must equal something in between $-1$ and $-9$ inclusive.
And finally,
The maximum value of $Boxtimes Box$ must be $12$... uh oh. That doesn't seem right... it's a contradiction; THERE EXISTS NO SOLUTION (without lateral thinking)!
Proof:
If we get the maximum value the LHS can equal (namely, $4$) and subtract the minimum value of $Box-Box$ (namely, $-9$), we get $$4-(-9)=4+9=13.$$ Therefore, the maximum value of $Boxtimes Box$ is $13$. But $13$ is prime, so the maximum value reduces to $12$.
But we now have a problem. The minimum product that can be made from the option numbers is $3times 6$ or $6times 3$ which is $18$. Therefore, the minimum value of $Boxtimes Box$ is $18$. $18>12$. This is a contradiction. We do not need to consider the minimum value of LHS anymore, now (can you guess why?).
Therefore,
THERE IS NO SOLUTION! (without lateral thinking).
This was tougher than the puzzle that inspired the OP... but I did it. In fact,
Invert $6$ to make $9$ and substitute in all the proofs (with minimum value of product being $7times 3$ or $3times 7$ which is $21$). You will still obtain a contradiction, so there is no solution with an inverted $6$ as well.
answered 4 hours ago
user477343
3,3181745
Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
– PotatoLatte
4 hours ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Partial Answer:
Well, well, well.
THERE EXISTS NO SOLUTION! (without lateral thinking, although inverting the $6$ to a $9$ has no solution either, in particular).
Let's call the numbers we can choose from, the Option Numbers.
The right hand side (RHS) will either equal $1$, $2$ or $4$ (without lateral thinking).
Proof:
This is our equation: $$Box-Box+BoxtimesBox=Box:/:Boxtag$smallrm given$$$ $7$ can only be in the fifth box if the sixth box is also $7$, as that is the only number out of the option numbers that divide $7$. Otherwise, the left hand side (LHS) would not be an integer. So, a possibility is that the RHS is equal to $7/7=1$.
Excluding $7$ now, $3$ cannot be in the fifth box because it is the lowest option number, and thus the fraction will not be an integer otherwise. That leaves $6$ and $12$.
So the fraction is either $12/6$, $12/3$ or $6/3$ which is $2$, $2$ or $4$ respectively. Since $2=2$, then the RHS is either equal to $2$ or $4$. But we mentioned that it can also be equal to $1$, so the possible values the RHS can equal are $$1,2,4.$$
Therefore, $$Box-Box+BoxtimesBox=1,2;textor;4.$$
$Box-Box$ equals a number in between $-14$ and $-83$ inclusive.
Proof:
From the previous proof, we know that the LHS is either $1$, $2$ or $4$. All the option numbers are positive, and the minimum product that can be made from them is $3times 6$ or $6times 3$ which is $18$; also, the maximum value that the RHS can equal is $4$. So the maximum value of $Box-Box$ is... $$Box-Box=4-18=-14.tag$smallrm as ; the ; maximum ; value$$$ Now, doing the opposite to find the minimum value, we find maximum product of option numbers and minimum value RHS can equal. That makes $84$ and $1$ respectively (since $84=12times 7$ or $7times 12$). So the minimum value of $Box-Box$ is... $$Box-Box=1-84=-83.tag$smallrm as ; the ; minimum ; value$$$
Therefore, the expression, $Box-Box$, must equal something in between $-14$ and $-83$ inclusive. Since it is negative, the first box must have an option number smaller than the second box. Now we know where to place numbers, we can make this range between $-14$ and $-83$ smaller!
$Box-Box$ must equal something in between $-1$ and $-9$ inclusive.
Proof:
Let the first box be the lowest option number (namely, $3$) and the second box be the highest option number (namely, $12$). We obtain that $$boxed3-boxed12=-9.tag$smallrm as ; the ; minimum ; value$$$ This is the minimum value because the pair $(3,12)$ is the furthest away from each other out of all the option numbers; and since we are dealing with negatives, the maximum (furthest away) turns to minimum (because positive becomes negative). Now look at the option numbers: $3$, $6$, $7$, $12$. Which pair is the closest to each other? That pair is $(6,7)$ which differs by only $1$. Therefore, $$boxed6-boxed7=-1.tag$smallrm as ; the ; maximum ; value$$$
And therefore, $Box-Box$ must equal something in between $-1$ and $-9$ inclusive.
And finally,
The maximum value of $Boxtimes Box$ must be $12$... uh oh. That doesn't seem right... it's a contradiction; THERE EXISTS NO SOLUTION (without lateral thinking)!
Proof:
If we get the maximum value the LHS can equal (namely, $4$) and subtract the minimum value of $Box-Box$ (namely, $-9$), we get $$4-(-9)=4+9=13.$$ Therefore, the maximum value of $Boxtimes Box$ is $13$. But $13$ is prime, so the maximum value reduces to $12$.
But we now have a problem. The minimum product that can be made from the option numbers is $3times 6$ or $6times 3$ which is $18$. Therefore, the minimum value of $Boxtimes Box$ is $18$. $18>12$. This is a contradiction. We do not need to consider the minimum value of LHS anymore, now (can you guess why?).
Therefore,
THERE IS NO SOLUTION! (without lateral thinking).
This was tougher than the puzzle that inspired the OP... but I did it. In fact,
Invert $6$ to make $9$ and substitute in all the proofs (with minimum value of product being $7times 3$ or $3times 7$ which is $21$). You will still obtain a contradiction, so there is no solution with an inverted $6$ as well.
answered 4 hours ago
user477343
3,3181745
Partial Answer:
Well, well, well.
THERE EXISTS NO SOLUTION! (without lateral thinking, although inverting the $6$ to a $9$ has no solution either, in particular).
Let's call the numbers we can choose from, the Option Numbers.
The right hand side (RHS) will either equal $1$, $2$ or $4$ (without lateral thinking).
Proof:
This is our equation: $$Box-Box+BoxtimesBox=Box:/:Boxtag$smallrm given$$$ $7$ can only be in the fifth box if the sixth box is also $7$, as that is the only number out of the option numbers that divide $7$. Otherwise, the left hand side (LHS) would not be an integer. So, a possibility is that the RHS is equal to $7/7=1$.
Excluding $7$ now, $3$ cannot be in the fifth box because it is the lowest option number, and thus the fraction will not be an integer otherwise. That leaves $6$ and $12$.
So the fraction is either $12/6$, $12/3$ or $6/3$ which is $2$, $2$ or $4$ respectively. Since $2=2$, then the RHS is either equal to $2$ or $4$. But we mentioned that it can also be equal to $1$, so the possible values the RHS can equal are $$1,2,4.$$
Therefore, $$Box-Box+BoxtimesBox=1,2;textor;4.$$
$Box-Box$ equals a number in between $-14$ and $-83$ inclusive.
Proof:
From the previous proof, we know that the LHS is either $1$, $2$ or $4$. All the option numbers are positive, and the minimum product that can be made from them is $3times 6$ or $6times 3$ which is $18$; also, the maximum value that the RHS can equal is $4$. So the maximum value of $Box-Box$ is... $$Box-Box=4-18=-14.tag$smallrm as ; the ; maximum ; value$$$ Now, doing the opposite to find the minimum value, we find maximum product of option numbers and minimum value RHS can equal. That makes $84$ and $1$ respectively (since $84=12times 7$ or $7times 12$). So the minimum value of $Box-Box$ is... $$Box-Box=1-84=-83.tag$smallrm as ; the ; minimum ; value$$$
Therefore, the expression, $Box-Box$, must equal something in between $-14$ and $-83$ inclusive. Since it is negative, the first box must have an option number smaller than the second box. Now we know where to place numbers, we can make this range between $-14$ and $-83$ smaller!
$Box-Box$ must equal something in between $-1$ and $-9$ inclusive.
Proof:
Let the first box be the lowest option number (namely, $3$) and the second box be the highest option number (namely, $12$). We obtain that $$boxed3-boxed12=-9.tag$smallrm as ; the ; minimum ; value$$$ This is the minimum value because the pair $(3,12)$ is the furthest away from each other out of all the option numbers; and since we are dealing with negatives, the maximum (furthest away) turns to minimum (because positive becomes negative). Now look at the option numbers: $3$, $6$, $7$, $12$. Which pair is the closest to each other? That pair is $(6,7)$ which differs by only $1$. Therefore, $$boxed6-boxed7=-1.tag$smallrm as ; the ; maximum ; value$$$
And therefore, $Box-Box$ must equal something in between $-1$ and $-9$ inclusive.
And finally,
The maximum value of $Boxtimes Box$ must be $12$... uh oh. That doesn't seem right... it's a contradiction; THERE EXISTS NO SOLUTION (without lateral thinking)!
Proof:
If we get the maximum value the LHS can equal (namely, $4$) and subtract the minimum value of $Box-Box$ (namely, $-9$), we get $$4-(-9)=4+9=13.$$ Therefore, the maximum value of $Boxtimes Box$ is $13$. But $13$ is prime, so the maximum value reduces to $12$.
But we now have a problem. The minimum product that can be made from the option numbers is $3times 6$ or $6times 3$ which is $18$. Therefore, the minimum value of $Boxtimes Box$ is $18$. $18>12$. This is a contradiction. We do not need to consider the minimum value of LHS anymore, now (can you guess why?).
Therefore,
THERE IS NO SOLUTION! (without lateral thinking).
This was tougher than the puzzle that inspired the OP... but I did it. In fact,
Invert $6$ to make $9$ and substitute in all the proofs (with minimum value of product being $7times 3$ or $3times 7$ which is $21$). You will still obtain a contradiction, so there is no solution with an inverted $6$ as well.
answered 4 hours ago
user477343
3,3181745
edited 3 hours ago
answered 4 hours ago
user477343
3,3181745
answered 4 hours ago
user477343
3,3181745
answered 4 hours ago
user477343
3,3181745
3,3181745
Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
– PotatoLatte
4 hours ago
add a comment |Â
Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
– PotatoLatte
4 hours ago
Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
– PotatoLatte
4 hours ago
Ah, but one of the tags is lateral thinking :P and his previous riddle also involved lateral thinking :P
– PotatoLatte
4 hours ago
add a comment |Â
up vote
0
down vote
OK, so:
$$ 12-7+6*3 = 2E_16/10_2 $$
You restrict the decimal numbers we can use to 3, 6, 7 and 12, but you don't restrict other numeric bases. Here, I'm using a base 16 number (2E, equivalent to 46 in base 10) and a binary number (10, equivalent to 2).
answered 4 hours ago
Racso
2886
New contributor
Racso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
OK, so:
$$ 12-7+6*3 = 2E_16/10_2 $$
You restrict the decimal numbers we can use to 3, 6, 7 and 12, but you don't restrict other numeric bases. Here, I'm using a base 16 number (2E, equivalent to 46 in base 10) and a binary number (10, equivalent to 2).
answered 4 hours ago
Racso
2886
New contributor
Racso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
OK, so:
$$ 12-7+6*3 = 2E_16/10_2 $$
You restrict the decimal numbers we can use to 3, 6, 7 and 12, but you don't restrict other numeric bases. Here, I'm using a base 16 number (2E, equivalent to 46 in base 10) and a binary number (10, equivalent to 2).
answered 4 hours ago
Racso
2886
New contributor
Racso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
OK, so:
$$ 12-7+6*3 = 2E_16/10_2 $$
You restrict the decimal numbers we can use to 3, 6, 7 and 12, but you don't restrict other numeric bases. Here, I'm using a base 16 number (2E, equivalent to 46 in base 10) and a binary number (10, equivalent to 2).
answered 4 hours ago
Racso
2886
New contributor
Racso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
answered 4 hours ago
Racso
2886
New contributor
Racso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
Racso
2886
answered 4 hours ago
Racso
2886
2886
New contributor
Racso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Racso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Racso is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
Roman Odaisky is a new contributor. Be nice, and check out our Code of Conduct.
Roman Odaisky is a new contributor. Be nice, and check out our Code of Conduct.
Roman Odaisky is a new contributor. Be nice, and check out our Code of Conduct.
Roman Odaisky is a new contributor. Be nice, and check out our Code of Conduct.
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1
Does this follow operator precedence rules? I mean, is the multiplication made before the addition, or after it?
– Racso
5 hours ago
@Rasco Of course it does, how else?
– Roman Odaisky
4 hours ago