Fill in the boxes to get the right equation v2

Question

My take on a recent puzzle. Place numbers into boxes to create a correct equation, using each number at least once:

$$\Box-\Box+\Box\times\Box=\Box\:/\:\Box$$

This time the four decimal numbers are 3, 6, 7, 12.

Answer 1

Edit: Got it.

-$7 + 3 * 3 = 12/6$ (Still don’t have to fill in every box.)

Great puzzle!

Answer 2

My answer is

$\boxed{12} - \boxed{12} + \boxed{\: 3 \: } \times \boxed{\: 3 \: } = \boxed{63} \: / \: \boxed{\: 7 \: }$

because

the question says "place numbers into boxes" so I placed two numbers into the 5th box.

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Note: There is one solution to the obvious question:

$\boxed{\: 7 \: } - \boxed{12} + \boxed{\: 3 \: } \times \boxed{\: 3 \: } = \boxed{12} \: / \: \boxed{\: 3 \: }$

But it is illegal because it does not use all the numbers.

^{Solved with pencil and paper only.}

Answer 3

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.

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The right hand side (RHS) will either equal $1$, $2$ or $4$ (without lateral thinking).

Proof:

This is our equation: $$\Box-\Box+\Box\times\Box=\Box\:/\:\Box\tag{$\small\rm given$}$$ $7$ 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, $$\Box-\Box+\Box\times\Box=1,2\;\text{or}\;4.$$

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$\Box-\Box$ 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 $3\times 6$ or $6\times 3$ which is $18$; also, the maximum value that the RHS can equal is $4$. So the maximum value of $\Box-\Box$ is... $$\Box-\Box=4-18=-14.\tag{$\small\rm as \; the \; maximum \; value$}$$ 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 $84=12\times 7$ or $7\times 12$). So the minimum value of $\Box-\Box$ is... $$\Box-\Box=1-84=-83.\tag{$\small\rm as \; the \; minimum \; value$}$$

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!

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$\Box-\Box$ 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 $$\boxed{3}-\boxed{12}=-9.\tag{$\small\rm as \; the \; minimum \; value$}$$ 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, $$\boxed{6}-\boxed{7}=-1.\tag{$\small\rm as \; the \; maximum \; value$}$$

And therefore, $\Box-\Box$ must equal something in between $-1$ and $-9$ inclusive.

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And finally,

The maximum value of $\Box\times \Box$ 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 $\Box\times \Box$ 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 $3\times 6$ or $6\times 3$ which is $18$. Therefore, the minimum value of $\Box\times \Box$ 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).

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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 $7\times 3$ or $3\times 7$ which is $21$). You will still obtain a contradiction, so there is no solution with an inverted $6$ as well.

Answer 4

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).