Here is a math puzzle I had a little bit of hard time with
这是一个数学难题,我有点困难
No computers please
请不要电脑
There is a solution without inverting 6 to 9
有一个没有反转6到9的解决方案
7 个解决方案
#1
17
The trick is that
诀窍在于
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.其中两个字母实际上是罗马数字。 D = 500和C = 100. $ 25 - 12 + D / C = 3 * 6 $ 13 + 5 = 18 $这使用了所有“下面的数字”一次。
#2
7
Partial Answer:
This answer follows by BODMAS or BEDMAS or PEDMAS.
这个答案由BODMAS或BEDMAS或PEDMAS提供。
Umm...
THERE IS NO SOLUTION! (without lateral thinking; without inverting the $6$, for example)
没有解决方案! (没有横向思考;例如,没有反转$ 6 $)
Let's call the numbers we can choose from, the Option Numbers.
我们可以选择我们可以选择的数字,即选项号码。
25 cannot be in the third and fourth box.
25不能在第三和第四个框中。
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 $$\begin{align}\Box - \Box + \boxed{25}\:/\:\boxed{25} &= \Box - \Box + 1 \\ &= \Box\times \Box.\end{align}$$
这是我们的等式:$$ \ Box- \ Box + \ Box \:/ \:\ Box = \ Box \ times \ Box。\ tag {$ \ small \ rm给予$} $$ $ 12 $,$ 6 $和$ 3 $不要划分25美元,所以如果第四个盒子是$ 25 $,第三个盒子只能是25美元。假设涉及一个解决方案。然后我们有$$ \ begin {align} \ Box - \ Box + \ boxed {25} \:/ \:\ boxed {25}&= \ Box - \ Box + 1 \\&= \ Box \ times \ Box \ {端对齐} $$
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 $18$. $$\Box-\Box+\Box\:/\:\Box=18.\tag*{$(3\times 6$ or $6\times 3)$}$$左手边的最大数字是$ 25-3 + 1 = 23 $所以右边不能超过$ 23 $。但23美元是最优惠的,22美元和21美元都有两个不同的素数因素(尽管没有期权数字是素数),因此RHS不能超过20美元。另外,20美元= 5 \次4 = 10 \次2美元也不使用任何选项号码,因为19美元是素数,这意味着RHS不能超过18美元,即3美元,6美元或6美元\次3 $。但是,严格涉及期权数量的其他所有产品都超过18美元,因此RHS也不能低于18美元。如果RHS不能大于或低于18美元,那么它等于18美元。 $$ \ Box- \ Box + \ Box \:/ \:\ Box = 18。\ tag * {$(3 \ times 6 $或$ 6 \ times 3)$} $$
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.现在$ 18 = 6 \ times 3 $,它使用两个选项号码。所以现在我们必须找到选项号,使得$$ \ Box- \ Box + 1 = \ boxed6 \ times \ boxed3 = 18 $$因此$ \ Box- \ Box = 18-1 = 17 $。当然第一个盒子的价值必须大于17美元,因为17美元是积极的,所有选项都是正数。唯一大于17美元的期权数字是25美元。所以$ \ boxed {25} - \ Box = 17 $。因此,第二个框的值为$ 25-17 = 8 $,但$ 8 $不是期权编号。这是一个矛盾,因此$ 25 $不能在第三个框中,因此也是第四个框。
$\Box\:/\:\Box=2$ or $4$.
$ \ Box \:/ \:\ Box = 2 $或$ 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 $4$.
现在$ \ Box \:/ \:\ Box $必须是一个整数,因为$ 18 $是一个整数,因此分子框(第三个)的选项编号大于分母框(第四个)。由于$ 3 $是最低的期权数字,因此$ 3 $不能在第三个框中。留下12美元或6美元,这样第四个盒子就是$ 6 $或$ 3 $。因此,这个分数必须等于$ 12/6 $,$ 6/3 $或$ 12/3 $ $ $ 2 $,$ 2 $或$ 4 $。由于$ 2 = 2 $,因此分数为$ 2 $或$ 4 $。
We thus have the equations: $$\begin{align}\Box-\Box+2&=18 \\ \small{\rm or} \quad \Box-\Box+4&=18.\end{align}$$ Therefore, $$\begin{align}\Box-\Box&=18-2=16 \\ \small{\rm or} \quad \Box-\Box&=18-4=12.\end{align}$$
因此,我们得到了等式:$$ \ begin {align} \ Box- \ Box + 2&= 18 \\ \ small {\ rm或} \ quad \ Box- \ Box + 4&= 18。\ end {align} $$因此,$$ \ begin {align} \ Box- \ Box&= 18-2 = 16 \\ \ small {\ rm或} \ quad \ Box- \ Box&= 18-4 = 12。\ end {align} $$
And finally,
From the previous proof, THERE EXISTS NO SOLUTION!
从之前的证据来看,这里没有解决方案!
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. $$\require{cancel}{\xcancel {\Box-\Box=16}}$$
现在考虑第一个方程式,第一个方框必须有一个大于$ 16 $的选项号。唯一的选项号码是$ 25 $。因此我们有$$ \ boxed {25} - \ Box = 16 $$因此$ \ Box = 25-16 = 9 $。但$ 9 $不是期权数字。这是一个矛盾,所以第一个方程式不可能存在。 $$ \ require {cancel} {\ xcancel {\ Box- \ Box = 16}} $$
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...
考虑到第二个等式,第一个框需要大于12美元。它不能是12美元,它必须大于12美元。同样,唯一大于12美元的期权数字是25美元。因此我们有$$ \ boxed {25} - \ Box = 12 $$因此$ \ Box = 25-12 = 13 $。但13美元不是期权数字。这是一个矛盾,所以第二个方程不可能存在。 $$ \ require {cancel} {\ xcancel {\ Box- \ Box = 12}} $$但如果两个方程都不存在,那么......
...THERE IS NO SOLUTION!
......没有解决方案!
Therefore,
Some lateral-thinking must be required, unless you do not follow by BODMAS or BEDMAS or PEDMAS.
除非你不遵守BODMAS或BEDMAS或PEDMAS,否则必须要有一些横向思维。
#3
4
There doesn't seem to be anything that says that only one number can be placed into each box. Thus
似乎没有任何内容表明每个盒子中只能放置一个数字。从而
$$12 - 25 + 66 \div 3 = 3 \times 3$$
$$ 12 - 25 + 66 \ div 3 = 3 \ times 3 $$
would be a valid solution.
将是一个有效的解决方案。
It just requires putting
它只需要推杆
two $6$s in the same box.
两个$ 6 $ s在同一个盒子里。
#4
4
The puzzle explicitly states: Each number from below must be used once at least once.
该谜题明确指出:下面的每个数字必须至少使用一次。
Our numbers are $12, 6, 25, 3$. Without changing any of the numbers, using integer math instead of decimals, and following the rule above:
我们的数字是12,6,25,3美元。在不更改任何数字的情况下,使用整数数学而不是小数,并遵循上述规则:
$12 - 3 + 6 / 25 = 3 * 3$
12美元 - 3 + 6/25 = 3 * 3 $
Following Order of Operations:
遵循运营顺序:
$3 * 3 = 9$
$6 / 25 = 0$
$3 + 0 = 3$
$12 - 3 = 9$
$9 = 9$$ 3 * 3 = 9 $ $ 6/25 = 0 $ $ 3 + 0 = 3 $ $ 12 - 3 = 9 $ $ 9 = 9 $
#5
3
how about
$25-9+12/6=3\times6$
to do that
要做到这一点
I rotated 6 into 9 as you suspected which is valid for the tag provided.
我怀疑你将6转为9,这对所提供的标签有效。
#6
2
My solution is
我的解决方案是
$25 - 12 + 25 / 3 = 3 \times 6$
$ 25 - 12 + 25/3 = 3 \次6 $
because
the numbers are octal base, and converting to decimal base
数字是八进制数,并转换为十进制数
gives
$21 - 10 + 21 / 3 = 3 \times 6$
$ 21 - 10 + 21/3 = 3 \次6 $
#7
0
Using the tag:
使用标签:
Each number must be used. It seems like there are 4 numbers: 12, 6, 25, 3. However, I'm guessing there are 6 numbers (lateral thinking): 1, 2, 6, 2, 5, 3. So one of the answers (there may be more with this logic): is
6 - 5 + 3 / 1 = 2 * 2
3 - 5 + 6 / 1 = 2 * 2 is another order必须使用每个号码。似乎有4个数字:12,6,25,3。但是,我猜测有6个数字(横向思维):1,2,6,2,5,3。所以其中一个答案(那里用这个逻辑可能更多):是6 - 5 + 3/1 = 2 * 2 3 - 5 + 6/1 = 2 * 2是另一个命令
#1
17
The trick is that
诀窍在于
Two of the letters are actually roman numerals. D = 500 and C = 100.
$25 - 12 + D / C = 3 * 6$
$13 + 5 = 18$
This uses all "numbers from below" once.其中两个字母实际上是罗马数字。 D = 500和C = 100. $ 25 - 12 + D / C = 3 * 6 $ 13 + 5 = 18 $这使用了所有“下面的数字”一次。
#2
7
Partial Answer:
This answer follows by BODMAS or BEDMAS or PEDMAS.
这个答案由BODMAS或BEDMAS或PEDMAS提供。
Umm...
THERE IS NO SOLUTION! (without lateral thinking; without inverting the $6$, for example)
没有解决方案! (没有横向思考;例如,没有反转$ 6 $)
Let's call the numbers we can choose from, the Option Numbers.
我们可以选择我们可以选择的数字,即选项号码。
25 cannot be in the third and fourth box.
25不能在第三和第四个框中。
Proof:
This is our equation: $$\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}$$
这是我们的等式:$$ \ Box- \ Box + \ Box \:/ \:\ Box = \ Box \ times \ Box。\ tag {$ \ small \ rm给予$} $$ $ 12 $,$ 6 $和$ 3 $不要划分25美元,所以如果第四个盒子是$ 25 $,第三个盒子只能是25美元。假设涉及一个解决方案。然后我们有$$ \ begin {align} \ Box - \ Box + \ boxed {25} \:/ \:\ boxed {25}&= \ Box - \ Box + 1 \\&= \ Box \ times \ Box \ {端对齐} $$
The largest number for the left hand side is $25-3+1=23$ so the right hand side cannot be greater than $23$. But $23$ is prime and both $22$ and $21$ have two distinct prime factors (although none of option numbers are prime), so the RHS cannot be greater than $20$.
Also, $20=5\times 4 = 10\times 2$ which uses none of the option numbers as well, and since $19$ is prime, that means the RHS cannot be greater than $18$ which is $3\times 6$ or $6\times 3$. But also, every other product strictly involving the option numbers is greater than $18$, so the RHS cannot be lower than $18$ either.
If the RHS cannot be greater or lower than $18$, then it is equal to $18$. $$\Box-\Box+\Box\:/\:\Box=18.\tag*{$(3\times 6$ or $6\times 3)$}$$左手边的最大数字是$ 25-3 + 1 = 23 $所以右边不能超过$ 23 $。但23美元是最优惠的,22美元和21美元都有两个不同的素数因素(尽管没有期权数字是素数),因此RHS不能超过20美元。另外,20美元= 5 \次4 = 10 \次2美元也不使用任何选项号码,因为19美元是素数,这意味着RHS不能超过18美元,即3美元,6美元或6美元\次3 $。但是,严格涉及期权数量的其他所有产品都超过18美元,因此RHS也不能低于18美元。如果RHS不能大于或低于18美元,那么它等于18美元。 $$ \ Box- \ Box + \ Box \:/ \:\ Box = 18。\ tag * {$(3 \ times 6 $或$ 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$$ Therefore $\Box-\Box=18-1=17$. Of course the first box has to have a bigger value than $17$, because $17$ is positive and all the option numbers are positive. The only option number bigger than $17$ is $25$. So $\boxed{25}-\Box=17$. Therefore the second box has a value of $25-17=8$ but $8$ is not an option number.
This is a contradiction, so $25$ cannot be in the third box, and thus fourth one, too.现在$ 18 = 6 \ times 3 $,它使用两个选项号码。所以现在我们必须找到选项号,使得$$ \ Box- \ Box + 1 = \ boxed6 \ times \ boxed3 = 18 $$因此$ \ Box- \ Box = 18-1 = 17 $。当然第一个盒子的价值必须大于17美元,因为17美元是积极的,所有选项都是正数。唯一大于17美元的期权数字是25美元。所以$ \ boxed {25} - \ Box = 17 $。因此,第二个框的值为$ 25-17 = 8 $,但$ 8 $不是期权编号。这是一个矛盾,因此$ 25 $不能在第三个框中,因此也是第四个框。
$\Box\:/\:\Box=2$ or $4$.
$ \ Box \:/ \:\ Box = 2 $或$ 4 $。
Proof:
Now $\Box\:/\: \Box$ has to be an integer since $18$ is an integer, therefore the numerator box (third one) has an option number greater than the denominator box (fourth one). Since $3$ is the lowest option number, then $3$ cannot be in the third box. That leaves $12$ or $6$, so that leaves the fourth box to be $6$ or $3$. Therefore, this fraction must be equal to $12/6$, $6/3$ or $12/3$ which is $2$, $2$ or $4$. And since $2=2$, then the fraction is either $2$ or $4$.
现在$ \ Box \:/ \:\ Box $必须是一个整数,因为$ 18 $是一个整数,因此分子框(第三个)的选项编号大于分母框(第四个)。由于$ 3 $是最低的期权数字,因此$ 3 $不能在第三个框中。留下12美元或6美元,这样第四个盒子就是$ 6 $或$ 3 $。因此,这个分数必须等于$ 12/6 $,$ 6/3 $或$ 12/3 $ $ $ 2 $,$ 2 $或$ 4 $。由于$ 2 = 2 $,因此分数为$ 2 $或$ 4 $。
We thus have the equations: $$\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}$$
因此,我们得到了等式:$$ \ begin {align} \ Box- \ Box + 2&= 18 \\ \ small {\ rm或} \ quad \ Box- \ Box + 4&= 18。\ end {align} $$因此,$$ \ begin {align} \ Box- \ Box&= 18-2 = 16 \\ \ small {\ rm或} \ quad \ Box- \ Box&= 18-4 = 12。\ end {align} $$
And finally,
From the previous proof, THERE EXISTS NO SOLUTION!
从之前的证据来看,这里没有解决方案!
Proof:
Now considering the first equation, the first box has to have an option number greater than $16$. The only option number like that is $25$. We thus have $$\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}}$$
现在考虑第一个方程式,第一个方框必须有一个大于$ 16 $的选项号。唯一的选项号码是$ 25 $。因此我们有$$ \ boxed {25} - \ Box = 16 $$因此$ \ Box = 25-16 = 9 $。但$ 9 $不是期权数字。这是一个矛盾,所以第一个方程式不可能存在。 $$ \ 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}}$$ But if both equations cannot exist, then...
考虑到第二个等式,第一个框需要大于12美元。它不能是12美元,它必须大于12美元。同样,唯一大于12美元的期权数字是25美元。因此我们有$$ \ boxed {25} - \ Box = 12 $$因此$ \ Box = 25-12 = 13 $。但13美元不是期权数字。这是一个矛盾,所以第二个方程不可能存在。 $$ \ require {cancel} {\ xcancel {\ Box- \ Box = 12}} $$但如果两个方程都不存在,那么......
...THERE IS NO SOLUTION!
......没有解决方案!
Therefore,
Some lateral-thinking must be required, unless you do not follow by BODMAS or BEDMAS or PEDMAS.
除非你不遵守BODMAS或BEDMAS或PEDMAS,否则必须要有一些横向思维。
#3
4
There doesn't seem to be anything that says that only one number can be placed into each box. Thus
似乎没有任何内容表明每个盒子中只能放置一个数字。从而
$$12 - 25 + 66 \div 3 = 3 \times 3$$
$$ 12 - 25 + 66 \ div 3 = 3 \ times 3 $$
would be a valid solution.
将是一个有效的解决方案。
It just requires putting
它只需要推杆
two $6$s in the same box.
两个$ 6 $ s在同一个盒子里。
#4
4
The puzzle explicitly states: Each number from below must be used once at least once.
该谜题明确指出:下面的每个数字必须至少使用一次。
Our numbers are $12, 6, 25, 3$. Without changing any of the numbers, using integer math instead of decimals, and following the rule above:
我们的数字是12,6,25,3美元。在不更改任何数字的情况下,使用整数数学而不是小数,并遵循上述规则:
$12 - 3 + 6 / 25 = 3 * 3$
12美元 - 3 + 6/25 = 3 * 3 $
Following Order of Operations:
遵循运营顺序:
$3 * 3 = 9$
$6 / 25 = 0$
$3 + 0 = 3$
$12 - 3 = 9$
$9 = 9$$ 3 * 3 = 9 $ $ 6/25 = 0 $ $ 3 + 0 = 3 $ $ 12 - 3 = 9 $ $ 9 = 9 $
#5
3
how about
$25-9+12/6=3\times6$
to do that
要做到这一点
I rotated 6 into 9 as you suspected which is valid for the tag provided.
我怀疑你将6转为9,这对所提供的标签有效。
#6
2
My solution is
我的解决方案是
$25 - 12 + 25 / 3 = 3 \times 6$
$ 25 - 12 + 25/3 = 3 \次6 $
because
the numbers are octal base, and converting to decimal base
数字是八进制数,并转换为十进制数
gives
$21 - 10 + 21 / 3 = 3 \times 6$
$ 21 - 10 + 21/3 = 3 \次6 $
#7
0
Using the tag:
使用标签:
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必须使用每个号码。似乎有4个数字:12,6,25,3。但是,我猜测有6个数字(横向思维):1,2,6,2,5,3。所以其中一个答案(那里用这个逻辑可能更多):是6 - 5 + 3/1 = 2 * 2 3 - 5 + 6/1 = 2 * 2是另一个命令