$\large 1 \, \square \, 2 \, \square \, 3 \, \square \, 4 = 10$

There are $4^3 = 64$ ways in which we can fill the squares with $+ , - , \times , \div$ .

How many ways would make the equation true?

**
Note:
**

You are not allowed to use parenthesis.

Obey the order of operations.

1
2
4
8

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Yes. It's always good to first reduce the number of possible cases to a minimum.

**
Bonus question
**
: Can you apply the same logic to the extension of this question?

$\large 1 \, \square \, 2 \, \square \, 3 \, \square 4 \, \square 5 = 15$

The second equation should be $1\times2\times3+4=10$

Pi Han Goh
- 5 years, 11 months ago

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Yes, thanks. Yet to have coffee this morning.

Chew-Seong Cheong
- 5 years, 11 months ago

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the third one is 1 x 2 - 3 x 4

Popescu Alexandru
- 5 years, 7 months ago

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@Popescu Alexandru – That would be $1 \times 2 - 3 \times 4 = -10$ .

Chew-Seong Cheong
- 5 years, 7 months ago

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@Chew-Seong Cheong – Even without parenthesis you're supposed to divide and multiply before you add and subtract. Please excuse my dear aunt sally.

Alex Hughes
- 5 years, 6 months ago

@Chew-Seong Cheong
–
can we write given qus in this form
3
*
4-1
*
2=22

Asif Khan
- 5 years, 6 months ago

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@Asif Khan – What do you mean? $34-12=22$ .

Chew-Seong Cheong
- 5 years, 6 months ago

there is actually three possible solutions -> 1+2+3+4=10 -> 1 * 2 * 3 + 4=10 -> 1 * -2 + 3 *4=10

haven't used parenthesis , and if u follow bodmas rule , solution 3 is correct

Sehrab Goyal
- 5 years, 6 months ago

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But we are required to follow BODMAS. See Note:

$\color{#D61F06}{\text{Obey the order of operations.}}$

Chew-Seong Cheong
- 5 years, 6 months ago

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Using bodmas, three solutions

தினேஸ் சாந்தி சீனிவாசன்
- 5 years, 6 months ago

its basically 1 * (-2) + 3 * 4 but u wont need parenthesis because even if u don't use them BODMAS says multiply 1 with -2 and 3 with 4 then add->subtract both :) basic school thing @Chew-Seong cheong , try google 1 * -2 + 3 * 4 it will say it as 10 , since the post haven't added constraint of using only 1 operator at a time

Sehrab Goyal
- 5 years, 6 months ago

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@Sehrab Goyal – It says fill the squares... u filled multiple and minus sign in, thats 2 signs... and also quotation marks count as operations right?

Ken Zhang
- 5 years, 5 months ago

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@Ken Zhang – I dont see using only 1 operator per Box instruction ? Since question have missed that i created the third solution using -2 , which is using - operator with 2 :)

Sehrab Goyal
- 5 years, 5 months ago

@Sehrab Goyal – Sorry, I got it wrong. But you are not allowed to use parenthesis for $(-2)$ and I guess $1 \times - 2$ does not make sense.

Chew-Seong Cheong
- 5 years, 5 months ago

I got 3 as well

Paul Goggin
- 5 years, 5 months ago

For the third one you're using (-2) rather than 2.

Matthew McCafferty
- 5 years, 6 months ago

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yes as there is not written to fill the box with only 1 operator ?

Sehrab Goyal
- 5 years, 5 months ago

i got three as well

Julija Golubovic
- 5 years, 5 months ago

Same as Chew-Seong Cheong

Rubel Hawladar
- 5 years, 6 months ago

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If we allow that the numbers could be in other base systems as well as decimal (not forbidden by the question), then there are many more solutions:

1 + 2 + 3 + 4 = 10 (base ten) 1 x 2 x 3 + 4 = 10 (base ten)

1 - 2 + 3 + 4 = 10 (base six) 1 / 2 x 3 x 4 = 10 (base six) 1 - 2 + 3 x 4 = 10 (base eleven) 1 + 2 x 3 + 4 = 10 (base eleven) 1 x 2 + 3 x 4 = 10 (base fourteen) 1 + 2 + 3 x 4 = 10 (base fifteen) 1 x 2 x 3 x 4 = 10 (base twenty-four) 1 + 2 x 3 x 4 = 10 (base twenty-five)

N.B. These equations all follow the BODMAS rule of operations.

Tony G

Tony Gillman
- 3 years, 1 month ago

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Why wasn't there a three option?

1+2+3+4=10

1×2×3+4=10

1×2-3×4=10

It didn't say you couldn't use the same sign multiple times

Desiree Brown
- 5 years, 6 months ago

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The third option, wouldn't that return -10 ?

Vignesh M
- 5 years, 6 months ago

1x2-3x4=10
| |

\/ \/
2 - 12 = -10

Kevalin Ketcham
- 2 years, 10 months ago

How is this table organized?

Oh wow, that's nice. I just listed out the 64 cases lol.

Chung Kevin
- 5 years, 11 months ago

It's interesting that if you reverse the arguments (4 3 2 1) there are FIVE solutions!

Mike Torr
- 2 years, 11 months ago

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That's a great observation! Can you post that version of the problem?

it should be NOTED that one should do multiplication and division before adding and subtraction, even though we all do it in life in that order

Nik Gibson
- 2 years, 9 months ago

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Yes, that's why the problem says "Obey the order of operations".

Chung Kevin
- 2 years, 9 months ago

There is only 2 possible solutions, BUT there is 8 ways to fill them, other wise wouldnt have 64 that we can fill the squares.. '* * +' and '* * +' looks the same, but the * switch places, same for the possible solution '+ + +', that have 3! ways to fill, so more 6. Making in total 8 possible ways to fill it. The question of the puzzle was miss made.

Mário Schuroff Rodrigues
- 5 years, 5 months ago

-1+3+2*4=2+8=10

ዓይጋ ዊንታ
- 5 years, 3 months ago

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You can't change the order of the numbers, and also you are only allowed to place the operations in the boxes only, not outside the box

Ong TH
- 2 years, 10 months ago

The solution is incorrect, if you consider the possibilities normally, then 1. all can be added, 2. divide 2 by 1 and then multiply by 3 and add 4, 3. multiply 1 2 and 3 and add 4 , and i can list all the possibilities but 2 would therefore be incorrect because there are more than 2 correct possibilities in which the order of operations without parenthesis can be used

Fumito Azama
- 5 years, 7 months ago

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Your 2nd case is not possible. You cannot divide 2 by 1 since the expression is $1\square 2.$

1+2+3+4=10 & 1 x 2 x 3+4=10

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1) 1+2+3+4 = 10 (2) 1 x 2 x 3 +4 =10

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a) 1x2x3+4=10 b)1+2+3+4=10

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simply 1+2+3+4 or 1X2X3+4

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- First look at the last operator:
- subtract gives 1?2?3=14 (10+4) and this has no solutions (maximum value is obtained by 1x2x3=6)
- multiplication gives 1?2?3 =2.5 (10/4) and this has no solutions
- divides gives 1?2?3 =40 and this has no solution (see first point) so the only possible answer must have plus as the last operator
- this reduces the problem to 1?2?3=6
- we can rule out divide and subtract since the maximum value is 6 (from above)
- this leads us to the two solutions of 1+2+3+4 and 1x2x3+4

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1 x 2x3 +4+5

Ahmed Elgamoudi
- 1 month ago

What about -1x2+3x4=10

Uday Mishra
- 3 years, 12 months ago

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Let $N = 1 \space \square \space 2 \space \square \space 3 \space \square \space 4$ .

We note that all $N$ involving $\div$ as an operator are fractions except $N = 1 \div 2 \times 3 \times 4 = 6 \ne 10$ which is not a solution, therefore, $\div$ is not an operator in the equation.

Now consider the nine cases of the first two operators.

$\quad \quad \begin{array} {rrrr} & 1 + 2 \space \square \space 3 & 1 - 2 \space \square \space 3 & 1 \times 2 \space \square \space 3 \\ 1 \space \square \space 2 + 3 & 3+3 & -1+3 & 2 + 3 \\ 1 \space \square \space 2 - 3 & 3-3 & -1-3 & 2 - 3 \\ 1 \space \square \space 2 \times 3 & 1+6 & 1-6 & 6 \end{array}$

$\quad \quad N = 6 + 4 = \begin{cases} 1+2+3+4 = 10 \\ 1 \times 2 \times 3 + 4 = 10 \end{cases}$

For $1 \space \square \space 2 \space \square \space 3 \space \square \space 4 \space \square \space 5 = 15$ , again there is no solution involving $\div$ as an operator. The necessary $N$ are:

$\begin{cases} N + 5 = 15 \Longrightarrow N = 10 \Longrightarrow \begin{cases} 1+2+3+4+5 = 15 \\ 1 \times 2 \times 3 + 4 + 5 = 15 \end{cases} \\ N - 5 = 15 \Longrightarrow N = 20 \Longrightarrow \color{#D61F06}{\text{No solution}} \\ 1 \square 2 \square 3 \square 4 \times 5 = 15 \Longrightarrow 1-2\times 3+4\times 5 = 15 \end{cases}$