1 and 2 and 3 and 4

Logic Level 2

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

There are 4 3 = 64 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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11 solutions

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

  • We note that all N N involving ÷ \div as an operator are fractions except N = 1 ÷ 2 × 3 × 4 = 6 10 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.

1 + 2 3 1 2 3 1 × 2 3 1 2 + 3 3 + 3 1 + 3 2 + 3 1 2 3 3 3 1 3 2 3 1 2 × 3 1 + 6 1 6 6 \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}

  • From the above nine cases, we see there are only 2 \boxed{2} solutions, when

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

For 1 2 3 4 5 = 15 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 N are:

{ N + 5 = 15 N = 10 { 1 + 2 + 3 + 4 + 5 = 15 1 × 2 × 3 + 4 + 5 = 15 N 5 = 15 N = 20 No solution 1 2 3 4 × 5 = 15 1 2 × 3 + 4 × 5 = 15 \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}

Moderator note:

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?

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

The second equation should be 1 × 2 × 3 + 4 = 10 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 × 2 3 × 4 = 10 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 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:

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

Chew-Seong Cheong - 5 years, 6 months ago

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Using bodmas, three solutions

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 ) (-2) and I guess 1 × 2 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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That will require you to indicate that the equations are in different bases. IE 1 2 + 3 + 4 = 1 0 6 1 - 2 + 3 + 4 = 10 _ 6 .

Calvin Lin Staff - 3 years, 1 month ago

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

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Yes, it would

Rory Follin - 2 years, 7 months ago

1x2-3x4=10 | |
\/ \/ 2 - 12 = -10

Kevalin Ketcham - 2 years, 10 months ago

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Absolute signs are not available.

Chung Kevin - 2 years, 10 months ago

How is this table organized?

Calvin Lin Staff - 5 years, 11 months ago

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?

Calvin Lin Staff - 2 years, 11 months ago

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

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it does indeed, i agree

Nik Gibson - 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 2. 1\square 2.

Eli Ross Staff - 5 years, 6 months ago
Saddam Hossain
Dec 5, 2015

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

Sadasiva Panicker
Nov 30, 2015

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

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

Hasan Feroz
Dec 28, 2016

1+2+3+4=10 1x2x3+4

Rajorshi Koyal
Dec 17, 2015

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

Ervyn Manuyag
Nov 29, 2018

1x2x3+4=1+2+3+4

Bryan House
Sep 6, 2018
  • 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
Gia Hoàng Phạm
Aug 16, 2018

In 1 2 3 4 = 10 1\square 2\square 3\square 4=10 ,we have 1 + 2 + 3 + 4 = 10 1+2+3+4=10 or 1 × 2 × 3 + 4 = 10 1 \times 2 \times 3+4=10 which have 2 \boxed{\large{2}} ways

Ferdy Harahap
Apr 9, 2018

(1 x 2) x 3 + 4 = 10

Divya Mg
Dec 16, 2015

Ans. 1x2x3+4=10

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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There is no box in front of the 1

Brian Mackenzie - 3 years, 3 months ago

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