The number within the four sectors of the outer circle is equal to the sum of the three numbers next to its sector. The numbers in the individual circles can only be 1 to 9 and each number can be used only once. One number has been provided to get you started. Find the remaining four numbers.

### Solution

### Solution explanation

Comments on solving the puzzle; a process of eliminating possible values as answers:

Value 3 is provided and can not be reused.

Starting at Equation (1), unknowns A & B can only be equal to 1, 2, 5 or 6 as shown. All other combination include a 3 value.

Equation (2): unknown B can now be equal to 1, 2 or 5 only as shown Then C can only be equal to 4, 7, or 8. Comparing with Equation (1), 6 is eliminated as a possible value for B, A can now only be equal to 2, 5 or 6.

Equation (3): unknowns C & D can only be equal 8 or 9 as shown. Therefore comparing with Equation (2), C must be equal to 8 & D equal to 9. With C equal to 8 in Equation (2), B must then be equal to 1 and with B equal to 1 in Equation (1), A must then be equal to 6.

Equation (4) confirms that if D is equal to 9 and A is equal to 6.

Now insert your final answers into the original puzzle and verify the resulting sums.

This article may be of interest to you! https://www.theguardian.com/science/2017/may/22/can-you-solve-it-the-maths-problem-for-5-year-olds-stumping-the-web

## Feedback

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also clockwise : 3,4,5,12 ?

Se puede usar sólo del 1 al 9

del 1 al 9

1 al 9

Top: 6

Left: 9

Right: 1

Bottom: 8

correcto

1 al 9

3, 4, 1.6 and 1,3

(18-10)/2

(18-12)/2

(18-2)/10

(18-2)/12

The original math problem does not have 20 in it, it has the number 2, nor does have any limitation for the criteria in the individual circles, but there you go, they are 1 to 9, and you stated nowhere they can not be decimal numbers.

I think the ‘hoax’ is simply a teacher having a typo and not checking their work.

For the solution, a less technical explanation is simply to look at the 20 quandrant. Numbers have to be 8 and 9. The 9 cant be to the right because 9+3+X>12. Once you have these two, the other two are easy.

I found an easier solution which considers the constraint that the numbers be natural numbers from 1 to 9. I was surprised that no one got that.

Consider the numbers as N,E,W,S representing the directions.

Set 1:

20=W+S+3

12=E+S+3

Set 2:

18=W+N+3

10=E+N+3

Consider wither of the 2 sets, we get:

W-E=8

Now only possible solution considering the above constraint is:

W=9,E=1

Substituting back in set 1 and 2, we get N=6, S=8.

There is yet another reasonable solution. I believe much less elegant than the correct one, but it has the advantage of being invariant under the change of 20 with 2 (that is in fact the version of the problem I first happened to encounter, and tried to solve). Just apply “casting out nines” to the sum of forward and backward quadrants of each circle and you’ll get (clockwise, starting on top): 1-4-5-2. The number 3 might seem an exception, but you can either obtain it by summing up the SW-NE quadrants (20+10=30 => 3, which would be the same if it was 2+10) or the SE-NW quadrants (18+12=>3), depending on how you rotate the picture to pick the “backward-forward” couple. In this way, the sum of all numbers in vertical or horizontal circles will be (guess what?) 9 and, nicely enough, circles are filled by all numbers from 1 to 5.

Sorry, I realize only now there were explicit rules. I only saw the version without any explanation other than a vague “fill the empty circles”.

Clockwise from the top: 6,1,8,9. I did not peek at the solution first, but I’m headed there next. I was referred here by a news story which featured a variant version which had a 2 in place of the 20, the solution of which did not involve actual math, although it cleverly seemed to present itself as a math problem.

I was interested in how other people solved it… here is my way a bit of deduction and luck.

I labeled all the empty circles as N,E,S,W.

Then I made a simple equation

3+N+E=10

3+E+S=12

3+S+W=20

3+W+N=18

Now I realized that north, west and south had to be pretty large

N+E=7

E+S=9

S+W=17

W+N=15

So even though I did not follow the instructions I solved for E to equal 0 after all we are only playing with additions

So I got

N=7

E=0

S=9

W=8

Afterwards realizing the diagram works with E=0, I changed it to E=1 and had to subtract from N and S and add 1 to W.

And that’s the way I solved it… a bit of luck…

The original picture has 2 as 20. Why was it changed?

The Grade One problem has introduced me to Gordon Burgin’s work. I will introduce it to my kids. Thank you.

Top: 6

Down: 8

Left: 9

Right:1

Top: 1

Down: 3

Left: 14

Right: 6

2. Solution Possibility! (found by my 11y old brother 😀 )

Do you have any other petite circle sum puzzles?

Solve the linear system of 4 equations with 4 unknowns:

a+b=7,a+d=15,b+c=9,c+d=17

(To save time, you can feed into https://www.symbolab.com/solver/system-of-equations-calculator)

The general solution is:

d=17-c, b=9-c, a=c-2

Now, a,b,c,d must all be between 1-9, and also d=17-c implies c=8 or 9. Testing both candidate c values reveals the only solution can be:

a=6,b=1,c=8,d=9

At first the formal description of the task.

Given:

The number in the middle: m=3

The sumnumber of each quandrant: q11=18, q12=10, q21=20, q22=12

Wantet:

Parts of the sum equations: x,y,z,w

Equations:

x+w=q11-m

x+y=q21-m

y+z=q22-m

z+w=q12-m

Four linear equations with four unknowns (x,y,z,w). I will transform into a matrix equation.

|1 0 0 1| |x| |q11-m|

|1 1 0 0|*|y|=|q21-m|

|0 0 1 1| |z| |q22-m|

|0 1 0 1| |w| |q12-m|

M*S=Q

This system of equations has only an unuiqe solution if det(M)!=0. We will check this.

det(M) = det(M1) – det(M2) = 2

|1 0 0| |0 0 1|

M1=|0 1 1| M2=|0 1 1| det(M1)=1*(1*1-0*1)+1*(0*1-0*1)=1 det(M2)=1*(0*1-1*1)=-1

|1 0 1| |1 0 1|

Thiese equations should have an unique solution. We will calculate it now with this example.

|1 0 0 1| |x| |15|

|1 1 0 0|*|y|=|7 |

|0 0 1 1| |z| |17|

|0 1 0 1| |w| |9 |

det(Mx)=

|15 0 0 1|

| 7 1 0 0|=13 det(Mx)=13 det(My)=1 det(Mz)=17 det(Mw)=17

|17 0 1 1|

| 9 1 0 1|

x=det(Mx)/det(M)=6,5

y=det(My)/det(M)=0,5

z=det(Mz)/det(M)=8,5

w=det(Mw)/det(M)=8,5

Checking the solution:

6,5+8,5+3=18 O.K.

6,5+0,5+3=10 O.K.

8,5+8,5+3=20 O.K.

0,5+8,5+3=12 O.K.

In the case q21=2:

x=det(Mx)/det(M)=6,5

y=det(My)/det(M)=0,5

z=det(Mz)/det(M)=-9,5

w=det(Mw)/det(M)=8,5

Checking the solution:

6,5+8,5+3=18 O.K.

6,5+0,5+3=10 O.K.

8,5+-9,5+3=2 O.K.

0,5+8,5+3=12 O.K.

This task has no Solutions with natural numbers. The set of rational numbers is necessary.

Oh dear, Cramer’s rule is not the easiest approach. Y

ou made an error in calculating det(M). Actually det(m)=0, so there are an infinite number of solutions. Limiting the answers to unique integers from 1 to 9 reduces the number of answers. In this case, there is only one answer, as shown elegantly by previous contributors.

Tambien funcionaria usar como las agujas del reloj (7, 0, 9, 8)

sencillo en cada circulo va de la siguiente manera (inicias con el 5, 2, 7, 10), el inicio es de acuerdo al sentido de las manecillas del reloj

EL EJERCICIO ES PARA NIÑOS DE 1° AÑO BASICO, LUEGO TODO ESE “ESTUDIO” NO ES POSIBLE PARA CHICOS DE 6 O 7 AÑOS. A LO SUMO SABEN SUMAR. POR LO QUE EL RESULTADO PODRIA SER:

31

23 25

17

S´lo se puede usar los números del 1 al 9, no se puede usar 10, 12, 15 etc

5 3 4 6

You all are making this way complicated. I simply subtracted the 3 from each quadrant to take it out of the way so you are only looking at 2 numbers for the sum of each. Then I looked at the 2 numbers that could make up the sum for 7. Can’t be 3 and 4 because you can’t reuse the 3. So that leaves 2 and 5 or 1 and 6. If you put any number but the 6 in the circle adjoining the 15 quadrant, the other number for the 15 sum would be more than 9. So therefore the 7 sum must be 6 and 1 with the 6 at the top and the 1 on the right. Therefore, a 9 must be on the left and an 8 on the bottom for the rest to work.

This is why it is a first grade problem. They are teaching them to work with numbers flexibly, not to do algebra. I used to teach elementary math.

That’s interesting – you are using logic to solve the puzzle rather than algebra. I guess this can be applied to the circle-sums puzzles? I kind of liked the algebraic way as it brushed up my algebra a little!

Same logic, simpler: left + bottom = 17, only combination is 9 and 8.

Right + bottom = 9, meaning bottom must be less than 9, therefore left = 9, bottom = 8, right = 1

Use these to get top = 6.

Of course, assuming the given rules and numbers are correct and not the original typo version.

Good logic and correct solution. This solution satisfies the original (& correct) version of the puzzle. A modified ‘typo’ version was distributed by others at a later date with no valid rules or solution(s).

Using ONLY subtraction and addition: the circles contain the differences between the adjacent wedges, ie, N=8, E=2, S=10 and W=16. Intrinsic check using addition, vertical and horizontal sum of circles are equal.(8+3+10=21=16+3+2). This implicit check is unlikely to be coincidencidental. Nothing fancy and pure 1st grade math: subtraction and addition.

Re-posting with formatting:

There is a much better answer to this that has yet to be broached by any party.

If we label the missing bubbles; starting from 9 on a clock; we get:

. . . . B

. .18 . . . 10

A . . . 3 . . . C

. .20 . . . 12

. . . . D

And:

If instead of focusing on individual solutions, and solve for a general solution, which can be done by setting A,B,C or D to be a generic number n, we get:

. . . . . . . (n)

. . . . . 18 . . . 10

(15 – n) . . . 3 . . . (7 – n)

. . . . . 20 . . . 12

. . . . . . (n + 2)

for all n from (-oo,oo) where n cannot be 1,3,4, or 12.

Hence, recognizing an infinite set of possible solutions.

Thank you Elliott for your input. I’m very happy that you have responded! Can you check this? I agree with your number B (top) = to n & your number C (Right) = to (7-n) & your number A (left) = to (15-n). However, since the values for A, B, C & D can only be positive whole numbers from 1 to 9 with no duplicates, then for A, n can only be = to 6, 7, 8 or 9. For C, n can only be = to 1, 2, 3, 4, 5 or 6. Therefore, to satisfy B & D, then n (A) can only be = to 6 and not an infinite set of possible solutions. Gordon

As this solution is general and works for all real numbers, this will yield *every* possible solution to the puzzle. This problem however, states that it is only interested in having the resulting values be between [1,9].

By setting

.

15-n=9

n=6

.

for our lower boundary

and

.

7-n=1

n=6

.

for our upper boundary, we find that the interval for n in which all values of A,B,C,D are between the interval of [1,9] is [6,6] or, simply, n=6.

I believe that Manfred was “overthinking” this problem. Larry M

its just the number of numbers with 2 numbers in it. 3 doublenumbered numbers are touching the middle circle. ezpz stupid math geniuses, greetings from germany.

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