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

There are more than one way of doing these puzzles and may well be more than one answer. Please let me and others know what alternatives you find by commenting below. We also welcome general comments on the subject and any feedback you’d like to give. If you have a question that need a response from me or you would like to contact me privately, please use the contact form.

## Get more puzzles!

**If you’ve enjoyed doing the puzzles, consider ordering the book; 150+ of the best puzzles in a handy pocket sized format. Click here for full details.**

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.

Very well presented. Every quote was awesome and thanks for sharing the content. Keep sharing and keep motivating others.

No matter if some one searches for his essential thing,

so he/she desires to be available that in detail, so that thing

is maintained over here.

Segera temukan solusi yang Anda butuhkan bersama obat Markastoto, solusi antiretroviral terkemuka untuk HIV.

Dikembangkan dengan teknologi terbaru dan formulasi ilmiah yang canggih, obat ini tidak cuma mengimbuhkan perlindungan optimal terhadap virus

HIV, tetapi juga tingkatkan mutu hidup Anda.

Jangan biarkan HIV mengendalikan hidup Anda.

Pilih obat Markastoto untuk melangkah maju menuju kebugaran dan kebahagiaan yang berkelanjutan. Percayakan perjalanan Anda terhadap keunggulan, percayakan terhadap obat

Markastoto.

Keunggulan Markastoto dalam Menangani HIV

Dalam dunia perawatan HIV, Markastoto sudah perlihatkan dirinya sebagai obat antiretroviral terkemuka yang tawarkan solusi

efektif. Dalam bagian ini, kita akan menjelajahi keistimewaan Markastoto yang membedakannya dari product sejenis.

Dengan formula inovatifnya, Markastoto tidak

cuma menghambat perkembangan virus HIV namun termasuk

menaikkan energi tahan tubuh. Dapatkan pertolongan maksimal dan mutu hidup yang lebih baik bersama keajaiban terapi Markastoto.

Pengalaman Pengguna yang Mencerahkan

Saksikan testimoni pengguna yang menginspirasi dan memberi tambahan gambaran nyata mengenai

bagaimana Markastoto udah mengubah hidup mereka.

Dari kisah kesembuhan sampai kualitas hidup yang lebih

tinggi, pengalaman nyata ini membuktikan efektivitas dan keamanan produk.

Temukan harapan dan keberanian lewat cerita-cerita privat ini,

mengilhami Anda untuk menyita langkah positif dalam mengatasi HIV bersama dengan Markastoto.

Mengapa Markastoto Pilihan Utama?

Dalam anggota ini, kami akan menjelajahi alasan mengapa Markastoto menjadi pilihan utama

dalam perawatan HIV. Dengan penekanan pada keamanan, efektivitas,

dan kenyamanan pengguna, Markastoto tidak hanya tawarkan obat, tapi termasuk kesejahteraan holistik.

Mari temukan bagaimana Markastoto tidak hanya mengobati gejala,

tetapi termasuk memberi tambahan kehidupan yang bermakna dan penuh harapan.

[url=http://zmkshop.ru/]ооо тулачермет сталь официальный сайт[/url]

I have read a few good stuff here. Certainly value bookmarking for revisiting.

I surprise how much attempt you place to create this sort of wonderful informative website.

Wonderful, what a web site it is! This weblog provides helpful facts to us, keep it up.

You are so awesome! I do not suppose I have read through something like

this before. So great to find another person with unique thoughts on this subject matter.

Seriously.. thank you for starting this up. This website is one thing that

is required on the internet, someone with some originality!

Feel free to visit my web site :: gaydating4singles.net

This is a good tip particularly to those fresh to the blogosphere.

Simple but very accurate information… Many thanks for sharing this one.

A must read post!

Also visit my blog :: japanesegirlcams.info

Hello colleagues, its impressive post concerning tutoringand fully explained, keep it up all the

time.

Feel free to visit my web blog; videomaturetube.com

I don’t know if it’s just me or if everybody else encountering problems

with your blog. It seems like some of the text in your

content are running off the screen. Can someone else please provide feedback

and let me know if this is happening to them as well? This could

be a issue with my internet browser because I’ve had this happen previously.

Kudos

Look at my webpage dashausb.de

Magnificent goods from you, man. I have understand your stuff previous to and you are just extremely fantastic.

I actually like what you have acquired here, certainly like what you are stating and the way in which you say it.

You make it enjoyable and you still care for to keep

it wise. I can not wait to read much more from you. This is actually

a terrific website.

my web page – sexyasscougars.info

Hi there, I enjoy reading through your article. I wanted to write a little comment to

support you.

Here is my web-site :: inc-diary.com

Just desire to say your article is as amazing.

The clarity in your post is just great and i coukd assume you’re an expert on this subject.

Well with your permission let me to grab your feed to keep up

to date with forthcoming post. Thanks a milllion and please continue the

gratifying work.

Hey there! Quick question that’s totally off topic. Do you know how to make your

site mobile friendly? My blog looks weird when browsing from my

iphone 4. I’m trying to find a theme or plugin that might be able to resolve this

issue. If you have any suggestions, please share.

Thank you!

My page – gay-3d-comix.com

This is very interesting, You are a very skilled blogger.

I have joined your rss feed and look forward to seeking more of your magnificent post.

Also, I have shared your website in my social networks!

Feel free to visit my page – bestpornoclips.info

Having read this I thought it was really enlightening.

I appreciate you taking the time and energy to put this informative

article together. I once again find myself personally spending way too much time both reading and posting comments.

But so what, it was still worthwhile!

Hi there! Someone in my Myspace group shared this site with

us so I came to check it out. I’m definitely enjoying the information. I’m bookmarking and will be tweeting this to my followers!

Great blog and superb style and design.

Hello! Thiis is my first visit to your blog!

We are a collectin of volunteers and starting a new prroject iin a community in tthe same

niche. Yoour blog provided us useful information to work on. Youu have done a marvellous job!

Thanks for great info. What trips can you recommend in 2024? Astro tourism, eco diving, home swapping, train stations are the new food destinations,sports tourism, coolcationing, gig tripping, private group travel?