Numbers in circles

Algebra Level 2

4 circles overlap and divide one another into a total of 10 parts, as indicated by the 10 question marks in the diagram. Replace each question mark with a distinct digit 0 to 9, such that the sum of the digits in each circle is equal to the same value x x .

Find the maximum possible value of x x .

21 22 23 24 25 26 27 28

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Since in each circle the sum is the same:

i + h + g + ( e + f + d + j ) = a + ( e + f + d + j ) i+h+g+(e+f+d+j)=a+(e+f+d+j) , so a = i + h + g a=i+h+g .

( i + h + d + j ) + g + f + e = b + ( i + h + d + j ) (i+h+d+j)+g+f+e=b+(i+h+d+j) , so b = g + f + e b=g+f+e .

( h + g + f + d ) + e + j + i = c + ( h + g + f + d ) (h+g+f+d)+e+j+i=c+(h+g+f+d) , so c = e + j + i . c=e+j+i.

The sum of all the digits is 45 45 . From these:

45 = a + b + c + d + e + f + g + h + i + j = ( i + h + g ) + ( g + f + e ) + ( e + j + i ) + d + e + f + g + h + i + j = d + 2 ( h + f + j ) + 3 ( e + i + g ) \begin{aligned} 45 & = a+b+c+d+e+f+g+h+i+j \\ & = (i+h+g)+(g+f+e)+(e+j+i)+d+e+f+g+h+i+j \\ & =d+2(h+f+j)+3(e+i+g) \end{aligned}

The sum in the central cirlce: x = d + e + f + g + h + i + j x=d+e+f+g+h+i+j , so 45 = 2 x d + ( e + i + g ) 45=2x-d+(e+i+g) From that: x = 45 + d ( e + i + g ) 2 x=\dfrac{45+d-(e+i+g)}{2} Since d d is not bigger, than 9 9 and ( e + i + g ) (e+i+g) minimum 0 + 1 + 2 = 3 0+1+2=3 , so x 45 + 9 3 2 = 25 x\leq\left \lfloor \dfrac{45+9-3}{2} \right \rfloor=25 . So the maximum value of the sum is 25 \boxed{25} , which is possible:

{ a = 7 b = 5 c = 8 d = 9 e = 3 f = 2 g = 0 h = 6 i = 1 j = 4 \large \begin{cases} a=7 \\ b=5 \\ c=8 \\ d=9 \\ e=3 \\ f=2 \\ g=0 \\ h=6 \\ i=1 \\ j=4 \end{cases}

If we write A = a + b + c A = a+b+c , B = e + i + g B = e+i+g , C = f + h + j C = f+h+j , D = d D = d , then the key equations become A + 2 B + 3 C + 4 D = 4 x A + B + C + D = 45 A + x = 45 B + C + D = x \begin{aligned} A + 2B + 3C + 4D & = 4x \\ A + B + C + D & = 45 \\ A + x & = 45 \\ B + C + D & = x \end{aligned} From which we deduce that A = 45 x B = 45 2 x + D C = 3 x 45 2 D A = 45 - x \hspace{1cm} B = 45 - 2x + D \hspace{1cm} C = 3x - 45 - 2D Since B 3 B \ge 3 and D 9 D \le 9 we obtain x 25.5 x \le 25.5 , and then find the specific solution for x = 25 x=25 .

Maybe, grouping some of the variables makes what is going on a little clearer?

Mark Hennings - 3 years, 11 months ago

Correct answer should be 27, the middle region of maximal overlap in all circles is valued at 9, while the non overlapping regions of the red, green, blue circles are also 9, finally, all other values are three. This allows the values of three to add up to 18 for the middle yellow circle, which also has the 9 from that center region (maximum overlapping region), which makes 27. The other three circles are symmetric in their overlapping nature and thus each gets a total of 9 from the regions of value 3 and gets 18 from the two regions of 9 (the very center plus the non-overlapping region). Yielding 27 for each circle. Since the question asked for the maximum value within one circle, 27 would replace 25 as the correct answer.

To be as clear as possible: "a,b,c,d" = 9 while the rest equal 3

Amun Patel - 3 years, 11 months ago

Log in to reply

Sorry just realized it said distinct digit!

Amun Patel - 3 years, 11 months ago

Log in to reply

I made the same mistake. I started with A + 3B + C = A + 6B, worked to C = 3B, so set C to 9, B to 3, and A (non-dependent) to 9 as well. That was my mistake. I reused the 9.

Doug Fraser - 3 years, 11 months ago

(In the first equation, the 2nd parenthesis should end with j, not i.) Great solution to a very messy problem.

Robert Beggs - 3 years, 11 months ago

Log in to reply

I will correct it, thanks!

Áron Bán-Szabó - 3 years, 11 months ago

ummm.. I think the actual answer is 28: Using your labelling system: a = 0 b = 1 c = 2 d = 9 e = 5 f = 6 g = 4 h = 7 i = 3 j = 8

Meaning x = 28.

Avish Kumar - 3 years, 11 months ago

Log in to reply

The example is not correct... In the center cirlce (colore yellow) the sum of the numbers equals 5 + 8 + 9 + 6 + 3 + 7 + 4 = 42 5+8+9+6+3+7+4=42 which is much bigger, than 28.

Áron Bán-Szabó - 3 years, 11 months ago

I got yellow circle 29.

MaryAnn Bruton - 3 years, 11 months ago

Log in to reply

Can you show us how it's done? I doubt it's possible (given what Mark and the author has already provided)

Pi Han Goh - 3 years, 11 months ago

But what if j = d = f = h = 7 j = d = f = h = 7 , e = i = g = 1 e = i = g = 1 and a = b = c = 9 a = b = c = 9 . Wouldn't x = 31 x = 31 be a valid solution bigger than 25 25 ? EDIT: Ok, now I see, they should be distinct...

Queijo Silva - 3 years, 9 months ago

By trial and error I found a different solution: a=7, b=9, c=4, d=8, e=1, f=6, g=2, h=5, i=0, j=3

I was worried that this wouldn't fit your equation for x because although (e+1+g)=3 as per your 'minimum', the central region (d) was only 8 instead of 9. However I then noticed your 'floor function' in the inequality and breathed a sigh of relief. My solution does give x=25.

I'm surprised that other comments are arguing for the minimum being 27 and 28 when (regardless of all the algebraic shenanigans) there clearly exists a solution for x=25.

Paul Cockburn - 2 years, 8 months ago

9+8+7 = 24 ??

Matt Ryan - 3 years, 11 months ago

Log in to reply

Oh nm haha

Matt Ryan - 3 years, 11 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...