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 .
Find the maximum possible value of x .
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If we write A = a + b + c , B = e + i + g , C = f + h + j , D = d , then the key equations become A + 2 B + 3 C + 4 D A + B + C + D A + x B + C + D = 4 x = 4 5 = 4 5 = x From which we deduce that A = 4 5 − x B = 4 5 − 2 x + D C = 3 x − 4 5 − 2 D Since B ≥ 3 and D ≤ 9 we obtain x ≤ 2 5 . 5 , and then find the specific solution for x = 2 5 .
Maybe, grouping some of the variables makes what is going on a little clearer?
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
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Sorry just realized it said distinct digit!
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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.
(In the first equation, the 2nd parenthesis should end with j, not i.) Great solution to a very messy problem.
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.
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The example is not correct... In the center cirlce (colore yellow) the sum of the numbers equals 5 + 8 + 9 + 6 + 3 + 7 + 4 = 4 2 which is much bigger, than 28.
I got yellow circle 29.
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Can you show us how it's done? I doubt it's possible (given what Mark and the author has already provided)
But what if j = d = f = h = 7 , e = i = g = 1 and a = b = c = 9 . Wouldn't x = 3 1 be a valid solution bigger than 2 5 ? EDIT: Ok, now I see, they should be distinct...
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.
9+8+7 = 24 ??
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Since in each circle the sum is the same:
i + h + g + ( e + f + d + j ) = a + ( e + f + d + j ) , so a = i + h + g .
( i + h + d + j ) + g + f + e = b + ( i + h + d + j ) , so b = g + f + e .
( h + g + f + d ) + e + j + i = c + ( h + g + f + d ) , so c = e + j + i .
The sum of all the digits is 4 5 . From these:
4 5 = 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 )
The sum in the central cirlce: x = d + e + f + g + h + i + j , so 4 5 = 2 x − d + ( e + i + g ) From that: x = 2 4 5 + d − ( e + i + g ) Since d is not bigger, than 9 and ( e + i + g ) minimum 0 + 1 + 2 = 3 , so x ≤ ⌊ 2 4 5 + 9 − 3 ⌋ = 2 5 . So the maximum value of the sum is 2 5 , which is possible:
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ a = 7 b = 5 c = 8 d = 9 e = 3 f = 2 g = 0 h = 6 i = 1 j = 4