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Replace a to f with the numbers 1 to 2 & 4 to 7, so that the sum inside each ring is the same. There is only one solution!

  • 2
    $\begingroup$ If there is more than one solution, your question will be flagged as too broad! $\endgroup$
    – A.D.
    Commented Jun 18, 2015 at 16:40
  • 1
    $\begingroup$ Here's a suggestion: Pick one of the solutions and think of an additional rule or pattern that they must follow. That would help narrow down the possibilities. $\endgroup$ Commented Jun 18, 2015 at 16:49
  • 3
    $\begingroup$ Or possibly ask for a generic solution which can be applied to find all possible solutions. $\endgroup$
    – Mark N
    Commented Jun 18, 2015 at 16:50
  • 1
    $\begingroup$ There were four equations to start with, and seven variables. You need to have at least one equation per variable in order to have a unique solution. The current version is 4 & 6, so there's still potentially multiple. $\endgroup$
    – Bobson
    Commented Jun 18, 2015 at 17:24
  • 2
    $\begingroup$ As far as I can tell, this is now a unique solution. I've voted to reopen. $\endgroup$
    – Bailey M
    Commented Jun 18, 2015 at 17:33

3 Answers 3


The answer is:

a=6; b=4; c=5; d=1; e=2; f=7


The sum is 10 in each circle


Blue circle equals purple, so

d+e=3 (they must equal 1 and 2)


  • e has a maximum value of 2
  • Only 4-7 remain, so b+c has a minimum value of 9 (4+5)
  • This constrains f to be 7 and e to be 2, leaving d to be 1.
  • The total per circle is then 10 and 6 is the only value left for a, so b is 4 and c is 5.

Giving us:

a=6, b=4, c=5, d=1, e=2, f=7


The unique answer is:

A, B,C, D,E, F--SUM
(6, 4, 5, 1, 2, 7) [10]


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