mensa calendar arithmetic puzzle (or, I'm not as smart as I think I am)

Factual Questions
1

So I have this Mensa Puzzle Calendar. Usually I can figure out the puzzles. Sometimes for the word puzzles I cheat, but I’ve never had a problem with the math puzzles. Until the one for Tuesday 6 October 2009. The only way I could solve it was by trail and error. I was thinking that there should be a smarter way using algebra, but I’m stumped. Does anyone else have an elegant solution?

Here is the puzzle (my diagram should appear in a monospace font in your browser window, if not, sorry, it will probably be impossible to read)



            o
           / \
Side      /   \    Side
 A       o-----o    B
        / \   / \
       /   \ /   \
      o-----o-----o
         Side C

Place the numbers 1-6 in the circles so that the following rules are obeyed:
[ul]
[li]The sum of the 3 inner circles = 1 more than the sum of the 3 circles on side A[/li][li]The sum of the 3 inner circles = 2 more than the sum of the 3 circles on side B[/li][li]The sum of the 3 circles on side C + the sum of the circles at the points of the big triangle = 1 more than the sum of the 3 inner circles.[/li][/ul]

2

I can’t come up with anything that doesn’t use any trial-and-error off the top of my head, but we can reduce it to two degrees of freedom: the three rules give us three linear equations that must be satisfied. The fact that we have numbers 1-6 means their sum is 21. So we have four linear equations. We can set up a matrix equation, solve it and find its null space, and work trial-and-error with that.

Unless someone else can tease out a couple more linear equations, I’m not sure that we’ll find an elegant solution to this one.

3

You can find some boundaries to the null space. The sum of side A has to be less than 15 since the max of the inner circle is 15. Similarly side B has to be less than 14.

4

4
56
321

5

Looks like a relatively simple algebra problem to me.

Label the 6 circles:
a
b c
d e f

Then your three rules turn into three equations:

  1. b+c+e = 1+ a+b+d
  2. b+c+e = 2 + a+c+f
  3. d+e+f + a+d+f = 1 + b+c+e

Now it’s just a matter of simplifying, substituting and solving for each letter. This part is messy and will probably take some time, but I it will get you an answer without any trial and error.

6

3 equations + 6 variables = no simple algebraic solution

7

Let T be the sum of the 3 inner circles. Let A, B, and C be the sum of the three circles on each side.

Translating the first two premises into equations is easy:

A+1 = T -> A = T-1
B+2 = T -> B = T-2

For the third premise, note that the sum of the circles at the points of the big triangle is equal to the sum of all circles in the diagram minus T. Since the circles are populated 1 thru 6, the sum of the circles at the points of the big triangle = 21-T. So, the third premise is:

C+(21-T) = T+1 -> C = 2T-20

Finally, note that the sum A+B+C+T counts every circle twice (e.g., the sum for A and the sum for B each count the value in the topmost circle twice, etc.). So A+B+C+T=42. Plugging in the values for A, B, and C from the premises about:

(T-1) + (T-2) + (2T-20) +T = 42 -> 5T - 23 = 42 -> T = 13.

So T must be composed of 6,5,2 or 6,4,3. Various other sums of digits in the figure are easy to calculate:

A=12
B=11
C=2T-20, C=6, so 1,2,3 must be on the bottom row.
Sum of the circles at the points of the big triangle=(1/2)[A+B+C-T] = 8

From all this information we can conclude that the middle digit of C must be 2 or 3. This choice also forces the topmost circle (shared by A and B) to be either 5 or 4. Once these circles are fixed, the remainder are easily sorted out:

(1) If the middle circle of C is 3, the following is the only solution:

5
64
132

(2) If the middle circle of C is 2, the following is the only solution:

4
56
321

Not surprisingly–given the inherent symmetry of the original requirements and the fact that this is a mensa problem–there are two possible solutions.

8

dracoi, I started exactly as you say (using the same variable names as you used)
a - c + d - e = -1
a - b - e + f = -2
a - b - c + 2d + 2f = 1
a <= 6
b <= 6
c <= 6
d <= 6
e <= 6
f <= 6

But after that, I couldn’t remember enough linear algebra to solve it, and as The Hamster King says, if you just use the first three equations, there is no algebraic solution. So I was stuck and resorted to trial and error.

By the way, pan1, you have a correct solution different than the one that the calendar gives (and the one I found). I guess there is more than on correct solution.

The solution on the calendar is



      5
     / \
    /   \
   6-----4
  / \   / \
 /   \ /   \
1-----3-----2
9

Thanks CJJ! Much shorter than the method I used to come up with my answer. I will ponder your solution later today.

10

here is a solution that I just solved.

let A = the sum of the circles on side a
let B = the sum of the circles on side b
let C = the sum of the circles on side c
let I = the sum of the inner circles
let T = the sum of the circle of the large triangle

From the circumstances given, we have 3 equations:
I=A+1, or A=I-1
I=B+2, or B=I-2
C+T=I+1, or C=I+1-T

Dr. Love reminds us that the sum of all 6 circle must be 21, so…
I+T=21 (the 3 inner circles plus the 3 outer circles)

But we need to find a fifth equation to help us solve for our 5 equations.
Notice that if we add A+B+C, we count all of the inner circles once, and all the outer circles twice, so…
A+B+C=I+2T

Now, substitute our solved values for A, B and C
I-1+I-2+I+1-T=I+2T
simplify and solve for T…
T=2I/3 - 2/3

Now back to Dr. Love’s equation…
I + 2I/3 - 2/3 = 21…
I = 13

Now, its easy to find all the other values by substitution…
A=12
B=11
C=6
T=8

from here, we can assign each circle their own variables, use the circumstances given to redfine our original equations, and solve for the new variables. we have more than nough equations to do so, and its just a matter of substitution to arrive at the solution.

11

too late to edit…

I admit, my last paragraph isn’t accurate, and CJJ has the right solution. I saw that I could logically determine that the bottom center circle must be either 2 or 3, but I sruggled to find an algebraic way to state that. My brain is fried after wasting over an hour on solving this…

Well done, CJJ

12

I didn’t use algebra or trial and error.

I realized that given the criteria, the 3 numbers in the middle were the largests of the defined sets, the numbers on the bottom had to be very small (had to be 1,2,3) and the numbers on the outside slightly bigger.

Putting the 3 largest numbers on the inside would make it impossible to come within 1 or 2. and since one of the inside numbers is in the bottom 3, that reinforced that theory.

Knowing that, the solution seemed obvious. But then again I missed one possible answer too.