Place the numbers 1 through 6 in the circles so that
the following rules are obeyed:
(1) The sum of the 3 inner circles = 1 less than
the sum of the 3 circles on side A.
(2) The sum of the 3 circles on side B = 3 less than
the sum of the 3 circles on side C.
(3) The number in the top circle = 3 more than
the number in the lower right circle.
Solution to the Problem:
Let x = number in the lower right circle.
Then x + 3 = number in the top circle (Clue #3).
Let w = number in the middle circle on side B.
Let y = number in the middle circle on side C.
Let v = number in the lower left circle.
Let z = number in the middle circle on side A.
Then from clue #1, w + y + z = v + z + (x + 3) - 1
From clue #2, (x + 3) + x + w = x + y + v - 3
Combining these two equations, we get:
y = x + 4 and w = v - 2.
So, x must equal 1 or 2.
Then x + 3 must equal 4 or 5.
So, y must equal 5 or 6.
Now, try x = 1.
Then x + 3 = 4 and y = 5.
But w cannot equal 1 because x already equals 1.
If w = 2, then v must equal 4 but x + 3 already equals 4.
If w = 3, then v must equal 5 but y already equals 5.
And w cannot equal 4 because x + 3 already equals 4.
And w cannot equal 5 because y already equals 5.
And w cannot equal 6 because v would have to equal 8.
So, x cannot equal 1. Therefore, x must equal 2.
So, x + 3 = 5. y = 6.
Then it is easy to show that w = 1, v = 3, and z = 4.
Correctly solved by:
| 1. James Alarie | Flint, Michigan |
| 2. Chad Fore |
Gate City High School, Gate City, Virginia |
| 3. Alex Roux | France |
| 4. David & Judy Dixon | Bennettsville, South Carolina |