+----+----+----+----+
| 23 | | | |
+----+----+----+----+
| | | 64 | |
+----+----+----+----+
| | N | | |
+----+----+----+----+
| | | |101 |
+----+----+----+----+
In the 4x4 grid provided above:
- Each the 16 values appearing in the 16 cells is a positive integer.
- The 4 values corresponding to each of the 4 rows are in arithmetic sequence.
- The 4 values corresponding to each of the 4 columns are in arithmetic sequence.
Determine the total number of
distinct positive integer values that N can assume.
Analytic Solution:
Without additional constraints, there is an infinite number of possible values for N, but N is given by the equation:
N = 135 - 5a,
where 'a' is the common difference for the top row.
There are 8 common differences, one for each row and column. Call them a,b,c,d for the rows and w,x,y,z for the columns.
Now label the 16 positions in the grid E through T:
wxyz
EFGH a
IJKL b
MNOP c
QRST d
+w +x +y +z
+----+----+----+----+
|E=23| F | G | H | +a
+----+----+----+----+
| I | J |K=64| L | +b
+----+----+----+----+
| M | N | O | P | +c
+----+----+----+----+
| Q | R | S |101 | +d
+----+----+----+----+
We are given that E=23, K=64, and T=101.
Wow, look at that, N is in exactly the right place!
E=23
F=23+a
G=23+2a
H=23+3a
y=K-G = 64 - (23+2a)
y = 41-2a
O=K+y = 64 + (41-2a)
O=105-2a
z=(101 - H)/3 = (101 - (23+3a))/3
z=26-a
P=101-z = 101 - (26-a)
P=75+a
c=P-O = (75+a) - (105-2a)
c=3a - 30
N=O-c = 105-2a - (3a - 30)
N = 135 - 5aI started with a spreadsheet solution and and then worked out the above method.
One symmetric solution is a=b=c=d=15 and w=x=y=z=11; in this case, N = 60 = 135 - 5*15
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Posted by Larry
on 2023-02-12 08:07:54 |
Solution
