Problems with Multiplicative Magic Square?
1. In a multiplicative magic square of order 3 (see illustration below), the product of the three entries in each line (row, column, diagonal) are equal to a magic constant Lamda. If all the entries in a magic square of order 3 are positive integers, what are all the possible values for the magic constant lamda that are in the interval [1000, 2008]?
a b c
d e f
g h i
2. The figure below is a multiplicative magic square. If all the entries are positive integers, what is the sum of all the possible values of g?
50| b | c
.d | e | f
.g | h | 2
- ☮ VašekLv 51 decade agoFavorite Answer
1) abc = def = ghi = adg = beh = cfi = aei = ceg = λ
Take all products containing e,
def = beh = aei = ceg = λ
df = bh = ai = cg = λ/e
acgi = abc * ghi / (beh) * e = λe
acgi = ai * cg = λ^2/e^2
λe = λ^2/e^2
e^3 = λ
Thus λ must be a cube of a positive integer number. There are only three such numbers in [1000, 2008]:
10^3 = 1000
11^3 = 1331
12^3 = 1728.
2) Using the previous result,
bh = cg = ai = 2*50 = 100
abcghi = abc * ghi = λ^2
abcghi = ai * bh * cg = 100^3 = 10^6
=> λ = 10^3 = 1000
e = 10
c = 100/g
d = 20/g
h = 500/g
so g must be a divisor of gcd(500,100,20) = 20 for all of these to be integer. Let's continue to find whether there are further constraints,
f = 100/d = 5*g
b = 100/h = g/5
so g must also be a multiple of 5.
Therefore, the only possible values of g are 5, 10 and 20. Their sum is 35.