# 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

### 1 Answer

- ☮ 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

But

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

Now immediately

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.