They are each missing something

If you multiply all the equations together, you get $abc\times bcd \times cda \times dab = 15\times30\times10\times6 \\ a^3 b^3 c^3 d^3 = 27000 \\ abcd = 30$ When you divide each equation by this result you get the respective solutions $d = 2 \\ a = 1 \\ b = 3 \\ c = 5$

Adding these gives $a+b+c+d = \boxed{11}$

factor LS into 3 ints:

abc = 15 = (1,3,5) bcd = 30 = (nah!) cda = 10 = (1,2,5) dab = 6 = (1,2,3)

it's obvious that 1 appears thrice, and also a, so a=1

bc = 15 bcd = 30

d = 2

cda = 10 cd = 10 c = 5

dab = 6 db = 6 b = 3

a=1, b=3, c=5, d=2 The answer is 1+3+5+2 :D

Equation 4 : 6=1 2 3 so a, b , d is either of 1,2 or 3. divide eequation 2 by 3 so b/a=3 this gives b=3 , a=1 thus d=2 put values in equation 1 gives c=5. Requires quantities is 11.

Using the first equation

$a*b*c=15$

we know that a, b, & c must all be factors of 15, so 1, 3, 5, or 15.

Suppose the case where a = 1. If this is true:

$1*b*c =15$

$b*c=15$ .

Then using the 2nd equation

$b*c*d=30$ , substitute in $15$ for $b*c$

$15*d=30$

$d=2$

If $d=2$ , then in the 3rd equation:

$c*d*a=10$ ,

$c*2*a=10$

$c*a=5$

Since we assumed $a=1$ to start, $c=5$ .

Plugging into any equation with $b$ , we arrive at $b=3$ .

$\begin{cases} 1\times 3\times 5=15 \\ 3\times 5\times 2=30 \\ 5\times 2\times 1=10 \\ 2\times 1\times 3=6 \end{cases}$

$a+b+c+d = 1+2+3+5 = 11$

I - cxd = 30/b

II - cxdxa = 10 --> a = b/3

III - axbxc = 15 --> (b/3)xbxc = 15 --> c = 45/b^2

IV - dxaxb = 6 --> dx(b/3)xb = 6 --> d = 18/b^2

V - cxdxa = 10 --> (45/b^2)x(18/b^2)x(b/3) = 10 --> 3b^3 = 81 --> b = 3

a = b/3 = 1

c = 45/b^2 = 5

d = 18/b^2 = 2

a + b + c + d = 11

$a \times b \times d = 6$ and $a \times b \times c = 15$ , so $a \times b$ must be a factor of both 6 and 15.

• $a \times b = 3$
• $3 \times c = 15$ , so $c = 5$
• $3 \times d = 6$ , so $d = 2$
• $a \times 5 \times 2 = 10$ , so $a = 1$
• $b \times 5 \times 2 = 30$ , so $b = 3$

$1 + 3 + 5 + 2 = \boxed{11}$

15 = 1 x 3 x 5; 30 = 1 x 2 x 3 x 5; 10 = 1 x 2 x 5; 6 = 1 x 2 x 3;

so, 1+2+3+5 = 11

I tried a different way and it works. First, look at the last equation,

$dab=6$

Only 3 distinct natural numbers satisfy this equation: $3,2$ and $1$ . From this we know either $d,a$ or $b$ equals $3, 2$ or $1$ .

Next, look at the third equation and the last one.

Both of them have $d$ and $a$ . Since both of the equations have an even number on the RHS, we may conclude that $d \times a = 2$ . So, either $d$ or $a$ equals $2$ or $1$ .

$2 \times 1 = 2$ ---- $(I)$

Substitute eqn^ $(I)$ into the last equation.

$dab=6$

$2b=6$

$b=3$

Substitute eqn^ $(I)$ into the third equation.

$cda=10$

$2c=10$

$c=5$

Substitute $b=3$ and $c=5$ into the second equation.

$bcd=30$

$3 \times 5 \times d=30$

$15d=30$

$d=2$

Since $d=2$ , $a$ must equal $1$ .

$a+b+c+d=1+3+5+2= \boxed{11}$

I'm not sure how mathematically sound this is, but logically:

Considering the three equations with a,

$abc=15$

$acd=10$

$abd=6$

a is a common factor on all of the left hand sides of the equations, so the possible values of a are the common factors of the values on the right hand sides of the equations, since a, b, c, and d are all natural numbers. The only common factor of these three numbers (15, 10, and 6) is 1. Therefore, $a=1$ .

Considering the three equations with b,

$abc=15$

$abd=6$

$bcd=30$

Likewise, the value of b must be a common factor of the numbers on the right hand sides of the equations. The only such common factor is 3. Therefore, $b=3$ .

We know that abc=15, and since a=1 and b=3:

$abc=15$

$1*3*c=15$

$3c=15$

$c=5$

And finally:

$abd=6$

$1*3*d=6$

$3d=6$

$d=2$

So the sum of a, b, c, and d:

$a+b+c+d=1+3+5+2=11$