Cube Face Sum

Let the integers on each face be x,y,z,a,b,c

the product on vertices will be xcz, xyz, xac, xay, bya, byz, bcz, abc

So xcz + xyz + xac + xay + bya + byz + bcz + abc = 1298 (given)

On factorisation

(c + y)(x + b)(a + z) = 1298

(c + y)(x + b)(a + z) = 59 x 11 x 2

So x + y + z + a + b + c = 59 + 11 + 2 = 72

Let the numbers on each face of the cube be a,b,c,d,e,f. The numbers on the vertices will be- b c e, a b c, b e f, e f d, a b f, d f a, d a c, d e c.

Given- bce+abc+bef+efd+abf+dfa+dac+dec=1298.

We have to calculate the sum of the numbers. a+b+c+d+e+f.

Solution- bc(a+e) + bf(a+e) + dc(a+e) + fd(a+e)=1298.

(bc+bf+dc+fd)(a+e)=1298.

b(f+c) + d(f+c) =1298.

(b+d)(f+c)(a+e)=1298. -[1]

Now, the number 1298 can be written as the product of 3 primes i.e. 2, 11 and 59.

equation[1] can be rewritten as- (b+d)(a+e)(c+f)=2x11x59.

So, a+b+c+d+e+f=2+11+59 => 72.

Let the numbers on all the six sides of the cube be (a,b,c,d,e,f). Then the numbers that will be written in all the six vertices are (abc,bcd,eab,ebd,efa,efd,caf,cdf). When they are added ,they give 1298., abc+bcd+eab+ebd+efa+efd+caf+cdf=1298. bc(a+d)+eb(a+d)+ef(a+d)+cf(a+d)=1298. (a+d)(bc+eb+ef+cf)=1298. (a+d)[b(c+e)+f(c+e)]=1298. Therefore,(a+d)(c+e)(b+f)=1298. Because of this separation of 1298 into factors, The prime factorisation of 1298 is done. From this 1298=(59 11 2). Thus (a+d)(c+e)(b+f)=(59)(11)(2). Therefore a+b+c+d+e+f=59+11+2=72.

let's assume $a , b , c , a', b', c'$ are 6 numbers written on the cube where $a , a'$ are on opposite sides of the cube . $b , b'$ and $c , c'$ are on opposite sides' number as well .
now , there are 8 vertices , $abc , abc' ,ab'c , ab'c' , a'bc , a'bc' , a'b'c , a'b'c'$ are numbers on vertices . $abc+ abc' + ab'c + ab'c' + a'bc + a'bc' + a'b'c + a'b'c' = 1298$ $\Rightarrow ab(c+c')+ab'(c+c')+a'b(c+c')+a'b'(c+c')=1298$ $\Rightarrow (c+c')(ab+ab'+a'b+a'b')=1298$ $\Rightarrow (a+a')(b+b')(c+c')=2.11.59$ as $a,a',b,b',c,c'$ are positive integers , the value of these 3 sums $(a+a'),(b+b'),(c+c')$ should be $2, 11,59$ [ doesn't not necessarily according to sequence ] therefore , $a+a'+b+b'+c+c'=2+11+59 = 72$

1298=59 11 2 each face be a b c d e f so (a+b)(c+d)(e+f)= 1298 so sum is= 11+2+59= 72

Let the vertices is ABCD.EFGH and let side ABFE = a, BCGF = b, CGHD = c, ADHE = d, EFGH = f, ABCD= e. it is said that each vertex, the product of 3 numbers on the faces containing the vertex is written,

so,

vertex A containing a,d,e vertex B containing a,b,e vertex C containing b,c,e vertex D containing c,d,e vertex E containing a,d,f vertex F containing a,b,f vertex G containing c,b,f vertex H containing d,c,f

so,

a.d.e + a.b.e + b.c.e + c.d.e + a.d.f + a.b.f + c.b.f + c.d.f = 1298

(a.d + a.b + b.c + c.d)e + (a.d + a.b + c.b + c.d)f = 1298

(a + c)(b+d)e + (a + c)(b+d)f = 1298

(a+c)(b+d)(e+f) = 2.59.11

note that 2,59,11 are coprime

so

a+b+c+d+e+f = 2+59+11 = 72

let the numbers written on six face be a,b,c,d,e,f

let a and f are written oposite.

a is in front side and f is in back side.

b is on left side of a, c is on right side of a, d is on up side of a and e is on down side of a.

this way b and c are on opposite faces & d and e are on opposite faces.

now consider from front side

now if we see from front left up corner of a has number (a X b X d)

right up corner has number up corner of a has number (a X c X d)

left below corner of a has number (a X b X e)

right below corner of a has number (a X c X e)

similarly if we consider back side i.e side conatining f we have

now if we see from front left up corner of a has number (f X b X d)

right up corner has number up corner of a has number (f X c X d)

left below corner of a has number (f X b X e)

right below corner of a has number (f X c X e)

summing all numbers on vertices we get

(a X b X d) + (a X c X d) + (a X b X e) + (a X c X e) + (f X b X d) +(f X b X d) + (f X b X e) + (f X c X e)

= a { (b X d) + (c X d) + (b X e) + (c X e) } + f { (b X d) + (c X d) + (b X e) + (c X e) }

= (a + f) X { (b X d) + (c X d) + (b X e) + (c X e) }

= (a + f) X { (b + c) X d + (b + c) X e }

= (a + f) X { (b + c) X (d + e) }

=(a + f) X (b + c) X (d + e) = 1298 = 11 X 2 X 59

AS ALL THREE OF THEM ARE PRIME NUMBERS THEREFORE CANNOT BE FACTORISED MORE

now (a + f) or (b + c) or (d + e) can be equal to any of 11 or 2 or 59 but for this question we don't have to worry because we have to find out sum of all six integers therefore

NOW WE HAVE TO FIND OUT a + f + b + c + d +e = 11 + 2 + 59 = 2 + 11 + 59 = 11 + 59 + 2 = 2 + 59 + 11 = 59 + 2 + 11 = 59 + 11 + 2 = 72 ( answer is same in all six cases)

answer is 72

Let the vertices be $A,B,C,D,E,F,G$ . Let the front side of the cube be given by the square $ABCD$ and denoted as face $\alpha$ . The face to the right of $\alpha$ be $\beta$ . Likewise, the face to the right of $\beta$ be $\gamma$ and so on until $\delta$ . The bottom face be denoted as $\phi$ and the top face as $\omega$ .

Given by the problem we know that: $A+B+C+D+E+F+G=1298$

On each vertex, the product of the 3 numbers on the faces containing the vertex is written:

$A=\alpha\delta\omega \\ B=\alpha\beta\omega \\ C=\beta\gamma\omega \\ D=\gamma\delta\omega \\ E=\alpha\delta\phi\\ F=\alpha\beta\phi \\ G=\beta\gamma\phi \\ H=\gamma\delta\phi$

Partially combining some of the terms show that: $A+B=\alpha\omega(\delta+\beta) \\ C+D=\gamma\omega(\delta+\beta) \\ E+F=\alpha\phi(\delta+\beta) \\ G+H=\gamma\phi(\delta+\beta)$

Again combining partially: $A+B+C+D=(\alpha+\gamma)\omega(\delta+\beta) \\ E+F+G+H=(\alpha+\gamma)\phi(\delta+\beta)$

Ultimately this leads to:

$A+B+C+D+E+F+G+H=(\alpha+\gamma)(\omega+\phi)(\delta+\beta)=1298$

This equation:

$(\alpha+\gamma)(\omega+\phi)(\delta+\beta)=1298$ can be reduced to: $xyz=1298$

Because $x=(\alpha+\gamma)$

we can say that $x \geq 2$ . Prime factorization of 1298 leads to: $2\cdot11\cdot59=1298=xyz$

So: $2+11+59=72$

Let the numbers on the faces be $A, B, C, D, E, F$ , where opposite pairs of faces are $(A, D), (B, E), (C, F)$ . Then, the sum of the vertices would be given by $(A+D)(B+E)(C+F) = 1298 = 2 \times 11 \times 59$ . Since all the faces have positive integers, thus $A+D \geq 2, B+E \geq 2$ and $C+F \geq 2$ . Also, $2, 11$ and $59$ are prime, thus $(A+D,B+E,C+F)$ can only be equal to a permutation of this set: $\{2, 11, 59\}$ . Therefore $A + D + B + E + C + F = 2 + 11 + 59 = 72$ .

Try to factor out 1298. You will get 2, 11, and 59, all of which are prime numbers. There are six vertices in a cube, which means there must be six factors. It then follows that the other three are 1, since we were only able to find three. 59 + 2 + 11 = 72.

In a cube, there will be six faces. From this, we can pick 6 variables to be assigned in each face of the cube.

Let a, b, c, d, e, and f. be the numbers(positive integers) on each face of the cube respectively.

Now, it is said in the problem that the vertices of the cube represents the product of the adjacent faces. And the sum of the products is said to be 1298.

Now, we can set up a cube with faces marked by the variables. From this point we can get different equations. What I got is

abe+abf+adf+aef+cbf+cbe+cdf+cde=1298 *note that equations will vary depending on the assignment of the variables in the cube.

Factoring out a and d, we get

a(be+bf+df+ef) + c(bf+be+df+de) = 1298

From here, we are pretty stucked. But considering the fact that the numbers may not be distinct, we can assume f = d.

Now we can continue on our factorization.

(a+c)(be+bf+df+ef)=1298

We can further factor be+bf+df+ef:

b(e+f)+f(e+d)

Since f=d:

->b(e+d)+f(e+d)

->(b+f)(e+d)

Now we can work on the number 1298. The prime factors of 1298 would be: 1, 2, 11, 59. We can ignore the factor 1 since it can be cancelled through multplicative/division property of equality. i.e.

(1)(a+c)(b+f)(e+d)=(1)(2)(11)(59) (a+c)(b+f)(e+d)=(2)(11)(59) 1 is "cancelled"

we can say that: a+c=2 b+f=11 e+d=59

Note that the assignment of the sum is arbitrary. For example, we can say that a+c=11, b+f=2, e+d=59. It doesn't matter.

Now, we can now solve a+b+c+d+e+f which is equal to 2+11+59 or 72.