Cube Dilemma

Number Theory Level 2

Once Hardy blindfolded Ramanujan and told him he wanted to give him a challenge.Hardy said he would present Ramanujan with a cube.On the $6$ faces of the cube would be written $6$ distinct positive integers.At each vertex where $3$ faces overlap Hardy would multiply the $3$ numbers of the $3$ faces and write them on the vertex.He would tell Ramanujan the sum of the numbers present on each vertex and based on this information Ramanujan would have to tell the sum of the numbers that were originally written on the faces of the cube.

If Hardy told Ramanujan that $1729$ was the sum of the $8$ numbers of the $8$ vertices ; What was the sum of the $6$ numbers originally written on the $6$ faces ?

Original !

The answer is 39.

1 solution

Chris Lewis
Oct 10, 2019

Nice problem! If the numbers on the faces are $a,b,c,a',b',c'$ (where $a'$ is on the face opposite $a$ , and so on), then the total of the vertex products is $T=abc+abc'+ab'c+ab'c'+a'bc+a'bc'+a'b'c+a'b'c'$ . This factorises as $T=(a+a')(b+b')(c+c')$ .

In the case $T=1729$ , the only way to write $T$ as a product of three integers each larger than $1$ is $1729=7\times13\times19$ . So, in some order, $\{a+a',b+b',c+c'\}=\{7,13,19\}$ and the total of the numbers on the faces is $a+b+c+a'+b'+c'=7+13+19=\boxed{39}$ .

Cube Dilemma

The answer is 39.

1 solution

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