GMB 8 / 1978

          La Rubrica REZOLVATORILOR de PROBLEME , pagina 361, apare si numele meu (am subliniat) :TIMISOARA , Lic. industrial nr. 2 cl XI Ciobanu Petru (146)

          In "2005 Tenure Test" , Problem II, point g) of the previous post, we have the very Sylvester Problem on page 323. It is exactly with the same demonstration. I took it from another source....(I don't know which one )

In the PROBLEM SOLVERS section, page 361, my name also appears (I underlined it):

TIMISOARA, Lic. industrial no. 2 cl XI Ciobanu Petru (146)

          Today we will play with a problem whose author we talked about here not long ago.

     " E : 6305.  Let there be 6 cubes with unequal sides, with the measures of the sides being natural numbers. Determine the volume of each cube, so that the 10 cubes together occupy a volume of 444 cubic meters.

{author :} Liviu PODGORNEI, student, Oltenița "

ANSWER CiP

The six cubes have sides of 1, 2, 3, 4, 5, and 6 meters respectively.

Solution CiP

          If we denote by  $a,\;b,\;c,\;d,\;e,\;f\;$ the dimensions (in meters) of the six cubes, then $1\leqslant a<b<c<d<e<f$  and 

$a^3+b^3+c^3+d^3+e^3+f^3=441 \tag{1}$

But, since the cubes have unequal sides, it results :

$a\geqslant 1\;,\;b\geqslant 2\;,\;c\geqslant 3\;,\;d\geqslant 4\;,\;e\geqslant 5\;,\;f\geqslant 6 \tag{2}$

Then, the sum of their volumes  $a^3+b^3+c^3+d^3+e^3+f^3$  is  $\geqslant 1^3+2^3+3^3+4^3+5^3+6^3=1+8+27+64+125+216=441$

     It means that in  (2)  we must have the equal sign everywhere.

$\blacksquare$