1. a) (3p) Calculate: $5^{3}+6^{3}+7^{3}+11^{3}$.

b) (4p) Show that the number $2015^{2014}$ can be written as a sum of four perfect cubes.

(Mathematical Gazette)
## Solution:

a) $5^{3}=125 ; 6^{3}=216 ; 7^{3}=343 ; 11^{3}=1331$

sum is: $125+216+343+1331=2015$
b) $2015^{2014}=2015^{2013} \cdot 2015=2015^{2013} \cdot\left(5^{3}+6^{3}+7^{3}+11^{3}\right)=2015^{2013} \cdot 5^{3}+2015^{2013} \cdot 6^{3}+$ $+2015^{2013} \cdot 7^{3}+2015^{2013} \cdot 11^{3}=\left(2015^{671} \cdot 5\right)^{3}+\left(2015^{671} \cdot 6\right)^{3}+\left(2015^{671} \cdot 7\right)^{3}+\left(2015^{671} \cdot 11\right)^{3}$.

Grading:

| {a) $\quad 5^{3}=125 ; 6^{3}=216 ; 7^{3}=343 ; 11^{3}=1331$  sum is: $125+216+343+1331=2015$} |  | $2 \mathrm{p} \quad$ |
| :---: | :---: | :---: |
|  |  | $1 \mathrm{p}$ |
| $\overline{\text { b) }}$ | $2015^{2014}=2015^{2013} \cdot 2015=2015^{2103} \cdot\left(5^{3}+6^{3}+7^{3}+11^{3}\right)$ | $1 \mathrm{p}$ |
|  | $2015^{2103} \cdot\left(5^{3}+6^{3}+7^{3}+11^{3}\right)=2015^{2013} \cdot 5^{3}+2015^{2013} \cdot 6^{3}++2015^{2013} \cdot 7^{3}+$  $2015^{2013} \cdot 11^{3}$ | $1 \mathrm{p}$ |
|  | $2015^{2013} \cdot 5^{3}+2015^{2013} \cdot 6^{3}++2015^{2013} \cdot 7^{3}+2015^{2013} \cdot 11^{3}=\left(2015^{671} \cdot 5\right)^{3}+$  $\left(2015^{671} \cdot 6\right)^{3}+\left(2015^{671} \cdot 7\right)^{3}+\left(2015^{671} \cdot 11\right)^{3}$ | $2 \mathrm{p}$ |