A "brand" problem Marius BURTEA from Alexandria

          This is problem  E:17125, from GMB 2/2025, proposed for 6th grade.  In translation

          

           "Determine the prime numbers  $p\geqslant 5$ for which 156 is written as the sum of

            three natural numbers, directly proportional to 2,3 and $p$." 

(Marius BURTEA, Alexandria)

ANSWER CiP

$p=7\:\;:\;26+39+91=156;$

$p=47\;:\;6+9+141=156;$

$p=73\;:\;4+6+146=156:$

$p=151:\;2+3+151=156.$

                                 Solution CiP

          $\frac{a}{2}=\frac{b}{3}=\frac{c}{p}=\frac{a+b+c}{p+5}=\frac{156}{p+5},\;\;a,b,c \in \mathbb{N}$.

From the above we obtain

 $a=\frac{312}{p+5},\;\;b=\frac{468}{p+5},\;\;c=\frac{156p}{p+5}=\frac{156p+780-780}{p+5}=156-\frac{780}{p+5}.$

For the numbers $a,\;b,\;c$ above to be natural, $p+5$ must divide all the numbers 312, 468, 780. So $p+5$ must divide their greatest common divisor, that is

 $p+5 \mid gcd(312,468,780)=156 \tag{D}$.

     Given that $p\geqslant 5$ is a prime number, and that $p+5$ is an even number, among the numbers that verify condition (D), we choose

$p+5 \in \{12,\;26,\;52,\;78,\;156\}.$

Therefore $p\in \{7,\;21,\;47,\;73,\;151\}$ from which we eliminate the inconvenient value 21.

$\blacksquare$