Let $p, q, r$ and $s$ be four prime numbers such that $$5 <p <q <r <s <p + 10.$$Prove that the sum of the four prime numbers is divisible by $60$. (Walther Janous)
Problem
Source: 2019 Austrian Mathematical Olympiad Junior Regional Competition , Problem 4
Tags: number theory, primes, divisible, divides
rafaello
18.12.2020 22:20
We have primes ending only with $1,3,7,9$ ($5$ does not work out), also if some ones digit is doubled (or more) then at least one prime is greater than $p+10$. Thus, their ones add up to $20$. Therefore, we have to show that others add up to $40\pmod{60}$. Thus our number decimals must be $10,40\pmod{60}$. For the sake of contradiction, assume $20,30,50,60\pmod{60}$. Notice that if we have decimals divisible by $60$, considering prime ending with $3$, it is divisible by $3$, therefore $60$ is excluded. Similarly $30\pmod{60}$, we have prime ending with $3$ being divisible by $3$. We now might have $20,50\pmod{60}$. Considering $20\pmod{60}$ and the prime ending with $7$, we have $3\mid 60k+20+7=60k+27$. Considering $50\pmod{60}$ and the prime ending with $7$, we have $3\mid 60k+50+7=60k+57$. Hence, our decimals can only be $10,40\pmod{60}$. And sum is $4\cdot 10+20=60\equiv 0\pmod{60}$ or $4\cdot 40+20=180\equiv 0\pmod{60}$. We are done.
Euler01
18.12.2020 23:31
$5 < p < q < r < s < p+10$.
(Note that each prime must be one less or one more than a multiple of $6$)
Case 1$p= 3k+1$, and let $3k $ be a multiple of $6$ ($3k+2$ can't be a multiple of $6$)
Out of
$ p, 3k+2 , 3k+3 , 3k +4, 3k +5 , 3k+6 , 3k+7 , 3k+8, 3k+9, 3k+10, p+10$.
only $3k+5 ,3k+7$ can be primes. But since we need to get three different primes( $q,r,s$) , this case gives no solution.
Case 2$p=3k+2$.
Now $3k+3$ must be a multiple of $6$.
So out of
$ p , 3k+3 , 3k +4, 3k +5 , 3k+6 , 3k+7 , 3k+8, 3k+9, 3k+10, 3k+11, p+10$
$3k+4 , 3k+8 \text{and} 3k+10$ must be the three primes $q,r,s$ respectively.
So , $(p,q,r,s) \equiv ( 3k+2, 3k+4, 3k+8, 3k+10)$.
$\implies p+ q + r +s = 12k +24 \implies 12 | p+q+r+s$.
Now to check divisibility by $5$:
$p= 5k_1 +j \implies p+q+r+s= 20k_1 +4j+16$
$j=1 \implies p+q+r+s=5k_2$.
$j=2 \implies s= 5k+10 , \text{not possible}$.
$j=3 \implies q= 5k + 5 , \text{not possible}$.
$j=4 \implies r= 5k+ 10 , \text{not possible}$.
$\implies 5 | p+ q+ r +s$.
$\implies 60 | p+q+r+s$.
Sadigly
10.11.2024 23:58
If $p=6k+1$, then $min(s)=6k+11=p+10$ Contradiction If $p=6k-1$, then the only primes are $(6k-1;6k+1;6k+5;6k+7)$ $p+q+r+s=24k+12$ which obviously divides 12, so I have to prove that $5\div 24k+12$, which is same as $k\equiv2\hspace{1mm}(mod\hspace{1mm}5)$ 1)$k=5m\Rightarrow r=6k+5=30m+5=5(6m+1)$. If $r$ is prime,then $m=0$ and $k=0$. Which means $p=-1$ which is not a prime 2)$k=5m+1\Rightarrow p=6k-1=30m+5=5(6m+1)$ $m=0$, $p=5$ but $5<p$ contradiction 3)$k=5m+3\Rightarrow s=6k+7=30m+25=5(6m+5)$ which is not a prime 4)$k=5m+4\Rightarrow q=6k+1=30m+25=5(6m+5)$ which is not a prime So, $k$ is in form of $5m+2$, which is what we've been trying to prove.