BMO1_2019_1

BMO1_2019_1

Part (i)
To solve the first one, I’ll first look at the list \pmod {2}.
Then the list becomes:
p, p, p, p, p
Now for these all to be prime, p has to be 1 \pmod {2}.
But this doesn’t really help us so on to \pmod {3}.

Reducing the list we get:
p, p+2, p, p+2, p
All of these have to be \neq 0 \pmod {3} because 3 is the only prime divisible by 3 and p=3 does not work.
For this to happen, p \equiv 2 \pmod {3}.

Now we look at \pmod {5}.
Our list simplifies to:
p, p+2, p+1, p+3, p+2.
This works when p \equiv 1 \pmod {5}, not counting the cases when p+2 \equiv 0 \pmod {5} or p \equiv 0 \pmod {5} (we will count those later)

So by CRT on 2 and 5 we know that p \equiv 1 \pmod {10}. And from here it is easy to check the possible values of p and get that p=11 and p=101 works.

But we only have two prime numbers.
Back to the case when p+2 \equiv 0 \pmod {5} or p \equiv 0 \pmod {5}!

When p+2 \equiv 0 \pmod {5}, p+2=5 and p=3.
But 3+6=9 is not prime so p=3 does not work.

When p=5 we get a sequence 5, 7, 11, 13, 17 that does work. (we have found three cases so part (i) is done).

Part (ii)
Now to show that we only have one sequence, q, q+2, q+6, q+8, q+12, q+14 that is all prime, we look at it \pmod {5}.

Simplifying we get
q, q+2, q+1, q+3, q+2, q+4
These are all the remainders \pmod {5}!

So we must have 5 somewhere in there.
The only two possible values that can equal 5 are q and q+2.

q+2 yields q=3 and as proven above, that does not work.
q=5 does indeed work and our only sequence is:
5, 7, 11, 13, 17, 19

We have proven that no other cases work (as q+6 > 5 for all positive q) so we are done.

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