Malcolm writes a positive integer on a piece of paper. Malcolm doubles this integer and subtracts 1, writing this second result on the same piece of paper. Malcolm then doubles the second integer and adds 1, writing this third integer on the paper. If all of the numbers Malcolm writes down are prime, determine all possible values for the first integer.
Problem
Source: Canada Repêchage 2017/1
Tags: number theory
Not_a_Username
13.04.2017 15:38
The three integers are x, 2x-1, 4x-1
If $x\equiv 1 \mod 3, 3|4x-1$
If $x\equiv 2 \mod 3, 3|2x-1$
If $2x-1=3$, $x=2$, and the three numbers are (2, 3, 5)
The only other case is $x=3$, we can verify that (3, 5, 11) satisfies, so 2 and 3 and are the only possible values.
kk108
13.04.2017 15:49
Not_a_Username wrote:
The three integers are x, 2x-1, 4x-1
If $x\equiv 1 \mod 3, 3|4x-1$
If $x\equiv 2 \mod 3, 3|2x-1$
The only other case is $x=3$, we can verify that (3, 5, 11) satisfies, so 3 is the only possible value.
Nice sol... I was thinking along the same lines but you got the sol!
p-2 also works.
Not_a_Username
13.04.2017 16:15
kk108 wrote: Not_a_Username wrote:
The three integers are x, 2x-1, 4x-1
If $x\equiv 1 \mod 3, 3|4x-1$
If $x\equiv 2 \mod 3, 3|2x-1$
The only other case is $x=3$, we can verify that (3, 5, 11) satisfies, so 3 is the only possible value.
Nice sol... I was thinking along the same lines but you got the sol!
p-2 also works.
Oops, you're right
harapan57
13.07.2017 10:59
Once we get that the three integers are $n$, $2n-1$ and $4n-1$, we need to check four cases for $n$, which are $2,3,6k+1$ and $6k-1$. Clearly only $2$ and $3$ work.