Prove that all numbers of the form 12008, 120308, 1203308, ... are divisible by 19.
Since AT has already submitted the "simple proof" to this puzzle, all that remains for me is to posit a somewhat harder and lengthier proof by way of conventional methodologies.
Let the given sequence be defined as {F(n)}, where:
F(1) =12008
F(2) = 120308
F(3) = 1203308
..............
..............
F(n) = 12033.........3308
n-1 3s
Then
F(n) = 120 * (10^(n+1)) + {(10^(n-1) - 1)/3} * 100 +8
or, 3*F(n) = 360 * 10^(n+1) + {10^(n-1)-1} * 100 + 24
= 361 * 10(n+1) -76
= 19{19 * 10^(n+1) - 4).....,(#)
Since gcd(3, 19) =1, it follows that F(n) is divisible by 19.
Consequently , each member of {F(n)} is divisible by 19.
So, the proof by conventional methods is now
COMPLETE.
..........................................
However, I will now derive the general form of the Quotient obtained upon dividing F(n) by 19.
Denoting The said quotient by Q(n), we observe from (#) that:
3*Q(n) = 19* 10^(n+1) - 4 = 18* 10^(n+1) - 3 + 10^(n+1) -1
so, Q(n) = 6* 10^(n+1) - 1 + 3333......33
n+1 3s
or, Q(n) = 6333....332
n 3s
Consequently, the required quotients corresponding to the members of the sequence are :
Q(1) = 632
Q(2) = 6332
Q(3) = 63332
.............
............
Q(n) = 6333....332
n 3s
Edited on June 6, 2022, 3:33 am
a conventional proof
