Prove every number in the sequence 38, 381, 3811, 38111, 381111, ... is composite.
CASE 1
------------
Suppose the #1s in the given sequence be of the form 3m-2, where m is any positive integer.
Then the number must have the form:
10^(3m-2) - 1
38 * 10^(3m-2) + ---------------------
9
= 38 * 10^(3m-2) + 111........111
3m -2 1s
Thus, the sum of the digits of the number is
3+8 + 1*(3m-2) = 3m+9, which is divisible by 3
Accordingly, by divisibility rules, the given number is also divisible by 3.
CASE 2
------------
Suppose, the # 1s in the given sequence is of the form 3m-1 for any positive integer m.
Then, the number must have the form:
10^(3m-1) - 1
38 *10^(3m-1) + ---------------------
9
10^(3m-1) - 1
= 37 * 10^(3m-1) + 10^(3m-1) + -----------------------
9
10^(3m) - 1
= 37 *10^(3m-1) + -------------------
9
= 37 * 10^(3m-1) + 111.......111
3m 1s
There are m blocks of [111]. Since each block of [111] is divisible by 37, it follows that 111.....111(3m 1s) is divisible by 37.
Accordingly, the number must be divisible by 37.
CASE 3
------------
Suppose the number is of the form 3m, for any given positive integer m.
Then, the number must have the form:
38 * 10^(3m) + 111......111
3m 1s
10^(3m) - 1
= 38 * 10^(3m) + .......................
9
343 * 10(3m) - 1 { 7 *10^(m)}^3 - 1^3
= ------------------------------- = ---------------------------------
9 9
{7 * (10)^m -1) ({7 * (10^m)}^2 + 7*10^m +1)}
= -----------------------------------------------------------------------
9
[/EDIT]
7* 10^m - 1 7 * 10 ^2m + 7 * 10^m + 1
= ----------------------- * ---------------------------------------------
3 3
Now,
(7 *10^m - 1)/3
= (6 * 10^m)/3 + (10^m - 1)/3
= 2 * 10^m + (10^m - 1)/3
= 2 *10^m + 33......33
m 3s
= 2333......333 .................. (i)
m 3s
Also,
{(7 * 10^2m + 7 * 10^m + 1)/3}
= { 7*(10^2m - 1)/3 + { 7*(10^m - 1)/3} + 15/3
= 7 * 333......333 + 7 * 333....333 + 5
2m 3s m 3s
= 2333......3335666...6667 ..............(ii)
m -1 3s m-1 6s
Accordingly, the given number corresponds to the product of the two positive integers given in (i) and (ii) and:
therefore the number is composite.
Consequently, in terms of the three abovementioned cases, NO MEMBER of the given sequence can ever be a PRIME NUMBER.
Edited on July 29, 2022, 10:06 am
Puzzle Solution (Proof)
