magical cook wrote:If x is a positive integer, is the remainder 0 when (3^x+1)/10?

1. x = 3n + 2

2. x>4



Hi, someone said if 1) was 4n+2 instead of 3n+2, that answer would have been A) but I can not figure out what it means by that. Can anyone possible explain the difference if 1) is 4n+2?

OK let me start by jotting down some rules of indices
1.) a^(m+n) = a^m*a^n..so 3^(3n+2) = 3^3n*3^2= 3^3n9
2.) a^(m*n)= (a^m)^n .. so 3^3n= 27^n

Now, any number will be divisible by 10 when the last number is a 0 .. so for any number of the form a+1 to be divisible by 10 a should have unit digit of 9 which in turn would make the unit digit of " a+1 " = 0

Now, consider the first statement

it says x = 3n+2 so we have 3^x+1 = 3^(3n+2) +1 , which according to the first rule of indices mentioned above = 3^3n*3^2+1 and this according to the second statement mentioned above = 27^n*9+1 , now 27^n*9+1 as mentioned will be divisible by 10 if 27^n*9 has a unit digit of 9 .. which means that 27^n should end with a one in its unit place ..

Now, let us investigate 27^n further by varying the value of n

let n = 0 in which case 27^n = 1
let n=1 in which case 27^n = 27
let n = 2 in which case 27^n = 729
let n = 3 in which case 27^n= 19683
let n=4 in which case 27^n = 531441 ...

So as can be seen 27^n ends in a 1 if n is a multiple of 4, but since the statement says nothing about whether n is a multiple of 4 or not statement 1 is insufficient.

Now, take a look at the second statement it says x > 4, which again does not help us in deciding whether 27^n ends in a 1 or not so this statement is insufficient.

Combine the two and we have x = 3n+2 and x> 4 this means n could be any integer greater than 1 so both the statement together are also not enuf ..

now what is the difference between x = 3n+2 and x= 4n+2?

when x = 4n+2, 3^x+1= 3^4n+2+1 which according to the first rule of indices mentioned = 3^4n*3^2+1 and this according to the 2nd rule = 81^n*9+1

Now 81^n will always end with one that means 81^n*9 will always have 9 as its unit digit and that would also mean that 81^n*9+1 will end with a 0 and hence the expression will be always be divisble by 10 .. hope the explanation helped

Regards.

Gabriel -- thank you very much for the detailed explanation! I finally got it since the pattern of the unit digit differ depending on n when 3n+1, on the other hand, the unit digit is always same when 4n+1, thus 4n+1 is sufficient.