3^x + 1 is divisible by 10 when 3^x ends with 9 (unit place will add to 9+1 =10 then and thus divisible by 10)
b) x>4 x=5 -> 3^5 = 243.. not true.
x = 6 -> 3^6 ends with 9. thus divisible.
Insufficient
a) x=3n+2. or x= 2,5,8,..
3^5 not true. 3^2 true.. Thus insufficient.

a&b) x=5,8,11,14...
unit digits of 3's power = 3,9,7,1,3,9,7,1..
thus powers repeat as 2,6,10,14,18.....
thus for 14 its true and for 5 its not. Insufficient.

Is there any shortcut to eliminate C?

because when 3^14 the unit digit is 9, do I really have to calculate that in order from 5 to 14?? Or any shortcut to know?

cans wrote:3^x + 1 is divisible by 10 when 3^x ends with 9 (unit place will add to 9+1 =10 then and thus divisible by 10)
b) x>4 x=5 -> 3^5 = 243.. not true.
x = 6 -> 3^6 ends with 9. thus divisible.
Insufficient
a) x=3n+2. or x= 2,5,8,..
3^5 not true. 3^2 true.. Thus insufficient.

a&b) x=5,8,11,14...
unit digits of 3's power = 3,9,7,1,3,9,7,1..
thus powers repeat as 2,6,10,14,18.....
thus for 14 its true and for 5 its not. Insufficient.

tracyyahoo wrote:Is there any shortcut to eliminate C?

because when 3^14 the unit digit is 9, do I really have to calculate that in order from 5 to 14?? Or any shortcut to know?

Yup, remmeber the cyclicity of 3 is 4. i'e
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243

If u raise to any more powers u will notice that the units digit will vary in patterns of 4.

So 3^14 is of the form 3n+2 where n=4. From the table above, we will get units digit of 3^14 as 9 which makes the problem statement true and any other value makes the stat false.

This is a problem that looks impossible at first glance because it appears you have to do tons of calculations - in this kind of problem look for a pattern to help you. The rule that units digits are found by simply dealing with units digits will help you here (becuase we need to find a time when the number achieved by 3^n ends in a 9. we know 3^2 ends in a 9, then:

3^3 = 27
3^4 = 81
3^5 = ends in a three (because 1(3)= 3
3^6- ends in a 9 because 3(3) = 9

Thus we can see a continuing pattern in the units digit that will create a 9 every 4th time after teh second so n can equal, 2, 6, 10, 14 ....

1) this statement is insufficient becuase the answer could be x = 5, 8, 11, or 14 - and only 14 is a value that works. Eliminate AD.

2) this statement is insufficient becuase our pattern tells us we will have many values of x that will work when x > 4 but we will also have many that won't work.

together they are insufficient becuase both x = 5 and x = 14 can be extrapolated from the information above - one works and one doesn't, therefore it is insufficient.

But the problem is that 3^8, the unit digit is not 9, but it is supposed to be having the unit digit 9 based on the cicle formular you give me.

bblast wrote:

tracyyahoo wrote:Is there any shortcut to eliminate C?

because when 3^14 the unit digit is 9, do I really have to calculate that in order from 5 to 14?? Or any shortcut to know?

Yup, remmeber the cyclicity of 3 is 4. i'e
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243

If u raise to any more powers u will notice that the units digit will vary in patterns of 4.

So 3^14 is of the form 3n+2 where n=4. From the table above, we will get units digit of 3^14 as 9 which makes the problem statement true and any other value makes the stat false.

tracyyahoo wrote:But the problem is that 3^8, the unit digit is not 9, but it is supposed to be having the unit digit 9 based on the cicle formular you give me.

bblast wrote:

tracyyahoo wrote:Is there any shortcut to eliminate C?

because when 3^14 the unit digit is 9, do I really have to calculate that in order from 5 to 14?? Or any shortcut to know?

Yup, remmeber the cyclicity of 3 is 4. i'e
3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243

If u raise to any more powers u will notice that the units digit will vary in patterns of 4.

So 3^14 is of the form 3n+2 where n=4. From the table above, we will get units digit of 3^14 as 9 which makes the problem statement true and any other value makes the stat false.

hey the cyclicity is 4, that means the cycle repeats after every 4 powers, so 3^12 will have same units digit as 3^8 and 3^4.

at another look is it (3^x)+1 is divided by 10?
if yes, then 3^x must end with 9 for the resultant above to be divisible by 10 without a remainder (or remainder=0)
st(1) 3^(3n+2) and n {integer}>0 is equivalent to 9* 3^(3n). A number multiplied by 9 will end with
9 only if the units digit of this number is 1, that is 3^(3n) should have it's units digit 1. It's obvious even without solving that it can and cannot be unit's digit 1 at all times. Not Sufficient
st(2)x>4 and (3^x) should end with 9. Again it's obvious with even less time thinking that not always 3^x will return the last number 9. Not Sufficient

what was difficult in this

tracyyahoo wrote:If x is a positive integer, is the remainder 0 when 3^x + 1 is divided by 10?
(1) x = 3n + 2, where n is a positive integer.
(2) x > 4

Pls explain this, and give an example, I'm really stuck in here