If x is a positive integer, does the remainder, when (7^x + 1) is divi

Question

If x is a positive integer, does the remainder, when  (7^x + 1)  is divided by 100, have 0 as the units’ digit?

(1) x = 4n + 2, where n is a positive integer.
(2) x > 5

GMATinsight · Answer

CONCEPT: Remainder when any number is divided by hundred are the last two digits of the number e.g. 137 divided by 100 leaves 37 remainder and likewise 923 divided by 100 leaves 23 remainder etc

i.e. we only need to know if the unit digit of  (7^x + 1)  is Zero or NOT

i.e. we need to know if  7^x  has unit digit 9 or not which is possible when  x = 4a+2  because cyclicity of 7 is 4 and  7^2  and  7^6  and  7^10  etc have unit digit 9

Statement 1 : x = 4n + 2, where n is a positive integer. 
i.e.  7^x  will always have unit digit 9 hence
SUFFICIENT

Statement 2 : x > 5 
x may be 6 or 7 etc. hence unit digit of  7^x  may or may not be 9 hence
NOT SUFFICIENT

Answer: Option A

Archit3110 · Answer

(7^x + 1) 
#1
x=4n+2
value of x will be even always ; x=2,6,10 at n=0,1,2
by cyclicity of 7 we know 
7^1=7
7^2=9
7^3=3
7^4=1
so we see x= 2,6,10.. we get unit digit as 9 so 9+1 ;10 unit digit is always 0

#2
 x > 5
not sufficient as value of unit digit of 7^x will vary
IMO A

auradediligodo · Answer

7^x+1=100q + m

Now m will have a 0 as its unit digit when 7^x+1 will end in 0 (10, 20, 30.....)

Also 7^ something can have different digit units since it has a ciclicity=4

7
9
3
1
7

These are the units digits of the resulting numbers from the application of powers to number 7.

Statement one gives us the following values for x---> 2, 6, 10, 14
As you can see they are evenly spaced and if plugged in the original 7^x the result will give always a number ending in 9. Hence 7^x+1 will always end up having a unit digit = 0.
  Hence statement 1 is sufficient.

Statement 2 on the other end gives us multiple remainders for the original equation.  Hence not sufficient.

Option A

If x is a positive integer, does the remainder, when (7^x + 1) is divi

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