Is 8^x+8 divisible by 10?

Question

Is  8^x+8  divisible by 10?
1)  x=8y+3 , where  y  is a positive integer
2)  x  is a positive multiple of 11

yashikaaggarwal · Answer

To find: Is 8^x+8 divisible by 10?
or 8( 8^{x-1} +1) is divisible by 10

Solution: Cyclicity of 8, (unit digit of 8)
8^1 = 8
8^2 = 4
8^3 = 2
8^4 = 6
8^5 = 8...............
so cyclicity repeats after 4th power

1) x=8y+3, where y is a positive integer
y = 1,2,3,4.............
X = 11,19,27........

case 1: x = 11
8( 8^{11-1} +1)
8( 8^{10} +1)
unit digit of 8^10
8(4+1) = 8(5) which is divisible by 10

case 2: x = 19
8( 8^{19-1} +1)
8( 8^{18} +1)
unit digit of 8^18
8(4+1) = 8(5) which is divisible by 10

case 3: x = 27
8( 8^{27-1} +1)
8( 8^{26} +1)
unit digit of 8^26
8(4+1) = 8(5) which is divisible by 10 
(Sufficient)

2) x is a positive multiple of 11
x = 11,22,33,..............

case 1: x = 11
8( 8^{11-1} +1)
8( 8^{10} +1)
unit digit of 8^10
8(4+1) = 8(5) which is divisible by 10

case 2: x = 22
8( 8^{22-1} +1)
8( 8^{21} +1)
unit digit of 8^21
8(8+1) = 8(9) which is not divisible by 10 
(Insufficient)

Answer is A

TestPrepUnlimited · Answer

To be divisible by 10 we need the last digit to be 0.
Finally, since the expression is  8^x+8  we need a 2 + 8 to end on a 0. In order to end on a 2 from  8^x , we can observe the pattern of last digits from  8^x . It repeats in a cycle of 4, 8 -> 4 -> 2 -> 6 -> 8. Thus we can conclude we need  x = 4*i + 3  to land on the last digit of 2, where  i  is a nonnegative integer.

Statement 1:

We have  x = 8y + 3  which is a subset of  x = 4*i + 3 , sufficient.

Statement 2:

The last digit of powers of 8 follows a cycle of 4. Multiples of 11 do not cooperate with 4's so we can get all last digit scenarios above. Insufficient. (Or we can see 11 is an element of  x = 4*i + 3  but 22 is not).