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(1) a^2 – 36 = 0
Then a^2=36...I.e a=6....
If b is positive...
Units digit of a^b will be 6 always
If b is 0
Units digit of a^b will be 1
If b is negative
Units digit of a^b will be different
So A is insufficient


(2) b is a positive integer
Clearly insuff...Without the value of a


CombiNing both will get units digit as 6

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If a^b = c, what is the units digit of c?

(1) a^2 – 36 = 0
a = 6, -6
but we do not know anything about b. If b =1, unit of c (if a =6, a = -6) will be 6
However, if b = 0, unit of c = 1
INSUFFICIENT

(2) b is a positive integer
we do not know anything about a. - insufficient

taken (1) + (2):
we now know that b is always positive
so either a = 6, -6
unit of c will be 6.

Therefore, C is the correct answer
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If a^b = c, what is the units digit of c?

(1) a^2 – 36 = 0
a = {6,-6}
Since value of b is unknown
NOT SUFFICIENT

(2) b is a positive integer
Since value of a is unknown
NOT SUFFICIENT

(1) + (2)
(1) a^2 – 36 = 0
a = {6,-6}
(2) b is a positive integer
Unit digit of c will be 6
SUFFICIENT

IMO C
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(1) a^2 – 36 = 0
--> a = ±6
But we do not know whether b is an integer or a fraction
So, we cannot have a definite unit digit --> Insufficient

(2) b is a positive integer
Does not talk anything about 'a' --> Insufficient

Combining (1) & (2),
Since Cyclicity of 6 is 1 and b is a positive integer
--> a^b will have end in 6
--> Unit digit of a^b = c is 6 --> Sufficient

IMO Option C
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#1
a^2 – 36 = 0
a=+/-6
so any power of b except 0 will give unit digit of c as 6; but b is not know it can be 0 at which c =1
insufficient
#2 b is a positive integer
insufficent
a not know
from 1 &2
b>0 and a = +/-6 ; unit digit of c = 6 always
IMO C


If a^b = c, what is the units digit of c?

(1) a^2 – 36 = 0
(2) b is a positive integer
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Quote:

If a^b = c, what is the units digit of c?

(1) a^2 – 36 = 0
(2) b is a positive integer

(1) a^2 – 36 = 0 insufic.
\(a^2 – 36 = 0…a^2=36…|a|=(6,-6)\)
\(b=0:a^0 = c…units=1\)
\(b≠0:(6,-6)^b = c…units=6\)

(2) b is a positive integer insufic.

(1 & 2) sufic.
\(b≠0,a=(6,-6):(6,-6)^{≠0} = c…units=6\)

Answer (C)
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a^{b} = c—> what is the units digit of c?

(Statement1): a^{2}—36=0
(a—6)(a+6)= 0
a= 6, a=—6
But no info about what b is.
—> if b=0, then 6^{0}=1 and (—6)^{0}= 1 —> units digit of c is equal to 1

—> if b = 2, then 6^{2}=36 and (—6)^{2}= 36 —> units digit of c is equal to 6

Insufficient

(Statement2) b is a positive integer.
No info about what a is.
Clearly insufficient.

Taken together 1&2,
—> 6 or (—6) to the power of any positive integer end with ...6 .

Units digit of c is equal to 6
Sufficient

The answer is C

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Quote:
If \(a^b\) = c, what is the units digit of c?

(1) \(a^2\) – 36 = 0
(2) b is a positive integer

(1) \(a^2\)=36
=>a=+6 or -6
From this we don't get a unique value of unit digit of \(a^b\) because for a=6 and b=2, \(a^b\)=\(2^6\)=36, but for a=6 and b=1/2, \(a^b=6^{1/2}\)=2.45
Therefore, insufficient.

(2) No information about value of a, so we don't get a unique value of \(a^b\).
Therefore, insufficient.

From (1) and (2), we get that a=+6 or -6 and that b is a positive integer.
So, for all the above values of a and b, \(a^b\) will have 6 as unit digit, for example, \(6^2=36, -6^3=-216\), and so on.
Thus, sufficient.

Therefore, the correct answer is option C.
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If a^b = c, what is the units digit of c?

(1) a^2 – 36 = 0
(2) b is a positive integer

Statement 1: a^2=36, a=+6 or -6. if a^2=36 units digit is 6.

If a^3= units digit is 2. Hence cannot determine.

Statement 2: b is positive. It could be any number 1 to 3 hence we cannot determine.

when combining statements 1 and 2 still even if b is positive we cannot determine exactly the units digit.

IMO E
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Given: a^b= c
a) a^2 - 36= 0
=> a= 6, a= -6

No info about b. so, Insufficient

b) b is a positive integer.
no info about a. Insufficient

a) + b) => cyclicity of 6 is 1 and b is positive integer.
so, unit digit of c is 6.

Correct answer: C
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