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by sunnyjohn » Wed Nov 25, 2009 12:21 am
IMO : B

overall you should know that every number from 0-9, has recurring format of unit digit when multiplied by itself.
for eg:
3^1 = 3 <-- unit digit of mulplication
3^2 = 9
3^3 = 7
3^4 = 1
3^5 = 3
similarly... 9 7 1 3 9 7 3 1...

so over all we are only interested to know the unit digit of 'a'. If we know that we can tell the unit digit of a^36.

Option A)
it give us two possible unit digit of a ==> 7 and 3 : INSUFFICIENT

Option B)
it give us only one option : 7 ==> Hence SUFFICIENT

so Answer should be B.
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by palvarez » Wed Nov 25, 2009 1:26 pm
heshamelaziry wrote:Q. What is the units digit of a^36?

a) a^2 has 9 as the units digit
b) a^3 has 3 as the units digit

Rephrase

What is a^36 (mod 10)

(a) a^2 (mod 10) = 9
(b) a^3 (mod 10) = 3


a. a^2 = 9 (mod 10); a^36 (mod 10) = 9^18 (mod 10) = (-1)^18 = 1 Sufficient
2. a^3 = 3 (mod 10); a^36 (mod 10) = 3^12 (mod 10) = 1 sufficient
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by palvarez » Wed Nov 25, 2009 1:28 pm
sunnyjohn wrote:IMO : B

overall you should know that every number from 0-9, has recurring format of unit digit when multiplied by itself.
for eg:
3^1 = 3 <-- unit digit of mulplication
3^2 = 9
3^3 = 7
3^4 = 1
3^5 = 3
similarly... 9 7 1 3 9 7 3 1...

so over all we are only interested to know the unit digit of 'a'. If we know that we can tell the unit digit of a^36.

Option A)
it give us two possible unit digit of a ==> 7 and 3 : INSUFFICIENT

Option B)
it give us only one option : 7 ==> Hence SUFFICIENT

so Answer should be B.

a^2 = 9 (mod 10)
a = 3 or -3 ( mod 10) = 3 or 7
Having 2 values for a (mod 10) doesn't entail that a^36 (mod 10) can have multiple values.
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by Stuart@KaplanGMAT » Wed Nov 25, 2009 1:59 pm
heshamelaziry wrote:Q. What is the units digit of a^36?

a) a^2 has 9 as the units digit
b) a^3 has 3 as the units digit
We can also solve quickly through algebra and common sense.

1) a^2 ends in 9

Well, a^36 = (a^2)^18

If we know that a^2 ends in 9, it's certainly possible to calculate the units digit of (...9)^18 - sufficient.

2) a^3 ends in 3

Well, a^36 = (a^3)^12

If we know that a^3 ends in 3, it's certainly possible to calculate the units digit of (...3)^12 - sufficient.

Each statement is sufficient alone: choose D.
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