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3^4n = 81^n
for Any value of n if it is integer be it even of odd (81^n+1) will not yield 0 as the last digit of the number will be 2
B is not a sufficient condition because n could be any number not necessarily integer.
For example n = 4.5 you have 81^4 * 81^0.5 + 1 which is divisible by 10
3 raised to anything will have the unit digit end in 3,9,7 or 1 and starts after every 4th exponent. Example 3^4 = 81, 3^8 = 6561.
From (a):
n is an integer and n > 1 or n = 1
4n will always end in 1. so ((3^4n) + 1)/10 will never have a remainder of 0.
From (b):
n > 4 and n may not be an integer so 3^4n may have unit digit = 9 and in this instance, (3^4n) + 1 will have unit digit = 0 which is divisible by 10.
You are in for big trouble. Watch the problem stem.
From A can you say for sure if the remainder is 0 or not 0? I think you can say for sure that the remainder is not zero
From B can you say for sure if the remainder is 0 or not 0? May be or may be not. The stem is not asking you to determine the value of n such that the remainder is 0.
You cannot say for sure because n could be any number not necessarily integer. n=5 does not give you remainder of 0 but n=4.5( >4) does give you a zero.
You are in for big trouble. Watch the problem stem.
From A can you say for sure if the remainder is 0 or not 0? I think you can say for sure that the remainder is not zero
From B can you say for sure if the remainder is 0 or not 0? May be or may be not. The stem is not asking you to determine the value of n such that the remainder is 0. You cannot say for sure because n could be any number not necessarily integer. n=5 does not give you remainder of 0 but n=4.5( >4) does give you a zero.
Anand.
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No, that is not what the question stem asks. It asks "is it POSSIBLE that the remainder is 0". You don't have to prove that it is. You need simply show that it is POSSIBLE (one example will suffice).
Yeah I get it now. is the remainder 0? is completely different from can the remainder be 0?
In that case answer is D.
Well now I am even more confused. If can is used then the answer for any DS question will always be D I believe.
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Now that I think about it, it would be difficult to come up with a question that would not automatically revert to D, i.e., the argument can be made that "everything is possible" unless you are know for sure that "it is not possible", hence, you can always answer the question -- even something like "Can God exist?" Hmmm.
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