x is a positive integer, is the remainder 0 when 3^x + 1 is divided by 10?
1> x = 4n + 2 2> x>4
Ans A
Powers of 3 follow a cyclicity of 4 and we need to find a definite remainder for which we need a definite units place ..
St 1 : since power of 3 has been expressed in terms of multiples of 4, units digit would be same as 3^2 ...Definite YES ...suff
St 2 : variable units place poss ...not suff
P.S : I have a wierd feeling that Ans is E ...coz sum only states that x is an integer, which means i cannot assume n to be an integer, then ans changes .. ( i mean for x to be an integer, n could be integer or multiple of 0.5, in which case ans changes )
Powers of 3 follow a cyclicity of 4 and we need to find a definite remainder for which we need a definite units place ..
St 1 : since power of 3 has been expressed in terms of multiples of 4, units digit would be same as 3^2 ...Definite YES ...suff
St 2 : variable units place poss ...not suff
P.S : I have a wierd feeling that Ans is E ...coz sum only states that x is an integer, which means i cannot assume n to be an integer, then ans changes .. ( i mean for x to be an integer, n could be integer or multiple of 0.5, in which case ans changes )
I totally agree with you bhavin. Because
n=0 then x=2 => reminder is 0. n=1/4 then x=3 => reminder is 8 n=1/2 then x=4 =. reminder is 2.
here we need not even consider -ve numbers to conclude that this option is insuff.
2) is also insuff and combining both we get no where near the answer.
I got this ques from another forum but there was no info on "n".
Perfect square always has an odd no of factors whose sum is also always odd ..
If required can post the proof for the same as well ..
Yes the answer is D.
Here is my expln: Consider 1)
here N has an even number of factors. Lets take an example 4: Factors of 4 are 1,2,4. take 9. Factors of 9 are 1,3,9. Take 16, again factors are 1,2,4,8,16.
So N is not a perfect square. Hence 1) alone is suff.
Consider 2) Sum of factors is even. Consider the above examples and add the factors.All the sums are odd.
(1) The number of distinct factors of N is even. (2) The sum of all distinct factors of N is even.
i think answer is E. take the example of 3, two distinct factors-1,3-even sum of factors-1+3=4 even but 3 is not a perfect square
Hey Gutsy ..
For all non perfect square integers (prime or composite) there are evn number of factors and perfect squares always have odd no of factors ..
So, if statement reads even no of distinct factors, it can never be a perfect square ...Hence sufficient ...And no surprise that 3 has even no of factors, all non perfect squares have to have even no of factors ..
Lets try to prove this :
Let us prime factorize any integer N and express as N=a^m*b^n*c^p*.......
Where a,b,c... are prime factors and m,n,p...are integers .. Then, No of factors is given as = (m+1)(n+1)(p+1).....
So, if N is a perfect square, prime factorizing leads to even powers for every prime factor .. i.e m,n,p are even when N is a perfect square
SO, (m+1) , (n+1), (p+1) are all odd and their product is odd as well ..
And when N is not a perfect square, atleast one of the powers is odd, hence atleast 1 term of the series (m+1)(n+1)(p+1) is even and product is even ...
So St 1 is sufficient
St 2 : Similarly, trying to think of mathematical proof as to why for perfect squares, sum of distinct factors is odd .. Though, this is a property of perfect sqaures and we can easily cross verify by trying first few perfect squares ..
Of the 60 animals in a farmhouse 2/3rd are either Pigs or Cows. How many of the animals are cows ?
1) The farmhouse has more than twice as many cows as it has pigs
2) The farm has more than 12 pigs
it has to be A. I believe cow and pig are the only two animals totalling 60. so if cow are more than double of pigs,2/3 will be cow only.
Both the statements are not sufficient to ans the q.
From the main statement 40 can be the no.of cows or pigs.But both the statements doesn't give any clue beyond this
Ans C ..
We, know C+P = 40
St 1: C>2P ...multiple possible solutions ..Not suff However, it is imp to realise that cut off value for no of pigs is 13 , ony then can the no of cows be more than double of no of pigs and yet add up to 40 .... In other words, no of pigs have to be 13 or lower ...
St 2 : Alone tells u nothing
Combined : Definite answer ...pigs = 13, cows =27 ( coz if pigs = 14, cows is 26 and 26 is not more than double of 14, so it violates the condition)
St 1: C>2P ...multiple possible solutions ..Not suff However, it is imp to realise that cut off value for no of pigs is 13 , ony then can the no of cows be more than double of no of pigs and yet add up to 40 .... In other words, no of pigs have to be 13 or lower ...
St 2 : Alone tells u nothing
Combined : Definite answer ...pigs = 13, cows =27 ( coz if pigs = 14, cows is 26 and 26 is not more than double of 14, so it violates the condition)
