ganesamurthy wrote:
Hi, What if I assume K is a fraction/decimal?
jamifahad wrote:
ruturaj wrote:
Is k2 odd?
(1) k - 1 is divisible by 2.
(2) The sum of k consecutive integers is divisible by k.
Question Is k^2 odd?
Stmt1: k-1 is divisible by 2.
(k-1)/2 = m where m is some integer
k-1 = 2m
k = 2m+1. Now this is the equation of any odd number
Hence k is odd. Therefore k^2 is odd.
Sufficient.
Stmt2: The sum of k consecutive integers is divisible by k
sum = k(k+1)/2
Sum/k = m where m is some integer
k(k+1)/2k = m
(k+1)/2 = m
k=2m-1. This again represents odd number.
Hence k is odd. Therefore k^2 is odd.
Sufficient.
OA D
Both statements imply that k is an integer.
Is k^2 odd?(1) k - 1 is divisible by 2 --> \(k-1=even\) --> \(k=even+1=odd=integer\) --> \(k^2=odd^2=odd\). Sufficient.
(2) The sum of k consecutive integers is divisible by k. Here k must be a positive integer because otherwise the statement does not make sense.
Properties of consecutive integers:
• If n is odd, the sum of n consecutive integers is always divisible by n. Given \(\{9,10,11\}\), we have \(n=3=odd\) consecutive integers. The sum is 9+10+11=30, which is divisible by 3.
• If n is even, the sum of n consecutive integers is never divisible by n. Given \(\{9,10,11,12\}\), we have \(n=4=even\) consecutive integers. The sum is 9+10+11+12=42, which is NOT divisible by 4.
The statement says that the sum of k consecutive integers is divisible by k, which, according to the above means that k is odd, therefore \(k^2=odd^2=odd\). Sufficient.
Answer: D.
For more check Number Theory chapter of our Math Book:
math-number-theory-88376.htmlHope it helps.
_________________