If N is a positive integer and TN is the sum of all the positive integers from 1 to N, inclusive, is N even?
(1) TN is even.
(2) T2N is even.
OA is 2
please help
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What do you mean by T2N is even?sana.noor wrote:If N is a positive integer and TN is the sum of all the positive integers from 1 to N, inclusive, is N even?
(1) TN is even.
(2) T2N is even.
OA is 2
Statement 1 is not sufficient alone, explanation is:
Tn = n * (n+1)/2 ( Number of terms * Average)
Tn is even means, either n or n+1 is divisible by 4, or n is of form 4t or 4t-1.
So, can't say if it is even or odd.
Please write statement 2 clearly.
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NOTE: the n and 2n are supposed to be in subscript.sana.noor wrote:If n is a positive integer and Tn is the sum of all the positive integers from 1 to n inclusive, is n even?
(1) Tn is even
(2) T2n is even
Target question: Is n even?
Given: Tn is the sum of all the positive integers from 1 to n inclusive
For example T3 (T subscript 3) = 1+2+3 = 6
Similarly, T7 = 1+2+3+4+5+6+7 = 28
Statement 1: Tn is even
There are several values of n that meet this condition. Here are two:
Case a: n = 3 (since T3 = 6, which is even). In this case, n is odd
Case b: n = 4 (since T4 = 10, which is even). In this case, n is even
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: T2n is even
There's a nice formula for finding sums of integers from 1 to some value:
The sum of integers from 1 to k = k(k+1)/2
Let's apply the above formula to the sum of numbers from 1 to 2n:
So, T2n = (2n)(2n+1)/2
= n(2n+1)
IMPORTANT: Statement 2 tells us that T2n is even.
This means that n(2n+1) is even.
Now, (2n+1) is an odd value for all integer values of n (in fact this is the definition of an odd number)
So, if (2n+1) is odd, and n(2n+1) is even, then it must be the case that n is even.
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Answer = B
Cheers,
Brent