even odd

[PAB2706]

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.

Note : the N & 2N are subscript.

[Frankenstein]

Hi,
Tk = k(k+1)/2
From(1):
For Tn=6, n = 3 (odd)
For Tn=10, n = 4 (even)
Not sufficient

From(2):
T2n = (2n)(2n+1)/2 = n(2n+1).
2n+1 is always odd. So, for n(2n+1) to be even, n has to be even.
Sufficient

Hence, B

[Brent@GMATPrepNow]

Hi,
Tk = k(k+1)/2
From(1):
For Tn=6, n = 3 (odd)
For Tn=10, n = 4 (even)
Not sufficient

From(2):
T2n = (2n)(2n+1)/2 = n(2n+1).
2n+1 is always odd. So, for n(2n+1) to be even, n has to be even.
Sufficient

Hence, B

Nice solution as usual, Frankenstein.

I thought I should point out to other viewers that Frankenstein is using a formula for finding the sum of all positive integers from 1 to k.

The formula goes like this: 1 + 2 + 3 + 4 + ...k = k(k+1)/2

Cheers,
Brent

[subhakam]

Hi,
Tk = k(k+1)/2
From(1):
For Tn=6, n = 3 (odd)
For Tn=10, n = 4 (even)
Not sufficient

From(2):
T2n = (2n)(2n+1)/2 = n(2n+1).
2n+1 is always odd. So, for n(2n+1) to be even, n has to be even.
Sufficient

Hence, B

Nice solution as usual, Frankenstein.

I thought I should point out to other viewers that Frankenstein is using a formula for finding the sum of all positive integers from 1 to k.

The formula goes like this: 1 + 2 + 3 + 4 + ...k = k(k+1)/2

Cheers,
Brent

Hello Brent - i did not understand the solution - can you explain please how the formula becomes 1 + 2 + 3 + 4 + ...k = k(k+1)/2?

Thank you very much!

[Brent@GMATPrepNow]

1 + 2 + 3 + 4 + ...+k = k(k+1)/2

So, for example, the sum 1+2+3+4+.....+39+40 = (40)(41)/2 = 820
Similarly, 1+2+3+4+.....+58+59 = (59)(60)/2 = 1770

Does that answer your question?

Cheers,
Brent

[subhakam]

Thank you Brent - this helps greatly.

Follow up question is - whether the below formula valid only for consecutive integers or for other evenly spaced sets as well where the difference between any 2 terms could be 5, 6 , -2 , 1/2 etc? What would be the formula for consecutive integers in decreasing order?

Lastly, the above does not hold valid (in my opinion) where some of the integers might be repeated in the set ? E.g. 1+2+3+3+3+4+4+5+5........+k ?

Many thanks once again!!
Subhakam

[Brent@GMATPrepNow]

Thank you Brent - this helps greatly.

Follow up question is - whether the below formula valid only for consecutive integers or for other evenly spaced sets as well where the difference between any 2 terms could be 5, 6 , -2 , 1/2 etc? What would be the formula for consecutive integers in decreasing order?

Lastly, the above does not hold valid (in my opinion) where some of the integers might be repeated in the set ? E.g. 1+2+3+3+3+4+4+5+5........+k ?

Many thanks once again!!
Subhakam

The formula k(k+1)/2 is specifically for the sum of consecutive integers from 1 to k.

Having said that, we can often make some adjustments to our sum in order to apply the formula.
For example:
7+14+21+28+....+203+210 = 7(1+2+3+4+....29+30)
= 7[(30)(31)/2]
= 7[465]
= 3255

Cheers,
Brent

[subhakam]

Thank you very much Brent!