BTGModeratorVI wrote: ↑

Wed Jan 06, 2021 8:16 am

If T is a set of 35 consecutive integers, of which 17 are negative, what is the sum of all the integers in T ?

A. -2
B. -1
C. 0
D. 1
E. 2

Answer: C
Source: official guide

If 17 integers are negative, then the first 17 integers of the sum look like this: (-17) + (-16) + (-15) + .... (-4) + (-3) + (-2) + (-1)

35 - 17 = 18, so there are still 18 numbers to add to our sum

The next number in the sum is 0, which means the remaining 17 integers must all be positive.
So our sum looks like this: (-17) + (-16) + (-15) + .... (-4) + (-3) + (-2) + (-1) + (0) + 1 + 2 + 3 + . . . + 16 + 17

As we can see, for every NEGATIVE value, we have a POSITIVE value of the same magnitude.
For example we have -16 & 16, and -8 & 8, etc.

When we add each pair (consisting of a NEGATIVE value and the POSITIVE value of the same magnitude), we get 0
For example, (-16) + 16 = 0, and (-7) + 7 = 0

When we do this, our sum reduces to the sum of a bunch of zeros: 0 + 0 + 0 + ..... + 0 + 0

Answer: C

Cheers,
Brent

BTGModeratorVI wrote: ↑

Wed Jan 06, 2021 8:16 am

If T is a set of 35 consecutive integers, of which 17 are negative, what is the sum of all the integers in T ?

A. -2
B. -1
C. 0
D. 1
E. 2

Answer: C
Source: official guide

Solution:

If 17 of the integers are negative, then the 18th integer is zero, and the remaining 17 integers are positive. Thus, the negative and positive integers cancel out, and the sum is zero.

Answer: C