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Re: If T is a set of 35 consecutive integers, of which 17 are negative [#permalink]
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Dav5599 wrote:
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

17 of 35 consecutive integers are negative.
(35 - 17) = 18 terms left.
The first number after -1 is 0.
(18 - 1) = 17 terms left: all positive.

\(-17, -16, ... -2, -1, 0, 1, 2 ... 16, 17\)
The 17 negative integers cancel the 17 positive integers; the sum of 34 of the numbers is 0. Add to the remaining number, 0. (0 + 0) = 0

Answer C

**If not sure, take fewer numbers, with an odd number of terms:
-3 + (-2) + (-1) + 0 + 1 + 2 + 3 = 0

Sum of negative: -6
Sum of positive: 6
Add 0: (-6 + 6 + 0) = 0
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Re: If T is a set of 35 consecutive integers, of which 17 are negative [#permalink]
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Dav5599 wrote:
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


If 17 of the numbers are negative, then 35 - 17 = 18 numbers are zero or greater. Thus, we have the consecutive integers from -17 to 17 (including 0), and thus, the sum is 0.

Answer: C
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Re: If T is a set of 35 consecutive integers, of which 17 are negative [#permalink]
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Total 35 consecutive numbers

{-17,-16,-15..........-1} = 17 numbers and the sum is -153
{0} = 1 number
{17,16,15.................1} = 17 numbers and the sum is 153

-153 + 0 + 153 = 0

Answer is C

(In fact this question can be answered in under 30 seconds logically without any calculations)
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Re: If T is a set of 35 consecutive integers, of which 17 are negative [#permalink]
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Dav5599 wrote:
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


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
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Re: If T is a set of 35 consecutive integers, of which 17 are negative [#permalink]
Expert Reply
Dav5599 wrote:
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


If 17 of the integers are negative, then one of the integers is zero, and the other 17 are positive. Thus, the negative and positive integers cancel out, and the sum is zero.

Answer: C
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Re: If T is a set of 35 consecutive integers, of which 17 are negative [#permalink]
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Re: If T is a set of 35 consecutive integers, of which 17 are negative [#permalink]
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