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Please explain why aren't we considering the possiblity of zero being one of the integers?

considering zero as a possible integer
1) a+b=11
a=0, b=1
c+d = 13
c = 1, d=23

in this scenario we have only 1 even integer.


Bunuel
The sum of the four integers a, b, c, and d is 24. How many of these integers are even?

(1) a + b = 11

(2) b + c = 12
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eeshajain
Please explain why aren't we considering the possiblity of zero being one of the integers?

considering zero as a possible integer
1) a+b=11
a=0, b=1
c+d = 13
c = 1, d=23

in this scenario we have only 1 even integer.




Your example has arithmetic mistakes. Also, note that 0 IS an even integer, so zero can be one of the integers.

If a + b = 11 and you set a = 0, then b must be 11, not 1.

Then c + d = 13 must be odd, so c and d must be one odd and one even. For example, c = 1 and d = 12.

So even with zero allowed, you still get exactly two even integers: a = 0 and d = 12.



ZERO:

1. Zero is an INTEGER.

2. Zero is an EVEN integer.

3. Zero is neither positive nor negative (the only one of this kind)

4. Zero is divisible by EVERY integer except 0 itself (\(\frac{0}{x} = 0\), so 0 is a divisible by every number, x).

5. Zero is a multiple of EVERY integer (\(x*0 = 0\), so 0 is a multiple of any number, x)

6. Zero is NOT a prime number (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed: anything/0 is undefined.

8. Any non-zero number to the power of 0 equals 1 (\(x^0 = 1\))

9. \(0^0\) case is NOT tested on the GMAT.

10. If the exponent n is positive (n > 0), \(0^n = 0\).

11. If the exponent n is negative (n < 0), \(0^n\) is undefined, because \(0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0}\), which is undefined. You CANNOT take 0 to the negative power.

12. \(0! = 1! = 1\).


Hope it helps.
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