If the sum of 5 consecutive integers is x, which of the

Question

If the sum of 5 consecutive integers is x, which of the following must be true?

I. x is an even number
II. x is an odd number
III. x is a multiple of 5

(A) I only

(B) II only

(C) III only

(D) I and III

(E) II and III

GyanOne · Answer

I. If you start off with an even number, the sum (x) is even, and if you start off with an odd number the sum (x) is odd. Therefore this is not always true. 
II. Same as above. This need not be always true. 
III. Say, the first number is p. Then the sum of the five numbers = p + (p+1) + (p+2)...... + (p+4) 
 = 5p + 10 = 5 (p+2) => divisible by 5. There this must be true in all cases.

Therefore C is the answer.

mayankdharwal · Answer

OA: C
there can be two cases
1. first no is odd---then sun of 5 consecutive no is odd
2. first no is even---then sun of 5 consecutive no is even
so we can't conclude 1 and second statement.
Sum of five consecutive no is always divisible by 5 hence option C

Bunuel · Answer

You can solve this question simply considering different sets of 5 consecutive integers.

I. x is an even number --> if the set is {-1, 0, 1, 2, 3}, then the sum is 5=odd. Thus this statement is not always true. Eliminate A and D.

II. x is an odd number --> if the set is {-2, -1, 0, 1, 2}, then the sum is 0=even. Thus this statement is not always true. Eliminate B and E.

Only option C remains.

Answer: C.

sdas · Answer

Sum of every odd number of consecutive integers is always divisible by the odd number
Eg: sum of 5 odd number of integers= 1+2+3+4+5 = 15/5=3

The same does not hold true for even number of consecutive integers
Thus IMO C

Statements 1 and 2 are not true, since we dont know what exactly is the value of x

veergmat · Answer

Assume the simplest series , I assumed following
a) 1,2,3,4,5

Sum=15 , Odd
b) 2,3,4,5,6
Sum=20, Even
c) 3,4,5,6,7
sum= 25, Odd

Took less than a minute to find such series will be multiple of 5.

Now see why ?

In another way this series can be re-written as (a-2),(a-1),a,(a+1),(a+2) , where a is an integer

Sum = 5a ...... Easy isnt it ? whatever may the value of 'a' be the sum of the given series will be a multiple of '5'

ScottTargetTestPrep · Answer

We can let our integers be the following:

n, n + 1, n + 2, n + 3, n + 4

Summing them, we have:

n + n + 1 + n + 2 + n + 3 + n + 4 = 5n + 10

Since the sum is 5n + 10, we see that if n were an odd number, then 5n + 10 would be odd; however, if n were even, 5n + 10 would be even. Thus, I or II do not have to be true. However, regardless of the value of n, we see that 5n + 10 is divisible by 5:

(5n + 10)/5 = 5n/5 + 10/5 = n + 2

Thus, III must be true.

Answer: C

Shiv2016 · Answer

The 5 nos. are:
a,(a+1),(a+2),(a+3),(a+4)
Add them, we get:
5a+10=  5(a+2)

- If a is even, the expression would be even. (5*even)
- If a is odd, the expression would be odd. (5*odd)
 But whatever the final answer would be, it would be a multiple of 5 because 5 is a factor here.

BrentGMATPrepNow · Answer

Let n = the smallest of the five integers
So, n + 1 = the next integer
So, n + 2 = the next integer
So, n + 3 = the next integer
So, n + 4 = the last integer

So the sum of all five integers = n + (n + 1) + (n + 2) + (n + 3) + (n + 4) =  5n + 20  
In other words,  x = 5n + 20

Now let's analyze each statement...

I. x is an even number
If n = 1, then  x = 5n + 20 = 5(1) + 20 = 25 , and 25 is not even. 
So, statement I is not necessarily true, which means we can eliminate answer choices A and D

II. x is an odd number
If n = 2, then  x = 5n + 20 = 5(2) + 20 = 30 , and 30 is not odd. 
So, statement II is not necessarily true, which means we can eliminate answer choices B and E

Important: As you can see, we've already determined that the correct answer is C, without even analyzing statement III. 
So, when answering Roman numeral questions such as this, be sure to eliminate answer choices along the way.

Let's analyze statement III for "fun".

III. x is a multiple of 5
We already determined that  x = 5n + 20  
We can factor the right hand side as follows:  x =  5 (n + 4) 
So, we can see that x is definitely divisible by   5  , which means statement III is true.

Answer: C

pepsi123 · Answer

Is this question not flawed because it does not take into account NEGATIVE integers?

i.e. -2, -1, 0, 1, 2 = 0.

0 is not a multiple of 5?

BrentGMATPrepNow 
 ScottTargetTestPrep

Bunuel · Answer

Check the  highlighted  part below:

PROPERTIES OF ZERO:

1. Zero is an INTEGER.

2. Zero is an EVEN integer. An even number is an   integer   that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

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

4. Zero is divisible by EVERY integer except 0 itself (x/0 = 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^n = 1/0^(-n) = 1/0. You CANNOT take 0 to the negative power.

12. 0! = 1! = 1.

P.S. Solution  HERE  does consider negative integers.

Hope it helps.

Abhishek009 · Answer

Plug in Nos and check....

Case  I

Let the  5  nos be  1, 2 , 3 , 4 , 5

Sum is  1 + 2 + 3 + 4 + 5 = 15

Case  II

Let the  5  nos be  2 , 3 , 4 , 5 , 6

Sum is  2 + 3 + 4 + 5 + 6 = 20

Check ,  x  can be both Odd and Even, but the result is always divisible by  5 , Answer must be (C) III only

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