If N different positive integers are added and the sum is denoted as S

Question

If N different positive integers are added and the sum is denoted as S, and if S=NX, is X an integer?

(1) N is odd.

(2) All N numbers are consecutive integers.

Kudos for a correct  solution .

gmat6nplus1 · Answer

statement 1: pick 1, 2, 3. Sum=6=3x and x is an integer. pick a set of odd prime numbers excluding 3 such as 7, 17, 19 or 13, 17, 5 sum is 35=3x and x is not an integer.

statement 2: pick 1, 2, 3, 4 sum=10=4x and x is not an integer. pick 11, 12, 13 sum=avg(n)=36=3x and x is an integer.

1+2) the average of an odd number of consecutive integer is equal to its median. In an evenly spaced set with odd number of elements the sum will always be a multiple of the average, which is always going to be an integer because the number of elements in the set is odd. So x is our average/median and it is always going to be integer.

Answer C.

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION

Statement (1): N is Odd. If the numbers are 1,2 and 3 then 6=3 * 2 here X is an integer. and if we take 2, 3, 5, then 10=3 * (10/3) here X is a fraction. Insufficient.

Statement (2): All the N numbers are consecutive integers.

If N is even, and if we take 1 and 2, then S=3 = 2 * 1.5 so X is a fraction. if N is odd, and if we take 1,2 and 3, then S = 6 = 3 * 2, so X is a integer.

Insufficient.

If we combine both the statements, then N is odd and all the numbers are consecutive integers, in that case X has to be positive integer. Hence Option C.

Lucky2783 · Answer

stmt 1: 
 1,2,3 
 6=3X; X=2 
 2,4,8
 14=3X; X=Fraction
 NS
stmt 2: 
 1,2,3 
 6=3X; X=2 
 7,8,9,10
 34=4X; X= Fraction
 NS

combine 1&&2 
If N is odd and we have consecutive integers 
2 possibilities:
sequence starts with a EVEN integer
eg:
2K + 2K+1 + 2K+2 = 6K+3 is divisible by 3 
sequence starts with a ODD integer
(2K-1) + (2K) + (2K+1) = 6K is divisible by 3

we can test this with N=4 as well.

Answer : C

Lucky2783 · Answer

Hi Bunuel,

could you please shed some light on the highlighted portion of your explanation for this question ?

If we combine both the statements, then N is odd and all the numbers are consecutive integers, in that case X has to be positive integer.  Hence Option C.

How do we know this is the case ? ( I know this is the case and i solved it using examples 2k, 2k+1.. etc. not sure how did you get this ).

Regards,
Lucky

Bunuel · Answer

The question basically asks whether S (the sum of N integers) is divisible by N (the number of integers).

Properties of consecutive integers:
• If n is odd, the sum of n consecutive integers is always divisible by n. Given \(\{9,10,11\}\), we have \(n=3=odd\) consecutive integers. The sum is 9+10+11=30, which is divisible by 3.
• If n is even, the sum of n consecutive integers is never divisible by n. Given \(\{9,10,11,12\}\), we have \(n=4=even\) consecutive integers. The sum is 9+10+11+12=42, which is NOT divisible by 4.

(1) says that n is odd and (2) says that the numbers are consecutive, hence the sum of N integers, S, must be divisible by the number of untergers, N.

Hope it's clear.

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Gotcha !!
thanks .

Lucky2783 · Answer

The question basically asks whether S (the sum of N integers) is divisible by N (the number of integers).

Properties of consecutive integers:
• If n is odd, the sum of n consecutive integers is always divisible by n. Given \(\{9,10,11\}\), we have \(n=3=odd\) consecutive integers. The sum is 9+10+11=30, which is divisible by 3.
• If n is even, the sum of n consecutive integers is never divisible by n. Given \(\{9,10,11,12\}\), we have \(n=4=even\) consecutive integers. The sum is 9+10+11+12=42, which is NOT divisible by 4.

(1) says that n is odd and (2) says that the numbers are consecutive, hence the sum of N integers, S, must be divisible by the number of untergers, N.

Hope it's clear.

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Gotcha !!
thanks .

matt882 · Answer

I hope this will make the solution faster and simpler, but remember the magic formula of a sum of consecutive numbers: #numbers*(num_min+num_max)/2

1) clearly not suff: S can be the sum of any number
2) 1+2=3 -> 3/2 not an integer II 1+2+3= 6 -> 6/3 integer

1&2

Since number consecutive S=N*(n1+nN)/2 = NX
Since N is odd n1+nN is even and so (n1+nN)/2= X = integer
Suff.

bumpbot · Answer

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If N different positive integers are added and the sum is denoted as S

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If N different positive integers are added and the sum is denoted as S

Prep Toolkit

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