Which of the following must be true?

Question

1) The sum of N consecutive integers is always divisible by N
2) If N is even then the sum of N consecutive integers is divisible by N.
3) If N is odd then the sum of N consecutive integers is divisible by N.
4) The Product of K consecutive integers is divisible by K.
5) The product of K consecutive integers is divisible by K!

A) 1,4,5
B) 3,4,5
C) 4 and 5
D) 1,2,3,4
E) only 4

Note => Its not directly a part of egmat module. But it sure as hell will clear some concepts.

bhatiaabhishek91 · Answer

Sum=n/2  sum is always divisible by n irrespective of n being even or odd.
Product=a(a+1)(a+2)...(a+k-1) = (a+k-1)!/(a-1)! Product of k consecutive integers contains k terms with at least one multiple of k
Answer: 1,2,3,4 = D

Sent from my ONE E1003 using  GMAT Club Forum mobile app

sachinb3 · Answer

My Answer :3,4,5 =B

Sent from my 2014818 using  GMAT Club Forum mobile app

mbaprep2016 · Answer

first check option 1 
lets take n= 4 and numbes 1 2 3 4 sum will be 10 answer is no . Directly delete A and D
now check option 3 if true answer will be B if not check option 5 if true answer will be C . You dont need to check option 4 as it is in every other option 
check for option 3 first.
take n as 3 5 or 7 , check any series it will be true 
Answer is B

KarishmaB · Answer

Let's look at stmnt 1.
1) The sum of N consecutive integers is always divisible by N
The sum of 2 consecutive integers is divisible by 2? 2+3 = 5. No.
Two consecutive integers will add up to give an odd number (because even + odd = odd). So the sum will not be divisible by 2.
1 may not be true. So (A) and (D) are out.

All of (B), (C) and (E) have 4 so it must be true. Ignore it. The options differ in stmnts 3 and 5.

Let's look at stmnt 3.
3) If N is odd then the sum of N consecutive integers is divisible by N.
Sum of 3 consecutive integers will be divisible by 3? 3 consecutive integers will be of the form 3a, 3a+1 and 3a + 2 (in any order). So they will add up to 9a + 3 = 3(3a + 1) which is divisible by 3.
Sum fo 5 consecutive integers will be divisible by 5? 5 consecutive integers will be of the form 5a, 5a+1, 5a+2, 5a + 3 and 5a+4. The last 4 terms have 1, 2, 3 and 4 extra. Each pair of terms at the extreme will add up to give 5:
1+4 = 5
2+3 = 5
So the total sum of all 5 terms will be 25a + 5 + 5 (divisible by 5).
The case will be the same for all odd integers. 
Hence 3 will be true.

Answer (B)

msk0657 · Answer

Which of the following must be true.?
1) The sum of N consecutive integers is always divisible by N
2) If N is even then the sum of N consecutive integers is divisible by N.
3) If N is odd then the sum of N consecutive integers is divisible by N.
4) The Product of K consecutive integers is divisible by K.
5) The product of K consecutive integers is divisible by K!

A) 1,4,5
B) 3,4,5
C) 4 and 5
D) 1,2,3,4
E) only 4

Taken 2 mins for this question

1) The sum of N consecutive integers is always divisible by N

Plug in some numbers:

Lets take N as 1,2,3 and total 3 number 6/3 - correct
 N as 1,2,3 and 4 and total 4 numbers = 10/4 - but not exactly..

2) If N is even then the sum of N consecutive integers is divisible by N.

Lets N take as even number of cases .... 1,2,3,4 = 10 and the number of numbers is 4 = 10/4 - Not possible.

3) If N is odd then the sum of N consecutive integers is divisible by N.

Let's take N = 1,2 and 3 = sum of 6 and the number is 3 = 6/3 = 2 
 N =1,2,3,4 and 5 = sum = 15 and number is 5 = 15/5 = 3 ...divisbile..

4) The Product of K consecutive integers is divisible by K.

Let's take K = 1,2 and 3 and K = 3 then 6/3 = 2
 K = 1,2,3 and 4 and K = 4 then 24/4 = 6...divisible

5) The product of K consecutive integers is divisible by K!

Let's take K = 1,2 and 3 and K = 3 then 6/3! = 1
 K = 1,2,3 and 4 and K = 4 then 24/4! = 1...divisible

So 3,4 and 5 options are fine...

energetics · Answer

1) This not always true because there's issues around 0. 
1+2+3 = 6/3 ... or 3n+6 / 3 ✔
0+1+2+3 = 6/4 ... or 4n+6 / 4 ✘
0+1+2+3+4 = 10/5 ... or 5n+10 / 5 ✔

2) The rule is that for any set of consecutive ints with an EVEN number of items, the Sum is NEVER a multiple of n.
i.e. Sum/n will not be an int. 
2+3 / 2 = 5/2 ... or n+(n+1)/n = 2n+1 / n ✘

3) This works because it's the opposite of the above. For any set of consec ints with an ODD number of items, the Sum is ALWAYS a multiple of the number of items. 
3+4+5 / 3 = 12/3 ... or 3n+12 /3 ✔

4) This one works because when we multiply, the first int will always be a factor of itself, so product will always be divisible by that first int.
3*4*5 / 3 = 60/30 ... or n(n+1)(n+2) / n ✔

5) This one works also because the definition of n! is the product of all positive ints ≤ n. So, if we are dividing by n!, it means n must be positive and the product of some n consecutive ints is always divisible by all the its ints. 
3*2*1 / 3! ... or n(n-1)(n-2) / n! ✔

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Which of the following must be true?

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Which of the following must be true?

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards