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

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sachinb3 · Answer

My Answer :3,4,5 =B

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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! ✔

onlyPlanA · Answer

IMO, the trap in this question lays in the fact that options 1,2,3 may not be an AP of "first" n consecutive numbers.

Which of the following must be true?

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Which of the following must be true?

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