The sum of the first hundred positive integers is divisible by?

Question

I. 2
II. 4
III. 8

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

The sum of first hundred numbers, 1 to 100 is divisible by?

yashikaaggarwal · Accepted Answer

=> Sum of first n positive integer is = n(n+1)/2 
=> Sum of first 100 positive integer is = 100(100+1)/2 
=> 100*101/2 
=> 5050

factors of 5050 = 2*5*5*101

among all factors only 2 is the only even prime factor of 5050,
therefore 5050 is divisible by 2 only

therefore, only statement 1 is true.

IMO A

rajatchopra1994 · Answer

Explanation:

Sum of First 100 Positive number: 100x{(1+100)/2} = 5050
Now checking, it is only divisible by 2.
So, IMO A

Kinshook · Answer

The sum of the first hundred positive integers is divisible by?

I. 2
II. 4
III. 8

The sum of the first hundred positive integers = 1 + 2 + ... + 100 = 100*101/2 = 50*101 = 5050 = 2*5*101*5

IMO A

GMATWhizTeam · Answer

Solution

• The sum of first hundred positive integer  = \frac{100*(100+1)}{2} = 50*101 = 2*5^2*101 
• Therefore, out of the given choices, the sum of first hundred positive integers is divisible by 2 only.

Thus, the correct answer is  Option A.

kawal27 · Answer

n= total terms in sequence . a= first term l= last term
Sn = n/2 ( a+l) 
 = 100/2(1+100)
 = 50(101)
 = 5050 
 
5050 is only divisible by 2
ANS:A

suchita2409 · Answer

IMO A

S = (n/2)(first term+lastterm)

(100+1)×100/2

= 5050.

divisible by 2 onlyy

A

Posted from my mobile device

GMATinsight · Answer

Some relevant properties
   CONCEPT:  Sum of n consecutive positive integers,  ∑n = (\frac{1}{2})n(n+1) 
Sum of n squares,  ∑n^2 = (\frac{1}{6})n(n+1)*(2n+1)

i.e. Sum of 1 to 100,  Sum = ∑100 = (\frac{1}{2})*100*(100+1) = 50*101

The sum of divisible by  2, 5^2 , and  101    but not by 4 and 8

Answer: Option A

rcgard · Answer

The sum of the first hundred positive integers is divisible by?

I. 2
II. 4
III. 8

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III
Solution:

Sum =  \frac{(100)(2+99)}{(2)}  = 5050

5050 only divisible by 2 out of given options

Answer choice A IMO

ScottTargetTestPrep · Answer

Solution:

The sum of the first hundred positive integers is:

(1 + 100)/2 x 100 = 101 x 50 = 5050

We see that 5050 is divisible by 2, but not by 4 and not by 8.

Answer: A

ArnauG · Answer

To determine if the sum of the first hundred positive integers is divisible by 2, 4, or 8, we can analyze the pattern of the sum.

The sum of the first n positive integers can be calculated using the formula: sum = n * (n + 1) / 2.

Let's consider each option:

I. 2: If a number is divisible by 2, it means it is even. The sum of the first hundred positive integers is 100 * (100 + 1) / 2 = 5050. Since 5050 is an even number, it is divisible by 2.

II. 4: If a number is divisible by 4, it means it is even and divisible by 2 twice. The sum of the first hundred positive integers is 5050. Dividing 5050 by 2 gives us 2525. Since 2525 is not divisible by 2, the sum is not divisible by 4.

III. 8: If a number is divisible by 8, it means it is even and divisible by 2 three times. The sum of the first hundred positive integers is 5050. Dividing 5050 by 2 repeatedly gives us 2525, 1262, and 631. Since 631 is not divisible by 2, the sum is not divisible by 8.

From the analysis, we can see that the sum of the first hundred positive integers is divisible by 2 (option I) but not divisible by 4 (option II) or 8 (option III).

Therefore, the answer is (A) I only.

Debaditya123 · Answer

how are we deriving the formula for Sum of first n positive integer is => n(n+1)/2?

MartyMurray · Answer

(Sum of the Integers in a Set) = (Number of Integers in the Set) × (Average of the Integers in the Set)

In the specific case of the first  n  positive integers, we have the following:

Number of Integers in the Set:  n

Average of the Integers in the Set: ( n  + 1)/2

The reason the average is ( n  + 1)/2 is that the first  n  integers are evenly spaced integers. So, the average of all of them is the same as the average of the first and last of them, which are 1 and  n .

Thus, in the specific case of the first  n  integers, we have the following:

(Sum of the First  n  Integers) = ( n ) × (The Average of the First  n  Integers) =  n  × ( n  + 1)/2

The sum of the first hundred positive integers is divisible by?

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

The sum of the first hundred positive integers is divisible by?

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