Last visit was: 03 Sep 2026, 15:05 It is currently 03 Sep 2026, 15:05
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
kevincan
User avatar
GMAT Instructor
Joined: 04 Jul 2006
Last visit: 03 Sep 2026
Posts: 2,462
Own Kudos:
4,868
 [16]
Given Kudos: 449
GMAT 1: 790 Q51 V51
GRE 1: Q170 V170
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
GMAT 1: 790 Q51 V51
GRE 1: Q170 V170
Posts: 2,462
Kudos: 4,868
 [16]
Kudos
Add Kudos
16
Bookmarks
Bookmark this Post
User avatar
Jash1666
Joined: 17 Jul 2024
Last visit: 01 Sep 2026
Posts: 8
Own Kudos:
Given Kudos: 57
Posts: 8
Kudos: 1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
kevincan
User avatar
GMAT Instructor
Joined: 04 Jul 2006
Last visit: 03 Sep 2026
Posts: 2,462
Own Kudos:
Given Kudos: 449
GMAT 1: 790 Q51 V51
GRE 1: Q170 V170
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
GMAT 1: 790 Q51 V51
GRE 1: Q170 V170
Posts: 2,462
Kudos: 4,868
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
kevincan
User avatar
GMAT Instructor
Joined: 04 Jul 2006
Last visit: 03 Sep 2026
Posts: 2,462
Own Kudos:
4,868
 [2]
Given Kudos: 449
GMAT 1: 790 Q51 V51
GRE 1: Q170 V170
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
GMAT 1: 790 Q51 V51
GRE 1: Q170 V170
Posts: 2,462
Kudos: 4,868
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
If a set consists of n integers, k of which are odd, how many subsets of the set contain at least one even integer?

There are \(2^n\) subsets in total. Of these, \(2^k\) contain no even integers, since such a subset can contain only the \(k\) odd integers.

Therefore, the number of subsets containing at least one even integer is

\(2^n-2^k\).

In the problem, there are between 300 and 400 such subsets, so

\(300<2^n-2^k<400\).

Since \(2^n>300\), \(n\ge9\). If \(n=9\), then

\(2^9-2^k=512-2^k\).

For this to be between 300 and 400, \(2^k\) must be 128, giving \(k=7\).

Thus, the set contains 7 odd integers and \(9-7=2\) even integers.

For a 3-element subset to have an even sum, it must contain either:

* \(\binom{2}{3}=0\) (3 even integers)
* \(\binom{2}{1}\binom{7}{2}=2(21)=42\) (1 even and 2 odd integers)

Therefore,

\(42\)
User avatar
Jash1666
Joined: 17 Jul 2024
Last visit: 01 Sep 2026
Posts: 8
Own Kudos:
1
 [1]
Given Kudos: 57
Posts: 8
Kudos: 1
 [1]
Kudos
Add Kudos
Bookmarks
Bookmark this Post
For a subset having 3 elements with an even sum, it should contain 2 odd and 1 even combination only right? Where am I going wrong

for eg : sums of 3 odds 1+3+5 is odd and 2 even 1 odd 2+4+7 is odd
kevincan
If a set consists of n integers, k of which are odd, how many subsets of the set contain at least one even integer?

There are \(2^n\) subsets in total. Of these, \(2^k\) contain no even integers, since such a subset can contain only the \(k\) odd integers.

Therefore, the number of subsets containing at least one even integer is

\(2^n-2^k\).

In the problem, there are between 300 and 400 such subsets, so

\(300<2^n-2^k<400\).

Since \(2^n>300\), \(n\ge9\). If \(n=9\), then

\(2^9-2^k=512-2^k\).

For this to be between 300 and 400, \(2^k\) must be 128, giving \(k=7\).

Thus, the set contains 7 odd integers and \(9-7=2\) even integers.

A 3-element subset has an even sum if it contains either 3 odd integers or 2 even integers and 1 odd integer.

The number of such subsets is

\(\binom{7}{3}+\binom{2}{2}\binom{7}{1}\)

\(=35+7=42\).

Therefore, the answer is \(\boxed{42}\), choice D.
User avatar
kevincan
User avatar
GMAT Instructor
Joined: 04 Jul 2006
Last visit: 03 Sep 2026
Posts: 2,462
Own Kudos:
Given Kudos: 449
GMAT 1: 790 Q51 V51
GRE 1: Q170 V170
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
GMAT 1: 790 Q51 V51
GRE 1: Q170 V170
Posts: 2,462
Kudos: 4,868
Kudos
Add Kudos
Bookmarks
Bookmark this Post
You are right ! I used AI to write out the solution and it reversed it .
User avatar
HarshavardhanR
Joined: 16 Mar 2023
Last visit: 03 Sep 2026
Posts: 649
Own Kudos:
756
 [1]
Given Kudos: 99
Status:Independent GMAT Tutor
Affiliations: Ex - Director, Subject Matter Expertise at e-GMAT
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 649
Kudos: 756
 [1]
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let's first visualize the set A (set of distinct integers).

n -> number of elements (integers) in the set A, overall
k -> number of odd integers in the set A
n - k -> number of even integers in the set A

A -> { (.....k odd integers.....) (.....n-k even integers.....) }

Number of subsets of the set = \(2^n\) (this includes the null set i.e., nC0).

Number of subsets of A that contain no even integers = \(2^k\) (we have to ignore the even numbers - from the odd numbers, how many sets can be created is what will yield this).

Hence,

The number of subsets of A that contain at least one even integer = \(2^n\) - \(2^k\)

Given:

300 < \(2^n\) - \(2^k\) < 400

Analyzing this further to see what values n and k can take:

\(2^n\) > 300 + \(2^k\)

=> n >=9

Can n be 10?

\(2^n\) < 400 + \(2^k\)

1024 < 400 + \(2^k\)

\(2^k\) > 624

which means k>=10.

But this is a problem. If n = 10, k has to be lesser than 10. k>=10 is impossible.

In other words, we can see that n cannot be 10.

We can even check whether n = 11 works. Then again, we observe that k has to be >=11. Which is impossible.

Key Insight:
We already established that n>=9. But we can also see that n cannot be >9. Hence, n = 9.

So, given that n = 9,

512 > 300 + \(2^k\)

=> k <=7

512 < 400 + 2^k

=> k >=7

=> k = 7

Thus, in set A, we have 9 integers in total, out of which we have 7 odd numbers and 2 even numbers.

How many 3-element subsets of A are such that the sum of the 3 integers in the subset is even?

The only way to get an even sum is if 2 odd numbers and 1 even number are picked.

How many such subsets can be created?

7C2 x 2C1 = 42. Choice D.

---
Harsha
Moderator:
Math Expert
113098 posts