Last visit was: 18 Nov 2025, 17:45 It is currently 18 Nov 2025, 17:45
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2   3 
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,073
 [1]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,073
 [1]
Fresh Meat!!! 18 Sep 2013, 03:34
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
muzammil
Bunuel

25*1=25 and 25*13=325;
25*3=75 and 25*11=275;
25*5=125 and 25*9=225.

Answer: C.

Hello, what about 175, 175 ? The question doesn't state they are distinct positive numbers.
Show more

The point is that if the two integers are 175 and 175, then their greatest common divisor going to be 175, not 25 as given in the stem.

Does this make sense?
User avatar
Transcendentalist
User avatar
Joined: 24 Nov 2012
Last visit: 04 Dec 2023
Posts: 130
Own Kudos:
1,059
Given Kudos: 73
Concentration: Sustainability, Entrepreneurship
Schools: CBS '17 (M) Kellogg '17 (A)
GMAT 1: 770 Q50 V44
WE:Business Development (Internet and New Media)
Schools: CBS '17 (M) Kellogg '17 (A)
GMAT 1: 770 Q50 V44
Posts: 130
Kudos: 1,059
Fresh Meat!!! 30 Sep 2013, 04:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.
Show more

I would like to request some some help with this question..

Could you please elaborate on the theory behind 2^n?
Isnt the null set considered to be a subset of this set?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,073
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,073
Fresh Meat!!! 30 Sep 2013, 05:41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Transcendentalist
Bunuel
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.

I would like to request some some help with this question..

Could you please elaborate on the theory behind 2^n?
Isnt the null set considered to be a subset of this set?
Show more

Consider simpler set: {a, b, c}. How many subsets does it have? Each of a, b, and c has two choices either to be included in subsets or not. Thus total of 2^38 subsets (including an empty set). The subsets are:

{a, b, c} --> each is included;
{a, b} --> a and b are included;
{a, c};
{b, c};
{a};
{b};
{c};
{} empty set, none is included.

2^3=8 subsets.

Harder question to practice about the same concept: how-many-subordinates-does-marcia-have-57169.html#p692676

Hope it helps.
User avatar
lindt123
User avatar
Joined: 19 Jul 2013
Last visit: 18 May 2014
Posts: 14
Own Kudos:
6
Given Kudos: 9
Location: United States
Concentration: Finance, International Business
GMAT 1: 340 Q27 V12
GPA: 3.33
GMAT 1: 340 Q27 V12
Posts: 14
Kudos: 6
Fresh Meat!!! 29 Dec 2013, 15:25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
bb
18! and 18!+1 are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Now, since we can factor out each 15, 17, 33=3*11, and 39=3*13 out of 18!, then 15, 17, 33 and 39 ARE factors of 18! and are NOT factors of 18!+1. Therefore only 19 could be a factor of 18!+1.

Answer: C
Show more

I Find it difficult to understand the explaination..is there any other way to solve this question... :(
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,073
 [19]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,073
 [19]
Fresh Meat!!! 30 Dec 2013, 01:39
4
Kudos
Add Kudos
15
Bookmarks
Bookmark this Post
lindt123
bb
18! and 18!+1 are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Now, since we can factor out each 15, 17, 33=3*11, and 39=3*13 out of 18!, then 15, 17, 33 and 39 ARE factors of 18! and are NOT factors of 18!+1. Therefore only 19 could be a factor of 18!+1.

Answer: C

I Find it difficult to understand the explaination..is there any other way to solve this question... :(
Show more

Frankly, solution provided is the easiest from my point of view.

We have that each 15, 17, 33, and 39 IS a factor of 18!, thus NOT a factor of 18!+1. Therefore, only 19 can be a factor of 18!+1. Since one of the options must be correct and we know that A, B, D and E are not, then C must be correct.

Does this make sense?

Check similar questions to practice:
if-a-and-b-are-odd-integers-a-b-represents-the-product-of-144714.html
is-x-1-a-factor-of-100740.html
if-x-and-y-are-positive-integers-what-is-the-gcf-of-x-and-y-144190.html
what-is-the-greatest-common-factor-of-positive-integers-a-126637.html
what-is-the-greatest-common-factor-of-x-and-y-109273.html
what-is-the-greatest-common-divisor-of-positive-integers-m-129802.html
x-and-y-are-positive-integers-such-that-x-8y-12-what-is-the-126743.html
gcd-of-a-b-126427.html
if-a-and-b-are-positive-integers-divisible-by-6-is-6-the-100324.html
if-a-and-b-are-positive-itegers-what-is-the-value-of-a-b-135199.html
is-x-1-a-factor-of-100740.html
find-the-number-that-divides-103251.html
if-n-is-the-least-of-3-consecutive-positive-integers-and-128102.html
if-both-x-and-y-are-positive-integers-that-are-divisible-by-119658.html
what-is-the-greatest-common-divisor-of-positive-integers-m-129802.html
for-every-positive-even-integer-n-the-function-h-n-is-defi-82314.html
if-a-b-and-c-are-positive-integers-with-a-b-c-85644.html
x-is-the-product-of-all-even-numbers-from-2-to-50-inclusive-156545.html
for-every-positive-even-integer-n-the-function-h-n-is-126691.html
for-every-positive-even-integer-n-the-function-h-n-149722.html
if-n-is-a-positive-integer-greater-than-1-then-p-n-represe-144553.html
if-x-y-and-z-are-3-different-prime-numbers-which-of-the-153338.html

Hope this helps.
User avatar
NGGMAT
User avatar
Joined: 20 Oct 2013
Last visit: 26 May 2014
Posts: 37
Own Kudos:
10
Given Kudos: 27
Posts: 37
Kudos: 10
Fresh Meat!!! 17 May 2014, 07:44
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.
Show more

Dear Bunnel

I didnt understand this.

y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,073
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,073
Fresh Meat!!! 17 May 2014, 07:51
Kudos
Add Kudos
Bookmarks
Bookmark this Post
NGGMAT
Bunuel
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.

Dear Bunnel

I didnt understand this.

y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from?
Show more

What do you mean by the red part?

As for 2^5: the number of subsets of n-element set is 2^n, thus the number of subsets of 5-element set {1, 2, 3, 4, 5} is 2^5 (note that this includes an empty set as well as the original set {1, 2, 3, 4, 5}). Now, all subsets of {1, 2, 3, 4, 5} are the subsets of {0, 1, 2, 3, 4, 5} and does not include 0.

Does this make sense?
User avatar
NGGMAT
User avatar
Joined: 20 Oct 2013
Last visit: 26 May 2014
Posts: 37
Own Kudos:
10
Given Kudos: 27
Posts: 37
Kudos: 10
Fresh Meat!!! 17 May 2014, 08:20
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
NGGMAT
Bunuel
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.

Dear Bunnel

I didnt understand this.

y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from?

What do you mean by the red part?

As for 2^5: the number of subsets of n-element set is 2^n, thus the number of subsets of 5-element set {1, 2, 3, 4, 5} is 2^5 (note that this includes an empty set as well as the original set {1, 2, 3, 4, 5}). Now, all subsets of {1, 2, 3, 4, 5} are the subsets of {0, 1, 2, 3, 4, 5} and does not include 0.

Does this make sense?
Show more

By red part i meant that y we havent solved it like we did the below qs:

2. Set S contains 7 different letters. How many subsets of set S, including an empty set, contain at most 3 letters?

A. 29
B. 56
C. 57
D. 63
E. 64

1 empty set;
C^1_7=7 sets with one element;
C^2_7=21 sets with two elements;
C^3_7=35 sets with three element.

Total 1+7+21+35=64 sets

y did 2^n come into qs 3 and not qs 2?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,073
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,073
Fresh Meat!!! 17 May 2014, 08:39
Kudos
Add Kudos
Bookmarks
Bookmark this Post
NGGMAT
Bunuel
NGGMAT

Dear Bunnel

I didnt understand this.

y didnt we take the set {1,2,3,4,5} into consideration and solve like the above qs? where did we get 2^5 from?

What do you mean by the red part?

As for 2^5: the number of subsets of n-element set is 2^n, thus the number of subsets of 5-element set {1, 2, 3, 4, 5} is 2^5 (note that this includes an empty set as well as the original set {1, 2, 3, 4, 5}). Now, all subsets of {1, 2, 3, 4, 5} are the subsets of {0, 1, 2, 3, 4, 5} and does not include 0.

Does this make sense?

By red part i meant that y we havent solved it like we did the below qs:

2. Set S contains 7 different letters. How many subsets of set S, including an empty set, contain at most 3 letters?

A. 29
B. 56
C. 57
D. 63
E. 64

1 empty set;
C^1_7=7 sets with one element;
C^2_7=21 sets with two elements;
C^3_7=35 sets with three element.

Total 1+7+21+35=64 sets

y did 2^n come into qs 3 and not qs 2?
Show more

We could use 2^n for the second question too:

{The number of subsets with 0, 1, 2, or 3 terms} = {The total # of subsets} - {Subsets with 4, 5, 6, or 7 elements} = \(2^7 - (C^4_7+C^5_7+C^6_7+C^7_7)=128-(35+21+7+1)=64\).

But as you can see this approach is longer than the one used in my solution for that question.
avatar
logophobic
avatar
Joined: 03 Sep 2014
Last visit: 09 Sep 2014
Posts: 5
Own Kudos:
14
Given Kudos: 10
Posts: 5
Kudos: 14
Fresh Meat!!! Updated on: 09 Sep 2014, 07:46
Originally posted by logophobic on 09 Sep 2014, 07:35.
Last edited by logophobic on 09 Sep 2014, 07:46, edited 3 times in total.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.
Show more

I used a choose method and got 31. With the binary method you used, there is a set which contains no numbers. This increases the number to 32. But the question doesn't say there can be a subset with no numbers because 0 is also not included... but I like the binary method & may use it on these kinds of problems & just subtract 1 if that is necessary so THANK YOU. :-D Thank you
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,073
 [1]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,073
 [1]
Fresh Meat!!! 09 Sep 2014, 07:38
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
logophobic
Bunuel
3. How many different subsets of the set {0, 1, 2, 3, 4, 5} do not contain 0?

A. 16
B. 27
C. 31
D. 32
E. 64

Consider the set without 0: {1, 2, 3, 4, 5}. Each out of 5 elements of the set {1, 2, 3, 4, 5} has TWO options: either to be included in the subset or not, so total number of subsets of this set is 2^5=32. Now, each such set will be a subset of {0, 1, 2, 3, 4, 5} and won't include 0.

Answer: D.

I used a choose method and got 31. With the binary method you used, there is a set which contains no numbers. This increases the number to 32. But the question doesn't say there can be a subset with no numbers because 0 is also not included...
Show more

Empty set is a subset of all non-empty sets.
User avatar
Sujithz001
User avatar
Joined: 09 Jun 2024
Last visit: 10 Nov 2025
Posts: 75
Own Kudos:
26
Given Kudos: 73
Posts: 75
Kudos: 26
Fresh Meat!!! 03 Oct 2024, 09:16
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hey Bunuel

I'm having a hard time understanding this line of your soln.,

"Since we can factor out each of the numbers 15, 17, 33 (3*11), and 39 (3*13) from \(18!\), these numbers are indeed factors of \(18!\) but cannot be factors of \(18!+1\). Therefore, only 19 could be a factor of \(18!+1\)."

Can you elaborate on this please?

Thanks a lot!
.
Bunuel
5. Which of the following is a factor of 18!+1?

A. 15
B. 17
C. 19
D. 33
E. 39

\(18!\) and \(18!+1\) are consecutive integers. Two consecutive integers are always co-prime, meaning that they don't share any common factors other than 1. For example, 20 and 21 are consecutive integers, and the only common factor they share is 1.

Since we can factor out each of the numbers 15, 17, 33 (3*11), and 39 (3*13) from \(18!\), these numbers are indeed factors of \(18!\) but cannot be factors of \(18!+1\). Therefore, only 19 could be a factor of \(18!+1\).

Answer: C.

Answer: C
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,073
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,073
Fresh Meat!!! 03 Oct 2024, 09:22
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Sujithz001
Hey Bunuel

I'm having a hard time understanding this line of your soln.,

"Since we can factor out each of the numbers 15, 17, 33 (3*11), and 39 (3*13) from \(18!\), these numbers are indeed factors of \(18!\) but cannot be factors of \(18!+1\). Therefore, only 19 could be a factor of \(18!+1\)."

Can you elaborate on this please?

Thanks a lot!
.
Bunuel
5. Which of the following is a factor of 18!+1?

A. 15
B. 17
C. 19
D. 33
E. 39

\(18!\) and \(18!+1\) are consecutive integers. Two consecutive integers are always co-prime, meaning that they don't share any common factors other than 1. For example, 20 and 21 are consecutive integers, and the only common factor they share is 1.

Since we can factor out each of the numbers 15, 17, 33 (3*11), and 39 (3*13) from \(18!\), these numbers are indeed factors of \(18!\) but cannot be factors of \(18!+1\). Therefore, only 19 could be a factor of \(18!+1\).

Answer: C.

Answer: C
Show more

I tried elaborating this here: https://gmatclub.com/forum/fresh-meat-1 ... l#p1311183

Please let me know if it helps.
User avatar
Nivev
User avatar
Joined: 07 Aug 2024
Last visit: 22 Feb 2025
Posts: 7
Own Kudos:
1
Given Kudos: 41
Posts: 7
Kudos: 1
Fresh Meat!!! 10 Jan 2025, 01:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
7. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

A. 1
B. 2
C. 3
D. 4
E. 5

We are told that the greatest common factor of two integers is 25. So, these integers are \(25x\) and \(25y\), for some positive integers \(x\) and \(y\). Notice that \(x\) and \(y\) must not share any common factor but 1, because if they do, then the greatest common factor of \(25x\) and \(25y\) will be more than 25.

Next, we know that \(25x+25y=350\). Reducing by 25 gives \(x+y=14\). Now, since \(x\) and \(y\) don't share any common factor but 1, the possible pairs \((x, y)\) can only be (1, 13), (3, 11), or (5, 9) (all other pairs like (2, 12), (4, 10), (6, 8), and (7, 7) share a common factor greater than 1).

So, there are only three pairs of such numbers possible:

\(25*1=25\) and \(25*13=325\);

\(25*3=75\) and \(25*11=275\);

\(25*5=125\) and \(25*9=225\).

Answer: C.
Show more
I have this genuine doubt. It has not mentioned x<y. So as far as my brain goes, if (1,13) is a pair, then (13,1) has to be one too which makes it a total of 6 pairs. I can feel I am missing something like a keyword or misconstruing the meaning of "how many such pairs". Kindly helps me with how to interpret this question or such questions.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,073
 [2]
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,073
 [2]
Fresh Meat!!! 10 Jan 2025, 01:29
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Nivev
Bunuel
7. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

A. 1
B. 2
C. 3
D. 4
E. 5

We are told that the greatest common factor of two integers is 25. So, these integers are \(25x\) and \(25y\), for some positive integers \(x\) and \(y\). Notice that \(x\) and \(y\) must not share any common factor but 1, because if they do, then the greatest common factor of \(25x\) and \(25y\) will be more than 25.

Next, we know that \(25x+25y=350\). Reducing by 25 gives \(x+y=14\). Now, since \(x\) and \(y\) don't share any common factor but 1, the possible pairs \((x, y)\) can only be (1, 13), (3, 11), or (5, 9) (all other pairs like (2, 12), (4, 10), (6, 8), and (7, 7) share a common factor greater than 1).

So, there are only three pairs of such numbers possible:

\(25*1=25\) and \(25*13=325\);

\(25*3=75\) and \(25*11=275\);

\(25*5=125\) and \(25*9=225\).

Answer: C.
I have this genuine doubt. It has not mentioned x<y. So as far as my brain goes, if (1,13) is a pair, then (13,1) has to be one too which makes it a total of 6 pairs. I can feel I am missing something like a keyword or misconstruing the meaning of "how many such pairs". Kindly helps me with how to interpret this question or such questions.
Show more

If the question had asked for "ordered pairs," then both (1, 13) and (13, 1) would be counted separately, resulting in 6 pairs. However, without that, (1, 13) and (13, 1) are essentially the same pair: 1 and 13.
User avatar
Krunaal
User avatar
Tuck School Moderator
Joined: 15 Feb 2021
Last visit: 17 Nov 2025
Posts: 805
Own Kudos:
850
 [1]
Given Kudos: 251
Status:Under the Square and Compass
Location: India
Schools: Booth '26 Kellogg '26 Johnson '26 Tuck '26 Duke '26 Ross '26 HBS '26 Stanford '26 CBS '26 Stern '26 Kellogg Darden '26 Yale '26 Wharton '26
GMAT Focus 1: 755 Q90 V90 DI82
GPA: 5.78
WE:Marketing (Consulting)
Products:
My Reviews
Schools: Booth '26 Kellogg '26 Johnson '26 Tuck '26 Duke '26 Ross '26 HBS '26 Stanford '26 CBS '26 Stern '26 Kellogg Darden '26 Yale '26 Wharton '26
GMAT Focus 1: 755 Q90 V90 DI82
Posts: 805
Kudos: 850
 [1]
Fresh Meat!!! 10 Jan 2025, 01:33
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Nivev
Bunuel
7. The greatest common divisor of two positive integers is 25. If the sum of the integers is 350, then how many such pairs are possible?

A. 1
B. 2
C. 3
D. 4
E. 5

We are told that the greatest common factor of two integers is 25. So, these integers are \(25x\) and \(25y\), for some positive integers \(x\) and \(y\). Notice that \(x\) and \(y\) must not share any common factor but 1, because if they do, then the greatest common factor of \(25x\) and \(25y\) will be more than 25.

Next, we know that \(25x+25y=350\). Reducing by 25 gives \(x+y=14\). Now, since \(x\) and \(y\) don't share any common factor but 1, the possible pairs \((x, y)\) can only be (1, 13), (3, 11), or (5, 9) (all other pairs like (2, 12), (4, 10), (6, 8), and (7, 7) share a common factor greater than 1).

So, there are only three pairs of such numbers possible:

\(25*1=25\) and \(25*13=325\);

\(25*3=75\) and \(25*11=275\);

\(25*5=125\) and \(25*9=225\).

Answer: C.
I have this genuine doubt. It has not mentioned x<y. So as far as my brain goes, if (1,13) is a pair, then (13,1) has to be one too which makes it a total of 6 pairs. I can feel I am missing something like a keyword or misconstruing the meaning of "how many such pairs". Kindly helps me with how to interpret this question or such questions.
Show more

Since the problem focuses on the resulting integers, not the intermediate pairs, only unique combinations are counted. The focus should be on the two integers formed at the end of the process, not on the intermediate values of x and y.
User avatar
argha3110
User avatar
Joined: 20 May 2024
Last visit: 18 Nov 2025
Posts: 6
Given Kudos: 56
Posts: 6
Kudos: 0
Fresh Meat!!! 27 Jul 2025, 01:01
Kudos
Add Kudos
Bookmarks
Bookmark this Post
In Q.10, shouldn't the answer be C since if we get either x=1 or y=0, we get x^y=1?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,073
Given Kudos: 99,964
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,355
Kudos: 778,073
Fresh Meat!!! 27 Jul 2025, 03:25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
argha3110
In Q.10, shouldn't the answer be C since if we get either x=1 or y=0, we get x^y=1?
Show more

No, the answer should be and is E. Check the solution here: https://gmatclub.com/forum/fresh-meat-1 ... l#p1215370 Hope it helps.
User avatar
WhitEngagePrep
User avatar
Joined: 12 Nov 2024
Last visit: 08 Oct 2025
Posts: 59
Own Kudos:
48
Given Kudos: 19
Location: United States
Expert
Expert reply
Posts: 59
Kudos: 48
Fresh Meat!!! 27 Jul 2025, 17:26
Kudos
Add Kudos
Bookmarks
Bookmark this Post
argha3110
In Q.10, shouldn't the answer be C since if we get either x=1 or y=0, we get x^y=1?
Show more
While the formal definition is fine, the fact that the question says must be true says that Testing Cases is also a great method for this problem.

So let's consider some different cases where X is not equal to 0, and x^y=1. There are some of the more obvious ones (and a scan of the roman numerals will give these to you.. base of 1 and power of 0). So imagine both of those cases as distinct since either would result in x^y = 1.

x = 1, y = 5, 1^5 = 1.
Eliminate II since y doesn't have to be 0

x = 5, y = 0, 5^0 = 1.
Eliminate I since x doesn't have to equal 1.

But are there any cases where x and y could both be something different, such that neither x = 1 and y = 0? What about negatives? Those aren't eliminated! And when you raise negatives to even powers it turns them positive!

x = -1, y = 2, (-1)^2 = 1.
Eliminate III since neither x = 1 nor y = 0.

The only choice left is E!

Hope this helps!
:)
Whit
   1   2   3 
Moderators:
Bunuel
Math Expert
105355 posts
Krunaal
Tuck School Moderator
805 posts