Last visit was: 19 Nov 2025, 14:22 It is currently 19 Nov 2025, 14:22
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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,361
 [32]
Given Kudos: 99,977
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: 105,390
Kudos: 778,361
 [32]
HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 28 Aug 2020, 08:00
1
Kudos
Add Kudos
31
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

85% (hard)

Question Stats:

50% (02:11) correct 50% (02:33) wrong based on 611 sessions

History

Date
Time
Result
Not Attempted Yet
When the digits of two-digit positive integer M are reversed, the result is two-digit positive integer N. Which of the following cannot be a factor of M^2+N^2?

A. 17
B. 29
C. 40
D. 73
E. 101


 

This question was provided by Veritas Prep
for the Heroes of Timers Competition

 

ShowHide Answer
Official Answer
E
Most Helpful Reply
User avatar
Saupayan
User avatar
Joined: 30 May 2013
Last visit: 23 May 2025
Posts: 108
Own Kudos:
137
 [13]
Given Kudos: 94
Status:Full-time employee
Affiliations: Apple Inc
Location: United States
Saupayan: Mazumdar
Concentration: Economics, Leadership
GMAT 1: 760 Q51 V41
GRE 1: Q170 V160
GPA: 3.89
WE:Engineering (Computer Hardware)
GMAT 1: 760 Q51 V41
GRE 1: Q170 V160
Posts: 108
Kudos: 137
 [13]
HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 28 Aug 2020, 08:33
9
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
Quote:
When the digits of two-digit positive integer M are reversed, the result is two-digit positive integer N. Which of the following cannot be a factor of M^2+N^2?

A. 17
B. 29
C. 40
D. 73
E. 101
Show more

Let M = 10x+y and N = 10y + x (where x and y are single digit whole numbers)
=> M^2 + N^2 = 101(x^2+y^2) + 40xy

Now, let's look at the answer choices. 40 and 101 stand out because they appear in our equation as well.
Let's consider 101... it divides the first term ( 101(x^2+y^2) )completely (without remainder). So, the only way the entire number will be divisible by 101 is if the 2nd term (40xy) is ALSO divisible by 101. 40 and 101 have no common factors because 101 is a prime number. Hence, the only way for 40xy to be divisible by 101 is if xy is divisible by 101. That will require x = 101 and y = 1 or vice-versa. Now, either of those options aren't possible because x and y are single digit numbers. So, 101 is definitely not a factor of M^2 + N^2.

If we do a similar test for 40... (this is just for understanding, and not a required step because we already found our answer)
The 2nd term is divisible.
For the first term to be divisible as well, (x^2 + y^2) has to be divisible by 40. A possibility is (2,6). Another is (6,2). Either of these will satisfy all the required conditions without violating any assumptions.
Hence, 40 can be a factor

For the other answer choices, neither 101 nor 40 is divisible directly. So, we may be able to find combination of x and y to satisfy the divisibility overall. But more importantly, we DON'T HAVE TO FIND ANY OF THOSE. Reason: This is a single option correct question! Once we found 101 is not a factor, we are done!!

Answer: E
User avatar
aviddd
User avatar
Joined: 26 Jul 2020
Last visit: 10 Mar 2021
Posts: 205
Own Kudos:
221
 [11]
Given Kudos: 66
Products:
My Reviews
Posts: 205
Kudos: 221
 [11]
HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 28 Aug 2020, 08:17
9
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Let M = xy, N = yx.

M = 10x + y. N = 10y + x

m^2 = (10x + y) ^ 2 = 100x^2 + y^2 + 20xy
n^2 = (10y + x) ^ 2 = 100y^2 + x^2 + 20xy

m^2 + n^ 2 = 101y^2 + 101x^2 + 40xy

The first two are multiples of 101. Since 40 is not a factor of 101, the third term can never be a multiple of 101. So, as a whole, this number cannot be divisible by 101.

Answer:E.

Bunuel, I enjoy your math problems. I only wish I was able to solve them in <2 minutes.
General Discussion
avatar
TarunKumar1234
avatar
Joined: 14 Jul 2020
Last visit: 28 Feb 2024
Posts: 1,107
Own Kudos:
1,348
 [2]
Given Kudos: 351
Location: India
Posts: 1,107
Kudos: 1,348
 [2]
HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte Updated on: 29 Aug 2020, 10:36
Originally posted by TarunKumar1234 on 28 Aug 2020, 08:21.
Last edited by TarunKumar1234 on 29 Aug 2020, 10:36, edited 1 time in total.
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
M= 10x+y and N= 10y+x

M^2 + N^2= (10x+y)^2 + (10y+x)^2= 100x^2 + y^2 + 20xy + 100y^2 + x^2 + 20xy = 101 (x^2 + y^2) + 40xy

So, we can see 40xy can't have 101 factor.

Ans. E
avatar
Tanushka77
avatar
Current Student
Joined: 10 Jul 2019
Last visit: 13 Apr 2021
Posts: 21
Own Kudos:
96
 [4]
Given Kudos: 30
Schools: Darden '23 (A)
Schools: Darden '23 (A)
Posts: 21
Kudos: 96
 [4]
HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 28 Aug 2020, 08:26
4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
When the digits of two-digit positive integer M are reversed, the result is two-digit positive integer N. Which of the following cannot be a factor of M^2+N^2?
Let \(M=10a+b\) and \(N=10b+a\)
Then \((10a+b)^2+(10b+a)^2=\\
=101b^2+101a^2+40ab=\\
=101(a^2+b^2)+40ab\)
As we can see the first part is dicisible by 101. However, the while sum cannot be divisible by because of 40ab. 101 is a prime and it is impossible to have in a product of 40ab since a and B are sinle digits.
The answer is E

Posted from my mobile device
User avatar
rajatchopra1994
User avatar
Joined: 16 Feb 2015
Last visit: 22 Jun 2024
Posts: 1,056
Own Kudos:
1,252
 [1]
Given Kudos: 30
Location: United States
Posts: 1,056
Kudos: 1,252
 [1]
HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 28 Aug 2020, 21:46
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Explanation:

Let, M=xy then N=yx
M = 10x + y
N = 10y + x

M^2 + N^2 = (10x+y)^2 + (10y+x)^2
= 100x^2 + y^2 + 20xy + 100y^2 + x^2 + 20xy
= 101x^2 + 101y^2 + 40xy
= 101(x^2 + y^2) + 40xy
= 101{(x+y)^2 - 2xy} + 40xy

But clearly we can see 101 cannot be factor

101 can be written as 10^2 + 1^2 (x,y: 10,1) but x,y has to be single integer no other values are possible.


Now we check all options

A. 17 : (4^2 ; 1^2) Yes possible
B. 29 : (5^2 ; 2^2) Yes possible
C. 40 : (6^2 ; 2^2) Yes possible
D. 73 : (8^2 ; 3^2) Yes possible
E. 101 not possible as explained above.

IMO-E

Posted from my mobile device
User avatar
exc4libur
User avatar
Joined: 24 Nov 2016
Last visit: 22 Mar 2022
Posts: 1,684
Own Kudos:
1,447
 [1]
Given Kudos: 607
Location: United States
Posts: 1,684
Kudos: 1,447
 [1]
HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 29 Aug 2020, 05:46
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Quote:
When the digits of two-digit positive integer M are reversed, the result is two-digit positive integer N. Which of the following cannot be a factor of M^2+N^2?

A. 17
B. 29
C. 40
D. 73
E. 101
Show more

m=10a+b, n=10b+a, (a,b)>0

(10a+b)^2+(10b+a)^2
100a^2+b^2+20ab+100b^2+a^2+20ab
101a^2+101b^2+40ab
101 cannot be a factor

(E)
User avatar
Crytiocanalyst
User avatar
Joined: 16 Jun 2021
Last visit: 27 May 2023
Posts: 950
Own Kudos:
208
Given Kudos: 309
Posts: 950
Kudos: 208
HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 27 Sep 2021, 07:09
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
When the digits of two-digit positive integer M are reversed, the result is two-digit positive integer N. Which of the following cannot be a factor of M^2+N^2?

A. 17
B. 29
C. 40
D. 73
E. 101


 

This question was provided by Veritas Prep
for the Heroes of Timers Competition

 

Show more

Simplification gives us 101(x^2 +y^2) + 40*x*y

101 is a primw number for the second term to be divisible we need x*y to be divisible by 101 which is not possible since either x and y needs to 101
which is not possible since x,y represents digits and it's single correct optionn we have arrived at the answer

Therefore IMO E
avatar
NamaySharma
avatar
Joined: 21 Mar 2021
Last visit: 31 Dec 2021
Posts: 24
Own Kudos:
21
Given Kudos: 26
Posts: 24
Kudos: 21
HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 30 Nov 2021, 22:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Saupayan
Quote:
When the digits of two-digit positive integer M are reversed, the result is two-digit positive integer N. Which of the following cannot be a factor of M^2+N^2?

A. 17
B. 29
C. 40
D. 73
E. 101

Let M = 10x+y and N = 10y + x (where x and y are single digit whole numbers)
=> M^2 + N^2 = 101(x^2+y^2) + 40xy

Now, let's look at the answer choices. 40 and 101 stand out because they appear in our equation as well.
Let's consider 101... it divides the first term ( 101(x^2+y^2) )completely (without remainder). So, the only way the entire number will be divisible by 101 is if the 2nd term (40xy) is ALSO divisible by 101. 40 and 101 have no common factors because 101 is a prime number. Hence, the only way for 40xy to be divisible by 101 is if xy is divisible by 101. That will require x = 101 and y = 1 or vice-versa. Now, either of those options aren't possible because x and y are single digit numbers. So, 101 is definitely not a factor of M^2 + N^2.

If we do a similar test for 40... (this is just for understanding, and not a required step because we already found our answer)
The 2nd term is divisible.
For the first term to be divisible as well, (x^2 + y^2) has to be divisible by 40. A possibility is (2,6). Another is (6,2). Either of these will satisfy all the required conditions without violating any assumptions.
Hence, 40 can be a factor

For the other answer choices, neither 101 nor 40 is divisible directly. So, we may be able to find combination of x and y to satisfy the divisibility overall. But more importantly, we DON'T HAVE TO FIND ANY OF THOSE. Reason: This is a single option correct question! Once we found 101 is not a factor, we are done!!

Answer: E
Show more
Hey, I had a similar conclusion for the question. I understand why 101 can't be a factor and that was my answer choice as well. My only doubt here is that the other options are prime numbers as well. Therefore xy can never be a product that is divisible by the other options. Further, in the first part x^2+y^2 will have to be the sum that equals the prime number. Lastly since Multiple + or - nonmultiple= non multiple, except 40 none of the answer choices can be a factor of the equation. Any clarity here will be of great help. Thanks!
User avatar
Regor60
User avatar
Joined: 21 Nov 2021
Last visit: 17 Nov 2025
Posts: 528
Own Kudos:
383
Given Kudos: 459
Posts: 528
Kudos: 383
HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 10 Jul 2024, 07:11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
NamaySharma
Saupayan
Quote:
When the digits of two-digit positive integer M are reversed, the result is two-digit positive integer N. Which of the following cannot be a factor of M^2+N^2?

A. 17
B. 29
C. 40
D. 73
E. 101

Let M = 10x+y and N = 10y + x (where x and y are single digit whole numbers)
=> M^2 + N^2 = 101(x^2+y^2) + 40xy

Now, let's look at the answer choices. 40 and 101 stand out because they appear in our equation as well.
Let's consider 101... it divides the first term ( 101(x^2+y^2) )completely (without remainder). So, the only way the entire number will be divisible by 101 is if the 2nd term (40xy) is ALSO divisible by 101. 40 and 101 have no common factors because 101 is a prime number. Hence, the only way for 40xy to be divisible by 101 is if xy is divisible by 101. That will require x = 101 and y = 1 or vice-versa. Now, either of those options aren't possible because x and y are single digit numbers. So, 101 is definitely not a factor of M^2 + N^2.

If we do a similar test for 40... (this is just for understanding, and not a required step because we already found our answer)
The 2nd term is divisible.
For the first term to be divisible as well, (x^2 + y^2) has to be divisible by 40. A possibility is (2,6). Another is (6,2). Either of these will satisfy all the required conditions without violating any assumptions.
Hence, 40 can be a factor

For the other answer choices, neither 101 nor 40 is divisible directly. So, we may be able to find combination of x and y to satisfy the divisibility overall. But more importantly, we DON'T HAVE TO FIND ANY OF THOSE. Reason: This is a single option correct question! Once we found 101 is not a factor, we are done!!

Answer: E
Hey, I had a similar conclusion for the question. I understand why 101 can't be a factor and that was my answer choice as well. My only doubt here is that the other options are prime numbers as well. Therefore xy can never be a product that is divisible by the other options. Further, in the first part x^2+y^2 will have to be the sum that equals the prime number. Lastly since Multiple + or - nonmultiple= non multiple, except 40 none of the answer choices can be a factor of the equation. Any clarity here will be of great help. Thanks!
Show more


You're correct. The question is faulty.

Posted from my mobile device
Moderators:
Bunuel
Math Expert
105390 posts
Krunaal
Tuck School Moderator
805 posts