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Bunuel
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HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 28 Aug 2020, 08:00
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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

 

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E
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HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 28 Aug 2020, 08:33
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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
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HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 28 Aug 2020, 08:17
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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.
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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.
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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
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HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 28 Aug 2020, 08:26
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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

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HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 28 Aug 2020, 21:46
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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

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HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 29 Aug 2020, 05:46
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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)
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HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 27 Sep 2021, 07:09
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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
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HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 30 Nov 2021, 22:34
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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!
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HOT Competition 28 Aug/8AM: When the digits of two-digit positive inte 10 Jul 2024, 07:11
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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.

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