Integer part problem. [closed]

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Find all positive integers which $fracn+2017[sqrtn+1]$ and $fracn+2018[sqrtn]$ are natural numbers.

algebra-precalculus elementary-number-theory contest-math divisibility

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edited Jul 30 at 12:37

greedoid

26.1k93473

asked Jul 30 at 12:10

mathematiciangrade8

313

closed as off-topic by Clarinetist, Michael Hoppe, 5xum, choco_addicted, Arthur Jul 30 at 12:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clarinetist, Michael Hoppe, 5xum, choco_addicted, Arthur

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That means that the numerator should be multiple of denominator...
  – MysteryGuy
  Jul 30 at 12:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You use the tag contest-math. What contest is it, and is it over?
  – Arthur
  Jul 30 at 12:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  this problem was at romanian math olympiad - the local stage subject
  – mathematiciangrade8
  Jul 30 at 12:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you use $[x]$ as the floor function or the nearest integer?
  – kvantour
  Jul 30 at 12:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  nearest integer
  – mathematiciangrade8
  Jul 30 at 12:40
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

Find all positive integers which $fracn+2017[sqrtn+1]$ and $fracn+2018[sqrtn]$ are natural numbers.

algebra-precalculus elementary-number-theory contest-math divisibility

share|cite|improve this question

edited Jul 30 at 12:37

greedoid

26.1k93473

asked Jul 30 at 12:10

mathematiciangrade8

313

closed as off-topic by Clarinetist, Michael Hoppe, 5xum, choco_addicted, Arthur Jul 30 at 12:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clarinetist, Michael Hoppe, 5xum, choco_addicted, Arthur

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That means that the numerator should be multiple of denominator...
  – MysteryGuy
  Jul 30 at 12:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You use the tag contest-math. What contest is it, and is it over?
  – Arthur
  Jul 30 at 12:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  this problem was at romanian math olympiad - the local stage subject
  – mathematiciangrade8
  Jul 30 at 12:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you use $[x]$ as the floor function or the nearest integer?
  – kvantour
  Jul 30 at 12:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  nearest integer
  – mathematiciangrade8
  Jul 30 at 12:40
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

Find all positive integers which $fracn+2017[sqrtn+1]$ and $fracn+2018[sqrtn]$ are natural numbers.

algebra-precalculus elementary-number-theory contest-math divisibility

share|cite|improve this question

edited Jul 30 at 12:37

greedoid

26.1k93473

asked Jul 30 at 12:10

mathematiciangrade8

313

Find all positive integers which $fracn+2017[sqrtn+1]$ and $fracn+2018[sqrtn]$ are natural numbers.

algebra-precalculus elementary-number-theory contest-math divisibility

share|cite|improve this question

edited Jul 30 at 12:37

greedoid

26.1k93473

asked Jul 30 at 12:10

mathematiciangrade8

313

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 12:37

greedoid

26.1k93473

edited Jul 30 at 12:37

greedoid

26.1k93473

edited Jul 30 at 12:37

greedoid

26.1k93473

26.1k93473

asked Jul 30 at 12:10

mathematiciangrade8

313

asked Jul 30 at 12:10

mathematiciangrade8

313

asked Jul 30 at 12:10

mathematiciangrade8

313

313

closed as off-topic by Clarinetist, Michael Hoppe, 5xum, choco_addicted, Arthur Jul 30 at 12:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clarinetist, Michael Hoppe, 5xum, choco_addicted, Arthur

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Clarinetist, Michael Hoppe, 5xum, choco_addicted, Arthur Jul 30 at 12:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Clarinetist, Michael Hoppe, 5xum, choco_addicted, Arthur

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That means that the numerator should be multiple of denominator...
  – MysteryGuy
  Jul 30 at 12:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You use the tag contest-math. What contest is it, and is it over?
  – Arthur
  Jul 30 at 12:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  this problem was at romanian math olympiad - the local stage subject
  – mathematiciangrade8
  Jul 30 at 12:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you use $[x]$ as the floor function or the nearest integer?
  – kvantour
  Jul 30 at 12:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  nearest integer
  – mathematiciangrade8
  Jul 30 at 12:40
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  That means that the numerator should be multiple of denominator...
  – MysteryGuy
  Jul 30 at 12:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You use the tag contest-math. What contest is it, and is it over?
  – Arthur
  Jul 30 at 12:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  this problem was at romanian math olympiad - the local stage subject
  – mathematiciangrade8
  Jul 30 at 12:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Do you use $[x]$ as the floor function or the nearest integer?
  – kvantour
  Jul 30 at 12:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  nearest integer
  – mathematiciangrade8
  Jul 30 at 12:40
  

  
  
  
  

That means that the numerator should be multiple of denominator...
– MysteryGuy
Jul 30 at 12:26

That means that the numerator should be multiple of denominator...
– MysteryGuy
Jul 30 at 12:26

You use the tag contest-math. What contest is it, and is it over?
– Arthur
Jul 30 at 12:29

You use the tag contest-math. What contest is it, and is it over?
– Arthur
Jul 30 at 12:29

this problem was at romanian math olympiad - the local stage subject
– mathematiciangrade8
Jul 30 at 12:35

this problem was at romanian math olympiad - the local stage subject
– mathematiciangrade8
Jul 30 at 12:35

Do you use $[x]$ as the floor function or the nearest integer?
– kvantour
Jul 30 at 12:40

Do you use $[x]$ as the floor function or the nearest integer?
– kvantour
Jul 30 at 12:40

nearest integer
– mathematiciangrade8
Jul 30 at 12:40

nearest integer
– mathematiciangrade8
Jul 30 at 12:40

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote

Let $k$ be such an integer, that $$k^2leq n< (k+1)^2implies [sqrtn]=k$$

So $[sqrtn+1] in k,k+1$.

Case 1: $[sqrtn+1] =k$ then we have $kmid n+2017$ and $kmid n+2018$ so $$kmid (n+2018)-(n+2017)=1implies k=1$$
Thus $1leq nleq 3$...(easy to check which one is good)

Case 2: $[sqrtn+1] =k+1$ then we have $n = k^2+2k$ and so $$k+1mid k^2+2k+2017 = (k+1)^2+2016implies k+1mid 2016;;;;(*)$$ and $$kmid k^2+2k+2018 implies kmid 2018 implies kin1,2,1009,2018$$

Because of $(*)$ $1009$ and $2018$ clearly doesn't work.

So the solution are $n=1,2,3,8$.

share|cite|improve this answer

edited Jul 30 at 12:41

answered Jul 30 at 12:17

greedoid

26.1k93473

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i found that n=8 it respects
  – mathematiciangrade8
  Jul 30 at 12:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I made an edit. If $k=2$ then $n=4+4=8$.
  – greedoid
  Jul 30 at 12:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  finally, u can write all solutions for n
  – mathematiciangrade8
  Jul 30 at 12:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i found that solution but i forgot to say that the fractions over the problem can be natural numbers but not togheter at once!
  – mathematiciangrade8
  Jul 30 at 12:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for example n=5 it respects the problem properties
  – mathematiciangrade8
  Jul 30 at 12:43
  

  
  
  
  

 | 
show 6 more comments

up vote
1
down vote

Let $t:=lfloor sqrtnrfloor$. Clearly, $tleq lfloor sqrtn+1rfloor leq t+1$. If $lfloor sqrtn+1rfloor=t$, then $tmid n+2017$ and $tmid n+2018$ implies that $tmid 1$, whence $t=1$. That is, $nin1,2$.

If $lfloorsqrtn+1rfloor=t+1$, then $n=(t+1)^2-1$. Thus, $$t+1mid n+2017=(t+1)^2+2016text and tmid n+2018=(t+1)^2+2017,$$

That is, $t+1mid 2016$ and $tmid 2018$. Since the divisors of $2018$ are $1$, $2$, $1009$, and $2018$, we conclude that $tin1,2,1009,2018$. What next?

share|cite|improve this answer

edited Jul 30 at 12:35

answered Jul 30 at 12:29

Batominovski

22.9k22777

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Let $k$ be such an integer, that $$k^2leq n< (k+1)^2implies [sqrtn]=k$$

So $[sqrtn+1] in k,k+1$.

Case 1: $[sqrtn+1] =k$ then we have $kmid n+2017$ and $kmid n+2018$ so $$kmid (n+2018)-(n+2017)=1implies k=1$$
Thus $1leq nleq 3$...(easy to check which one is good)

Case 2: $[sqrtn+1] =k+1$ then we have $n = k^2+2k$ and so $$k+1mid k^2+2k+2017 = (k+1)^2+2016implies k+1mid 2016;;;;(*)$$ and $$kmid k^2+2k+2018 implies kmid 2018 implies kin1,2,1009,2018$$

Because of $(*)$ $1009$ and $2018$ clearly doesn't work.

So the solution are $n=1,2,3,8$.

share|cite|improve this answer

edited Jul 30 at 12:41

answered Jul 30 at 12:17

greedoid

26.1k93473

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i found that n=8 it respects
  – mathematiciangrade8
  Jul 30 at 12:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I made an edit. If $k=2$ then $n=4+4=8$.
  – greedoid
  Jul 30 at 12:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  finally, u can write all solutions for n
  – mathematiciangrade8
  Jul 30 at 12:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i found that solution but i forgot to say that the fractions over the problem can be natural numbers but not togheter at once!
  – mathematiciangrade8
  Jul 30 at 12:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for example n=5 it respects the problem properties
  – mathematiciangrade8
  Jul 30 at 12:43
  

  
  
  
  

 | 
show 6 more comments

up vote
2
down vote

Let $k$ be such an integer, that $$k^2leq n< (k+1)^2implies [sqrtn]=k$$

So $[sqrtn+1] in k,k+1$.

Case 1: $[sqrtn+1] =k$ then we have $kmid n+2017$ and $kmid n+2018$ so $$kmid (n+2018)-(n+2017)=1implies k=1$$
Thus $1leq nleq 3$...(easy to check which one is good)

Case 2: $[sqrtn+1] =k+1$ then we have $n = k^2+2k$ and so $$k+1mid k^2+2k+2017 = (k+1)^2+2016implies k+1mid 2016;;;;(*)$$ and $$kmid k^2+2k+2018 implies kmid 2018 implies kin1,2,1009,2018$$

Because of $(*)$ $1009$ and $2018$ clearly doesn't work.

So the solution are $n=1,2,3,8$.

share|cite|improve this answer

edited Jul 30 at 12:41

answered Jul 30 at 12:17

greedoid

26.1k93473

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i found that n=8 it respects
  – mathematiciangrade8
  Jul 30 at 12:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I made an edit. If $k=2$ then $n=4+4=8$.
  – greedoid
  Jul 30 at 12:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  finally, u can write all solutions for n
  – mathematiciangrade8
  Jul 30 at 12:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i found that solution but i forgot to say that the fractions over the problem can be natural numbers but not togheter at once!
  – mathematiciangrade8
  Jul 30 at 12:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for example n=5 it respects the problem properties
  – mathematiciangrade8
  Jul 30 at 12:43
  

  
  
  
  

 | 
show 6 more comments

up vote
2
down vote

up vote
2
down vote

Let $k$ be such an integer, that $$k^2leq n< (k+1)^2implies [sqrtn]=k$$

So $[sqrtn+1] in k,k+1$.

Case 1: $[sqrtn+1] =k$ then we have $kmid n+2017$ and $kmid n+2018$ so $$kmid (n+2018)-(n+2017)=1implies k=1$$
Thus $1leq nleq 3$...(easy to check which one is good)

Case 2: $[sqrtn+1] =k+1$ then we have $n = k^2+2k$ and so $$k+1mid k^2+2k+2017 = (k+1)^2+2016implies k+1mid 2016;;;;(*)$$ and $$kmid k^2+2k+2018 implies kmid 2018 implies kin1,2,1009,2018$$

Because of $(*)$ $1009$ and $2018$ clearly doesn't work.

So the solution are $n=1,2,3,8$.

share|cite|improve this answer

edited Jul 30 at 12:41

answered Jul 30 at 12:17

greedoid

26.1k93473

Let $k$ be such an integer, that $$k^2leq n< (k+1)^2implies [sqrtn]=k$$

So $[sqrtn+1] in k,k+1$.

Case 1: $[sqrtn+1] =k$ then we have $kmid n+2017$ and $kmid n+2018$ so $$kmid (n+2018)-(n+2017)=1implies k=1$$
Thus $1leq nleq 3$...(easy to check which one is good)

Case 2: $[sqrtn+1] =k+1$ then we have $n = k^2+2k$ and so $$k+1mid k^2+2k+2017 = (k+1)^2+2016implies k+1mid 2016;;;;(*)$$ and $$kmid k^2+2k+2018 implies kmid 2018 implies kin1,2,1009,2018$$

Because of $(*)$ $1009$ and $2018$ clearly doesn't work.

So the solution are $n=1,2,3,8$.

share|cite|improve this answer

edited Jul 30 at 12:41

answered Jul 30 at 12:17

greedoid

26.1k93473

share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 12:41

edited Jul 30 at 12:41

edited Jul 30 at 12:41

answered Jul 30 at 12:17

greedoid

26.1k93473

answered Jul 30 at 12:17

greedoid

26.1k93473

answered Jul 30 at 12:17

greedoid

26.1k93473

26.1k93473

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i found that n=8 it respects
  – mathematiciangrade8
  Jul 30 at 12:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I made an edit. If $k=2$ then $n=4+4=8$.
  – greedoid
  Jul 30 at 12:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  finally, u can write all solutions for n
  – mathematiciangrade8
  Jul 30 at 12:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i found that solution but i forgot to say that the fractions over the problem can be natural numbers but not togheter at once!
  – mathematiciangrade8
  Jul 30 at 12:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for example n=5 it respects the problem properties
  – mathematiciangrade8
  Jul 30 at 12:43
  

  
  
  
  

 | 
show 6 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  i found that n=8 it respects
  – mathematiciangrade8
  Jul 30 at 12:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, I made an edit. If $k=2$ then $n=4+4=8$.
  – greedoid
  Jul 30 at 12:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  finally, u can write all solutions for n
  – mathematiciangrade8
  Jul 30 at 12:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  i found that solution but i forgot to say that the fractions over the problem can be natural numbers but not togheter at once!
  – mathematiciangrade8
  Jul 30 at 12:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  for example n=5 it respects the problem properties
  – mathematiciangrade8
  Jul 30 at 12:43
  

  
  
  
  

i found that n=8 it respects
– mathematiciangrade8
Jul 30 at 12:31

i found that n=8 it respects
– mathematiciangrade8
Jul 30 at 12:31

Yes, I made an edit. If $k=2$ then $n=4+4=8$.
– greedoid
Jul 30 at 12:32

Yes, I made an edit. If $k=2$ then $n=4+4=8$.
– greedoid
Jul 30 at 12:32

finally, u can write all solutions for n
– mathematiciangrade8
Jul 30 at 12:38

finally, u can write all solutions for n
– mathematiciangrade8
Jul 30 at 12:38

i found that solution but i forgot to say that the fractions over the problem can be natural numbers but not togheter at once!
– mathematiciangrade8
Jul 30 at 12:43

i found that solution but i forgot to say that the fractions over the problem can be natural numbers but not togheter at once!
– mathematiciangrade8
Jul 30 at 12:43

for example n=5 it respects the problem properties
– mathematiciangrade8
Jul 30 at 12:43

for example n=5 it respects the problem properties
– mathematiciangrade8
Jul 30 at 12:43

 | 
show 6 more comments

up vote
1
down vote

Let $t:=lfloor sqrtnrfloor$. Clearly, $tleq lfloor sqrtn+1rfloor leq t+1$. If $lfloor sqrtn+1rfloor=t$, then $tmid n+2017$ and $tmid n+2018$ implies that $tmid 1$, whence $t=1$. That is, $nin1,2$.

If $lfloorsqrtn+1rfloor=t+1$, then $n=(t+1)^2-1$. Thus, $$t+1mid n+2017=(t+1)^2+2016text and tmid n+2018=(t+1)^2+2017,$$

That is, $t+1mid 2016$ and $tmid 2018$. Since the divisors of $2018$ are $1$, $2$, $1009$, and $2018$, we conclude that $tin1,2,1009,2018$. What next?

share|cite|improve this answer

edited Jul 30 at 12:35

answered Jul 30 at 12:29

Batominovski

22.9k22777

add a comment | 

up vote
1
down vote

Let $t:=lfloor sqrtnrfloor$. Clearly, $tleq lfloor sqrtn+1rfloor leq t+1$. If $lfloor sqrtn+1rfloor=t$, then $tmid n+2017$ and $tmid n+2018$ implies that $tmid 1$, whence $t=1$. That is, $nin1,2$.

If $lfloorsqrtn+1rfloor=t+1$, then $n=(t+1)^2-1$. Thus, $$t+1mid n+2017=(t+1)^2+2016text and tmid n+2018=(t+1)^2+2017,$$

That is, $t+1mid 2016$ and $tmid 2018$. Since the divisors of $2018$ are $1$, $2$, $1009$, and $2018$, we conclude that $tin1,2,1009,2018$. What next?

share|cite|improve this answer

edited Jul 30 at 12:35

answered Jul 30 at 12:29

Batominovski

22.9k22777

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let $t:=lfloor sqrtnrfloor$. Clearly, $tleq lfloor sqrtn+1rfloor leq t+1$. If $lfloor sqrtn+1rfloor=t$, then $tmid n+2017$ and $tmid n+2018$ implies that $tmid 1$, whence $t=1$. That is, $nin1,2$.

If $lfloorsqrtn+1rfloor=t+1$, then $n=(t+1)^2-1$. Thus, $$t+1mid n+2017=(t+1)^2+2016text and tmid n+2018=(t+1)^2+2017,$$

That is, $t+1mid 2016$ and $tmid 2018$. Since the divisors of $2018$ are $1$, $2$, $1009$, and $2018$, we conclude that $tin1,2,1009,2018$. What next?

share|cite|improve this answer

edited Jul 30 at 12:35

answered Jul 30 at 12:29

Batominovski

22.9k22777

Let $t:=lfloor sqrtnrfloor$. Clearly, $tleq lfloor sqrtn+1rfloor leq t+1$. If $lfloor sqrtn+1rfloor=t$, then $tmid n+2017$ and $tmid n+2018$ implies that $tmid 1$, whence $t=1$. That is, $nin1,2$.

If $lfloorsqrtn+1rfloor=t+1$, then $n=(t+1)^2-1$. Thus, $$t+1mid n+2017=(t+1)^2+2016text and tmid n+2018=(t+1)^2+2017,$$

That is, $t+1mid 2016$ and $tmid 2018$. Since the divisors of $2018$ are $1$, $2$, $1009$, and $2018$, we conclude that $tin1,2,1009,2018$. What next?

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edited Jul 30 at 12:35

answered Jul 30 at 12:29

Batominovski

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share|cite|improve this answer

share|cite|improve this answer

edited Jul 30 at 12:35

edited Jul 30 at 12:35

edited Jul 30 at 12:35

answered Jul 30 at 12:29

Batominovski

22.9k22777

answered Jul 30 at 12:29

Batominovski

22.9k22777

answered Jul 30 at 12:29

Batominovski

22.9k22777

22.9k22777

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