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0

$begingroup$

Theatre Square in the capital city of Berland has a rectangular shape with the size n × m meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size $a times  a$.

What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.

Note: $a$, $n$ and $m$ are integers and the final answer should also be an integer.

Say that $n$ and $m$ are both $6$ and $a$ is $4$. Then the minimum no should be $4$.

arithmetic puzzle

share|cite|improve this question

edited Mar 21 at 0:57

dantopa

6,66442245

asked Mar 20 at 13:20

RaphXRaphX

167

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $a$ an integer number?
  $endgroup$
  – Berci
  Mar 20 at 13:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Edited the question!
  $endgroup$
  – RaphX
  Mar 20 at 13:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe you can consider the two dimensions separately, if the stones are laid in a rectangular grid. So you can ask: What is the smallest multiple of $a$ that's $geq n$ ? And the same for $m$ ...
  $endgroup$
  – Matti P.
  Mar 20 at 13:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The problem statement says "It's allowed to cover the surface larger than the Theatre Square". So what disallows a solution with one giant square flagstone with sides $ge max(m,n)$ ?
  $endgroup$
  – gandalf61
  Mar 20 at 13:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For equations, please use MathJax.
  $endgroup$
  – dantopa
  Mar 21 at 0:56
  

  
  
  
  

add a comment | 

-2

0

$begingroup$

Theatre Square in the capital city of Berland has a rectangular shape with the size n × m meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size $a times  a$.

What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.

Note: $a$, $n$ and $m$ are integers and the final answer should also be an integer.

Say that $n$ and $m$ are both $6$ and $a$ is $4$. Then the minimum no should be $4$.

arithmetic puzzle

share|cite|improve this question

edited Mar 21 at 0:57

dantopa

6,66442245

asked Mar 20 at 13:20

RaphXRaphX

167

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is $a$ an integer number?
  $endgroup$
  – Berci
  Mar 20 at 13:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Edited the question!
  $endgroup$
  – RaphX
  Mar 20 at 13:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe you can consider the two dimensions separately, if the stones are laid in a rectangular grid. So you can ask: What is the smallest multiple of $a$ that's $geq n$ ? And the same for $m$ ...
  $endgroup$
  – Matti P.
  Mar 20 at 13:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The problem statement says "It's allowed to cover the surface larger than the Theatre Square". So what disallows a solution with one giant square flagstone with sides $ge max(m,n)$ ?
  $endgroup$
  – gandalf61
  Mar 20 at 13:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For equations, please use MathJax.
  $endgroup$
  – dantopa
  Mar 21 at 0:56
  

  
  
  
  

add a comment | 

$begingroup$

Theatre Square in the capital city of Berland has a rectangular shape with the size n × m meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size $a times  a$.

What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.

Note: $a$, $n$ and $m$ are integers and the final answer should also be an integer.

Say that $n$ and $m$ are both $6$ and $a$ is $4$. Then the minimum no should be $4$.

arithmetic puzzle

share|cite|improve this question

edited Mar 21 at 0:57

dantopa

6,66442245

asked Mar 20 at 13:20

RaphXRaphX

167

$endgroup$

share|cite|improve this question

edited Mar 21 at 0:57

dantopa

6,66442245

asked Mar 20 at 13:20

RaphXRaphX

167

share|cite|improve this question

edited Mar 21 at 0:57

dantopa

6,66442245

asked Mar 20 at 13:20

RaphXRaphX

167

edited Mar 21 at 0:57

dantopa

6,66442245

edited Mar 21 at 0:57

dantopa

6,66442245

asked Mar 20 at 13:20

RaphXRaphX

167

asked Mar 20 at 13:20

RaphXRaphX

167

1 Answer
1

active

oldest

votes

3

$begingroup$

We can without loss generality dub n the height of the rectangle and m the width.

The total number of flagstones is equal to the number of flagstones needed to cover the width of theatre square times the number of flagstones needed to cover the height.

The number of flagstones that are needed to cover the height is equal to the height n divided by the height of a flagstone a, but if n is not divisible by a then that number is some fraction, but we still need to cover the same area as that fraction of a flagstone does, the solution the next integer higher than $frac{n}a$, the ceiling function.

As the same thing holds for theatre squares width m, the total number of flagstones must be

$lceilfrac{n}arceiltimeslceilfrac{m}arceil$

share|cite|improve this answer

edited Mar 20 at 15:55

answered Mar 20 at 14:10

TuesdaydeviceofmathdestructionTuesdaydeviceofmathdestruction

463

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
  $endgroup$
  – jgon
  Mar 20 at 14:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is this better? sorry, I am new to this website.
  $endgroup$
  – Tuesdaydeviceofmathdestruction
  Mar 20 at 16:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, your explanation makes sense!
  $endgroup$
  – RaphX
  Mar 21 at 3:01
  

  
  
  
  

add a comment | 

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3

$begingroup$

We can without loss generality dub n the height of the rectangle and m the width.

The total number of flagstones is equal to the number of flagstones needed to cover the width of theatre square times the number of flagstones needed to cover the height.

The number of flagstones that are needed to cover the height is equal to the height n divided by the height of a flagstone a, but if n is not divisible by a then that number is some fraction, but we still need to cover the same area as that fraction of a flagstone does, the solution the next integer higher than $frac{n}a$, the ceiling function.

As the same thing holds for theatre squares width m, the total number of flagstones must be

$lceilfrac{n}arceiltimeslceilfrac{m}arceil$

share|cite|improve this answer

edited Mar 20 at 15:55

answered Mar 20 at 14:10

TuesdaydeviceofmathdestructionTuesdaydeviceofmathdestruction

463

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
  $endgroup$
  – jgon
  Mar 20 at 14:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is this better? sorry, I am new to this website.
  $endgroup$
  – Tuesdaydeviceofmathdestruction
  Mar 20 at 16:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, your explanation makes sense!
  $endgroup$
  – RaphX
  Mar 21 at 3:01
  

  
  
  
  

add a comment | 

3

$begingroup$

We can without loss generality dub n the height of the rectangle and m the width.

The total number of flagstones is equal to the number of flagstones needed to cover the width of theatre square times the number of flagstones needed to cover the height.

The number of flagstones that are needed to cover the height is equal to the height n divided by the height of a flagstone a, but if n is not divisible by a then that number is some fraction, but we still need to cover the same area as that fraction of a flagstone does, the solution the next integer higher than $frac{n}a$, the ceiling function.

As the same thing holds for theatre squares width m, the total number of flagstones must be

$lceilfrac{n}arceiltimeslceilfrac{m}arceil$

share|cite|improve this answer

edited Mar 20 at 15:55

answered Mar 20 at 14:10

TuesdaydeviceofmathdestructionTuesdaydeviceofmathdestruction

463

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review
  $endgroup$
  – jgon
  Mar 20 at 14:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is this better? sorry, I am new to this website.
  $endgroup$
  – Tuesdaydeviceofmathdestruction
  Mar 20 at 16:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, your explanation makes sense!
  $endgroup$
  – RaphX
  Mar 21 at 3:01
  

  
  
  
  

add a comment | 

$begingroup$

We can without loss generality dub n the height of the rectangle and m the width.

The total number of flagstones is equal to the number of flagstones needed to cover the width of theatre square times the number of flagstones needed to cover the height.

The number of flagstones that are needed to cover the height is equal to the height n divided by the height of a flagstone a, but if n is not divisible by a then that number is some fraction, but we still need to cover the same area as that fraction of a flagstone does, the solution the next integer higher than $frac{n}a$, the ceiling function.

As the same thing holds for theatre squares width m, the total number of flagstones must be

$lceilfrac{n}arceiltimeslceilfrac{m}arceil$

share|cite|improve this answer

edited Mar 20 at 15:55

answered Mar 20 at 14:10

TuesdaydeviceofmathdestructionTuesdaydeviceofmathdestruction

463

$endgroup$

share|cite|improve this answer

edited Mar 20 at 15:55

answered Mar 20 at 14:10

TuesdaydeviceofmathdestructionTuesdaydeviceofmathdestruction

463

answered Mar 20 at 14:10

TuesdaydeviceofmathdestructionTuesdaydeviceofmathdestruction

463

answered Mar 20 at 14:10

TuesdaydeviceofmathdestructionTuesdaydeviceofmathdestruction

463