Let a (cm) be the largest possible square size

=> a = GCD (140 ; 240) = 20 

So the largest size of the square's edge is 20 cm.Then the smallest number of squares can be cut from the cardboard is :

140 x 240 : 202 = 84

Answer : 20 cm ; 84 squares

If a is odd,then a3 and a2 are also odd,so a3 + a2 + 1 is odd

If a is even,then a3 and a2 are also even,so a3 + a2 + 1 is odd

So a3 + a2 + 1 is always odd with any integer a

a) Assume that the square number 16 = 42 is equal to,so n = 48

(b) Assume that the cubic number 27 = 33 is equal to,so n = 135

Let a,b,c be the number of the correctly answered questions , unanswered questions , incorrectly answered questions of a participant respectively

Then the participant's total score is 5a + b - c with a + b + c = 30

If a is odd,then 5a and b + c are odd,so b - c is odd and 5a + b - c is even

If a is even,then 5a and b + c are even,so b - c is even and 5a + b - c is even

From 2 cases,we know that the total score of all participants is always an even number

Similary, we have:

   $\{\begin{array}{c}P\left(2\right)=0\\ P\left(-2\right)=0\\ P\left(-3\right)=0\end{array}$

  $\iff \{\begin{array}{c}{2}^5+2^4-{9.2}^2+4a+2b+c=0\\ {\left(-2\right)}^5+{\left(-2\right)}^4-9.{\left(-2\right)}^2+4a-2b+c=0\\ {\left(-3\right)}^5+{\left(-3\right)}^4-9.{(}^{}\end{array}$

The distance between the houses is: 

               3000: 100 = 30 (minute)
Homeownership is: 

              3000: 150 = 20 (minutes)
Average by John in a minute:

              3000: (20 + 30) = 60 (minute)
                       Answers:60 minute

The distance between the houses is:

               3000: 100 = 30 (minute)
Homeownership is:

              3000: 150 = 20 (minutes)
Average by John in a minute:

              3000: (20 + 30) = 60 (minute)
                       Answers:60 minute

We have : $\frac{4}{n}=\frac{1}{n}+\frac{3}{n}$

With n = 3k (k is the natural number greater than 1) , i have :

$\frac{4}{n}=\frac{1.\left(n+1\right)}{n.\left(n+1\right)}+\frac{3}{n}=\frac{1}{n+1}+\frac{1}{n.\left(n+1\right)}+\frac{3}{n}=\frac{1}{3k}+\frac{1}{3k.\left(3k+1\right)}+\frac{3}{3k}=\frac{1}{3k}+\frac{1}{9k^2+3k}+\frac{1}{k}$

With n = 3k + 1

$\frac{4}{n}=\frac{1}{n}+\frac{3}{n}=\frac{1}{n}+3(\frac{1}{n-1}-$

Replacing x = 1/2 we have: 4x - 3a = 0  

<=> 4.1/2 - 3a = 0

=> 2 - 3a = 0

=> 3a = 2

=> a = 2/3

Dress a = 2/3 good

Hi Steve Jobs! First; you can see : 3 . 5 = 15 ; 33 . 35 = 1155;

333 . 335 = 111555, ... It's the first step to solve ur problem.

Now, we have to prove that expression equal 333...3 . 333....5.

We have :

111...11111111111555...555555555111...11111111111555...555555555

( 2002 1s)                      (2002 5s)

=111.....11000....0 + 555.......5 

( 2002 1s) (2002 0s) (2002 5s)

= 1111.....111 . ( 10000...000 + 5 )

    ( 2002 1s)          ( 2002 0s)

= 111....111 .  10000...00005

    ( 2002 1s)       (2001 0s )

= 1111...1111 . ( 3 . 333...33335 )

  ( 2002 1s)               (2001 3s )

= 333......3333 . 333333...3335

       ( 2002 3s)     ( 2001 3s )

The sum of these 2 numbers is 6666......68

                                                  (2002 6s)

Have: n3 - 3n2 + 2 = n3 - n2 - 2n2 + 2

= n2(n - 1) - 2n2 + 2

It is easy to see that n2(n - 1) is an even number because it contains n(n - 1) which is the product of two consecutive natural numbers, so n2(n - 1) - 2n2 + 2 is an even number

But n3 - 3n2 + 2 = n2(n - 1) - 2n2 + 2 is a prime number so n3 - 3n2 + 2 = 2

=> n3 - 3n2 = 0 

<=> n2(n - 3) = 0

According to the topic, n should be positive n = 3

MCP 22/03 at 17:43

We have: 

⇒a=2017n−2016b⇒a=2017n−2016b

We have: 2a+2015b=2(2017n−2016b)+2015b=2.2017n−2017b⋮2017

We have: 

⇒a=2017n−2016b⇒a=2017n−2016b

We have: 2a+2015b=2(2017n−2016b)+2015b=2.2017n−2017b⋮20172a+2015b=2(2017n−2016b)+2015b=2.2017n−2017b⋮2017

Same as above. We hava: ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩3a+2014b=3.2017n−2.2017b⋮20174a+2013b=4.2017n−3.2017b⋮2017...2015a+2b=2015.2017n−2014.2017b{3a+2014b=3.2017n−2.2017b⋮20174a+2013b=4.2017n−3.2017b⋮2017...2015a+2b=2015.2017n−2014.2017b

So that: A⋮2017.2017...2017A⋮2017.2017...2017

⇒A⋮20172014

The distance between the houses is: 3000: 100 = 30 (minute)
Homeownership is: 3000: 150 = 20 (minutes)
Average by John in a minute: 3000: (20 + 30) = 60 (minute)

Answers:60 minute

( x + 1 ) . ( 1 + 1999 ) = 4000

 (x + 1 ) . 2000 = 4000

   x + 1 = 4000 : 2000

   x + 1 = 2 

   x = 2 - 1 

  x = 1 

We have : (x + 2)(x + 2) = 24(x + 2)(x + 2) = 24

⇒x2 + 4x + 4 =24

=> x2 + 4x = 20

=> x(x + 4) = 20

=> x ,x + 4 thuộc Ư(20) = {1;2;4;5;10;20}

 When x = 1 thì x + 4 = 20 => x = 16 

Not yes value satisfies

Number of female students is : 

   40 x 2525= 16 ( student )

 Number of  male students is :

    40 - 16 = 24 ( student )

         Answer : 24 student

Number of terms is :

  ( 1000 - 1 ) : 2 + 1 = 500,5 

The sum is :

  ( 1000 + 1 ) x 500,5 : 2 = 250500,25