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Exercise 1:

The three Euclidean divisions below are correct.

1.only 23 is a divisor of 368 because the remainder is zero.

The smallest multiple of 15 greater than 368 is 25×15=375.

3. The greatest multiple of 14 less than 368 is 26×14= 364.

Exercise 2:

The corresponding Euclidean divisions are :

• 475= 16 x 29 +11;
• 9 957 = 23 x 432 + 21;
• 456 = 41 x 11 +5;
• 781 = 27 x 28 + 25;
• 935 = 17 x 55 + 0

Exercise 3:

A day care center with 131 children organizes a “Sport Co” day with basketball, handball, soccer and rugby.

For each sport, how many teams can be formed?

How many children will be without a team?

131=32×4+3.

We can build 32 teams and 3 children will be without a team.

Exercise 4:

Write the list of divisors of the following numbers: 16; 20; 36; 90; 59; 33.

Divisors of 16 :1,2,4,8,16.

Divisors of 20 :1,2,4,5,10,20.

Divisors of 59:1.59.

Exercise 5:

Complete the table below.

Exercise 6:

1.Prove that the sum of two consecutive odd positive integers is a multiple of 4.

If n=2k+1 (with k a positive integer) is an odd positive number then the consecutive odd positive integer is n’=2k+3.

n+n’=2k+1+2k+3=4k+4=4(k+1)=4K with K=k+1 so the sum of two consecutive odd integers is a multiple of 4.

2.show that a multiple of 8 is also a multiple of 4.

Let n=8k (with k a positive integer) be a multiple of 8 then n=4x(2k)=4K with K=2k so n is also a multiple of 4.

Exercise 7:

Nori wants to make packs of marbles, dividing his 90 red marbles and 150 black marbles evenly.

How many packages will he be able to make?

Find the different possibilities.

Can there be 9 packages? 30 packs?

There cannot be 9 packets because 150 is not divisible by 9.

There can be 30 packages because 150 and 90 are divisible by 30.

Give the list of divisors of 90 and then of 150.

Divisors of 90 :1,2,3,5,6,9,10,15,18,30,45,90

Divisors of 150 :1,2,3,5,6,10,15,25,30,50,75,150

What are the different possibilities for the number of packages?

The possibilities are 1,2,3,5,6,10,15,30.

Exercise 8:

Give the list of all prime numbers less than 50.

The list is 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

Exercise 9:

Use the equations below to write the decompositions into prime factors

of the proposed numbers.

Exercise 10:

Write the prime factor decomposition of the following integers:

Exercise 11:

Find the number you are looking for.

The solutions are 101; 113; 137 and 149.

Exercise 12:

Use prime factor decompositions to make these fractions irreducible.

Make the following fractions irreducible: .

Exercise 13:

Make the following fractions irreducible: .

Exercise 14:

I have more than 400 cds but less than 450. Whether I group them by 2, by 3, by 4 or by 5, it’s always the same thing: there is only one left.
How many CDs does Nori have?

We are looking for an odd number that ends in 1.

There are 421 CDs.

Exercise 15:

1. Calculate the PGCD of 110 and 88.

GCD( 110 ; 88 ) = 22

2. A worker has metal plates 110 cm long and 88 cm wide.

He was instructed as follows:

” Cut out squares from these plates, all identical, as large as possible ,

so that there is no loss. ”

What will be the length of the side of the square?

The length of the side of the square will be 22 cm

3. How many squares will be obtained per plate?

110:22=5 and 88:22=4

5×4=20

There will be 20 squares.

Exercise 16:

1. Calculate the PGCD of 114 400 and 60 775.

GCD( 114400 ; 60775 ) = 3575

2. > Explain how, without using the “fraction” key on a calculator, to make the fraction irreducible

dividing by the gcd (114,400; 60,775)

3. Give the simplified writing of

.

Exercise 17:

Let the numbers A = and B = – .

1. Explain why the fraction A is not irreducible.

117 and 63 are divisible by 3 so their gcd is different from 1 so the fraction is reducible.

2. Simplify this fraction to make it irreducible.

GCD( 117 ; 63 ) = 9

3. Show, by indicating the calculation steps, that A – B is an integer.

so A-B is a whole number.

Exercise 18:

1. Prove that the numbers 65 and 42 are prime to each other.

PGCD( 65 ; 42 ) = 1 so these two integers are indeed prime between them.

2. Show that = .

GCD( 520 ; 336 ) = 8

Exercise 19:

1. Determine the PGCD of 108 and 135.

GCD( 135 ; 108 ) = 27

2. Mark has 108 red marbles and 135 black marbles.

He wants to make packages so that :

• all packages contain the same number of red marbles;
• all packages contain the same number of black marbles;
• all the red balls and black balls are used.

a. What is the maximum number of packages it can achieve?
It can make a maximum of 27 packages.
b. How many red and black marbles will there be in each package?
108:27=4 red balls

135:27=5 black balls.

Exercise 20:

1. Calculate the PGCD of 1 756 and 1 317 (the necessary calculations will be detailed).

GCD( 1756 ; 1317 ) = 439

2. A florist received 1,756 white roses and 1,317 red roses.

He wants to make identical bouquets

(i.e. with the same number of roses and the same

distribution between white and red roses) using all the flowers.
a. What will be the maximum number of identical bouquets? Justify the answer clearly.

It can create a maximum of 439 identical bouquets.

b. What will be the composition of each bouquet?
1756:439=4 white roses.
1317:439=3 black roses.

Exercise 21:

1) Show that PGCD( 578 ; 408 ) = 34

Show that PGCD( 2499 ; 1911 ) = 147

2) Show that PGCD( 252 ; 144 ) = 36 .

a. This association can form a maximum of 36 teams.

b. 144 : 36 = 4 and 252 : 36 = 7

There are 4 girls and 7 boys per team.

Exercise 23:

1. Are the numbers 682 and 352 prime between them ? Justify.

These are two even integers so they cannot be prime to each other because their gcd will be greater than or equal to 2.

2. Calculate the greatest common divisor (GCD) of 682 and 352.

Let’s use Euclid’s algorithm.

682=1×352+330

352=1×330+22

330=15×22+0

So gcd (352 ; 682 ) = 22

3. Make the fraction irreducible

clearly indicating the method used.

By dividing the numerator and denominator by the gcd, we get a fraction that is irreducible.

Exercise 24:

Calculate and give the result as an irreducible fraction:

.

Exercise 25:

Calculate and give the result in scientific notation:

Exercise 26:

1.

Calculate A and give the result as a fraction.

2. Write B in the form where b is an integer .

3.

Calculate C and give the scientific form of the result.

Exercise 27:

1. Determine the PGCD of 288 and 224.

Let’s use Euclid’s algorithm.

The PGCD being the last non-zero remainder, we deduce that

2 . Write the fraction in irreducible form.

is an irreducible fraction.

3 . A photographer has to make an exhibition by presenting his works on panels each containing the same number of landscape photos and the same number of portraits.

It has 224 landscape photos and 288 portraits.

a ) How many panels can he make using all the photos?

It will be able to make a maximum of 32 panels.

b) How many landscape and portrait photos are in each panel?

Each panel will contain 7 landscape photos and 9 portrait photos.

Exercise 29:

a. Are 255 and 154 prime between them?
PGCD( 255 ; 154 )
We use Euclid’s algorithm
And we group the results in a table.

Dividend	Divider	Remainder
Dividend	Divider	Remainder
255	154	101
154	101	53
101	53	48
53	48	5
48	5	3
5	3	2
3	2	1
2	1	0

Now, in Euclid’s algorithm the PGCD is the last non-zero remainder.
PGCD( 255 ; 154 ) = 1 so these two numbers are prime between them.
b. Are 609 and 465 prime between them?
GCD( 609 ; 465 )
We use Euclid’s algorithm
And we group the results in a table.

Dividend	Divider	Remainder
Dividend	Divider	Remainder
609	465	144
465	144	33
144	33	12
33	12	9
12	9	3
9	3	0

Now, in Euclid’s algorithm the PGCD is the last non-zero remainder.
PGCD( 609 ; 465 ) = 3 so these two numbers are not prime to each other.
c. Are 11,913 and 7,259 prime to each other?
GCD( 11913 ; 7259 )
We use Euclid’s algorithm
And we group the results in a table.

Dividend	Divider	Remainder
Dividend	Divider	Remainder
11913	7259	4654
7259	4654	2605
4654	2605	2049
2605	2049	556
2049	556	381
556	381	175
381	175	31
175	31	20
31	20	11
20	11	9
11	9	2
9	2	1
2	1	0

Now, in Euclid’s algorithm the PGCD is the last non-zero remainder.
PGCD( 11913 ; 7259 ) = 1 so these two numbers are prime between them.

Exercise 30:

1. By Euclid’s method:
481=2×234+13
234=18×13+0
the gcd being the last non-zero remainder, we deduce that gcd (481 , 234 ) =13.

Exercise 31:

1. By the method of Euclid’s algorithm:
137=3×41+14
41=2×14+13
14=1×13+1
13=1×13+0
the gcd being the last non-zero remainder, we deduce that gcd (137 , 41 ) =1.
2. These two integers are prime to each other because gcd (137 , 41 ) =1.
Note that Euclid’s algorithm is faster.

Exercise 32:

1. It is therefore necessary to calculate the pgcd (2 622,2 530).
Let’s use Euclid’s algorithm.
2622=1×2530+92
2530=27×92+46
92=2×46+0
the gcd being the last non-zero remainder, we deduce that gcd (137 , 41 ) = 46.
There will be 46 eggs and 46 fish in each package.
2. In a package there are 2×46=92 elements and in total we have 2622+2530=5152 elements.
There will be 56 packages (5 152:92=56).

Exercise 33:

1. 1. It is therefore necessary to calculate the gcd (161,133).
Let’s use Euclid’s algorithm.
161=1×133+28
133=4×28+21
28=1×21+7
21=3×7+0
the gcd being the last non-zero remainder, we deduce that gcd (161 , 133 ) = 7.
There will be 7 pencils in each package.
2. In a package there are 14 pencils of each color for a total of 294 pencils.
so there are a total of 21 packages ( 294:14 = 21).

Exercise 34:

For A:
Let’s calculate the gcd (945,595)
945=1×595+350
595=1×350+245
350=1×245+105
245=2×105+35
105=2×35+35
35=1×35+0
so gcd(945,595)=35
thus

For B:
Let’s calculate the gcd (1 771.736)
1 771=2×736+299
736=2×299+138
299=2×138+23
138=6×23+0
so gcd (1 771,736)=23
thus

Exercise 35:

Exercise 36:

1. We use Euclid’s algorithm

And we group the results in a table.

Dividend	Divider	Remainder
Dividend	Divider	Remainder
135	108	27
108	27	0

Now, in Euclid’s algorithm the PGCD is the last non-zero remainder.
GCD( 135 ; 108 ) = 27

2. a. What is the maximum number of packages it can achieve?

It can make a maximum of 27 packages.
b. How many red and black marbles will there be in each package?
108:27=4 red balls.
135:27=5 black balls.

Conclusion:
There will be, in each pack, 4 red marbles and 5 black marbles.

Exercise 37:

The number of Cd is not divisible by 2 and 5 so the number of CD does not end with 0,2,4,5,6,8.

So it ends with 1,3,7,9

The number sought is 421 by successive eliminations.

Exercise 38:

a) To the question: ” How many divisors does 48 have? “, John answers that there are 9, while Cedric finds 10.

Who is right?

Which method finds all the divisors of a number?

The divisors of 48 are 1;2;3;4;6;8;12;16;24;48 .

There are 10, Cedric is right.

b) An artist has a canvas measuring 60 cm by 48 cm.

He wants to paint a paving composed of identical squares, but of different colors. The side length of these squares is an integer.

What is the longest possible length for these squares (in cm) ?

The longest length corresponds to the gcd (60.48)

GCD( 60 ; 48 )
We use Euclid’s algorithm
And we group the results in a table.

Dividend	Divider	Remainder
Dividend	Divider	Remainder
60	48	12
48	12	0

Now, in Euclid’s algorithm the PGCD is the last non-zero remainder.
GCD( 60 ; 48 ) = 12

Conclusion : The longest length is 12 cm .

Exercise 39:

Are the following sentences true or false? Justify the answers.
a) 3 is a divisor of 43. FALSE g) The multiple of 24 is 240. TRUE
b) 132 is divisible by 11. TRUE h) 5 divides 450. TRUE
c) 7 is a divisor of 21. FALSE i) 8 is a divisor of 0. TRUE
d) 222 is a divisor of 31,024. FALSE j) 1 is a multiple of 67. FALSE
e) 31,024 is a multiple of 113. FALSE k) 1 divides 0. TRUE
f) 45 is divided by 5. TRUE l) 0 divides 15. FALSE

Exercise 40:

If the number has only two divisors then it is a prime number.

Just create a table with the excel software.

We observe that n=11, we obtain 121 .

Now 121 is divisible by 121 , 11 and 1 so has 3 divisors.

The statement is false.

Exercise 41:

1. Calculate the number of tarts.

Let’s calculate the gcd(411,685)

685=1×411 +274

411=1×274+137

274=2×137+0

so gcd(411,685)=137

2. Calculate the number of raspberries and strawberries in each tartlet.

411:137= 3

There will be 3 raspberries per tartlet.

685:137 = 5

There will be 5 strawberries per tartlet.

Exercise 42:

1. How many bouquets can the florist make?

Show that gcd (1105,935) = 85

2. What will be the constitution of each bouquet?

1105 : 85 = 13 and 935 : 85 = 11

Each bouquet consists of 13 carnations and 11 irises.

Exercise 43:

1) Are the numbers 3120 and 2760 prime to each other? Justify

These two numbers are divisible by 10 so they are not prime to each other because their gcd is different from 1.

2) Calculate the greatest common divisor of 3120 and 2760.

3120=1×2760 + 360

2760=7×360+240

360=1×240+120

240=2×120+0

The gcd being the last non-zero remainder, gcd(3120,2760)=120

3) Make the fraction irreducible.

4) A confectioner has 3120 pink dragees and 2760 white dragees, he wishes to make identical packages of pink and white dragees.

In order to make the maximum profit on these sales, the number of packages must be as large as possible and he must use all his jelly beans.

a) How many packages does the confectioner make?

Each package must contain the maximum number of dragées of each color

so it is like looking for the gcd(3120,2760) which is 120 packets.

b) What is the number in each package of pink jelly beans?

3120 : 120 =26.

Each package contains 26 pink dragées.

c) What is the number in each package of white jelly beans?

Each package contains 26 white dragees.

Exercise 1:

Where is the cat when you click on the block?

I click on but the program does not work. Why is this?

When you click on the block, the cat moves forward 100 pixels.

The program does not work because the brick containing is not inserted in the program.

Exercise 2:

Initially, the cat is located at x=0 and y= – 50.

What will happen if we throw it several times ?

How to solve this problem?

The cat moves forward 10×20=200 pixels so if you run this program several times , it will come out of the background.

We can solve this problem by inserting a brick at the start that gives the initial position of the sprite.

Exercise 3:

Program D has just been executed.

Exercise 4:

This program constructs a rectangle of length 100 pixels and width 50 pixels.

Exercise 5:

Which of these three programs has just been run?

The C program has just been executed.

Exercise 6:

The dog must go to his friend the frog for his birthday.

But first he must retrieve the gift while avoiding the lion.

Which of these three programs is appropriate?

Program C is suitable.

Exercise 7:

When the program starts, what will the lion do?

The lion will retrieve the green marble and return to its original position.

Exercise 8:

Which of these three programs has just been executed?

The C program has just been executed.

Exercise 9:

After the execution of one of the two programs and after having proposed the number 10, the cat announced 35.

If we had run the other program, what result would have been announced?

Program B has been executed.

If program A is executed by taking 10, the sprite would have announced (5+3)*10=8*10=80.

Exercise 10:

the cat is positioned in (0;0) and the tree in (70;0).

We launch the program.

What is the probability that the cat will reach the tree?

The probability is 0.7.

What is the probability that the cat will pass the tree?

The probability is 0.3.

Exercise 1:

What does the leprechaun announce at the end of the program?

This sprite announces 2*(10+1)=2*11=22.

Exercise 2:

What is the value of the number variable at the end of these two programs?

Program 1: 0+1+1+1+1+1+1+1+1+1+1+1=10

Program 2: 0+1+2+3+4+5+6+7+8+9+10=55

Exercise 3:

What plot do you get with this program?

What is the abscissa of the ball sprite once the program is executed?

The program executes plot 3.

Exercise 4:

What does the cat say at the end of the program?

The program quotes the numbers from the list.

Exercise 5:

We ran the program and entered 13 and 8.

What were the elf’s responses?

The elf’s responses are “That doesn’t work for me” and then “Thank you!”.

Exercise 6:

What does the cat say at the end of the program?

The cat says the whole list of numbers contained in maliste.

Exercise 7:

The user has entered the values 15 and 9.

What are the values of variables a and b at the end of each of these two programs?

Program 1: a=9 and b=15.

Program 2: a=9 and b= 9.

Program 3: a=9 and b= 9.

Exercise 8:

If the butterfly touches the bat then the game is lost!

But the program does not work.

Why is this?

The program does not work because of the “stop-all” brick.

Exercise 9:

Which instruction calculates 3 + 4×5 – 2?

This is instruction 4.

Exercise 10:

Is it possible that the cat announces 750?

What is the maximum number that can be obtained?

This program will provide as maximum result :

.

It is impossible for this program to provide 750.

Exercise 1

Associate each program with the corresponding output.

Program 1: plot 2.

Program 2: plot 3.

Program 3: plot 1.

Exercise 2

If a = 1 then the sprite moves forward by a number of steps between 1 and 20 pixels and if a = 2, the sprite moves backward by a number of steps between 1 and 20 pixels.

Exercise 3

What does this program do?

This program adds to the number result1: 2,3,4 etc… then to the number result2, it multiplies it by 2,3,4 etc…

Exercise 4

We ran this program and entered the number 45.

What is the content of my list at the end of the execution?

The content of maliste will be the list of divisors of the number 45.

Exercise 5

How to complete this program in order to add the entered number to the list only

in case the number is not yet included?

You have to add a condition “otherwise” and the brick “Add to list” the value “answer”.

Exercise 6
What code should be added to the two sprites so that the Bear1 sprite says hello to the cat
once the latter is near him?

It would be necessary to add a condition of the type :

If the x-position of the cat and the Bear1 sprite is less than 10 then both sprites say hello.

Exercise 7

We click on the chat.

After exactly how many seconds will the cat say hello?

1+2+1=4 s

The cat will say hello after 4 seconds.

Exercise 8

The scene features a Ball goblin.

Is it possible to win in this game?

It is impossible to win this game because each time the mouse moves closer to the sprite than 50 pixels, the sprite will move to an abscissa between -200 and 200 pixels.

Exercise 9

What does this program do?

We entered 12 and then 15. The cat announces NAN! Why?

How can the program be improved?

This program will calculate which does not exist!

Exercise 10

What happens if you click on the balloon once?

What happens if you click on the balloon several times quickly?

If you click on this balloon, another balloon will be created for 4 seconds and then deleted automatically.

This balloon will take a position whose abscissa is between -240 and +240 pixels and whose ordinate is zero.