Answer key for the maths brevet France 2017

Exercise 1: The sum of the probabilities of the outcomes is equal to [latex]\frac{1}{5}[/latex]1/5. 5/5-2/5=3/5. No, it will have as many probabilities because the ball is put back in the urn. 8 green balls for a probability of 2/5. 1/5 represents 4 balls and 3/5 represents 3×4=12 green balls. Exercise 2: The coordinates of the … Read more

Arithmetic and prime factor decomposition: answer key to 3rd grade math exercises in PDF.

Answer key for math exercises in 3rd grade on arithmetic and prime factor decomposition. To know how to do a Euclidean division and the notion of divisor and multiple, to make a fraction irreducible.

Exercise 1:

The three Euclidean divisions below are correct.

Euclidean division

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:

Euclidean division.

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?

sport-co

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.

Divisors and arithmetic.

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:

bead packNori 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.

a.36=4\times  \,9=2^2\times  \,3^2

b.18375=3\times  \,125\times  \,49=3\times  \,5^3\times  \,7^2

c.3872=32\times  \,121=2^5\times  \,11^2

d.1183=91\times  \,13

e.214375=625\times  \,343=5^4\times  \,7^3

Exercise 10:

Write the prime factor decomposition of the following integers:

180=2^2\times  \,3^2\times  \,5

63=3^2\times  \,7

1225=5^2\times  \,7^2

3672=2^3\times  \,3^3\times  \,17

416=2^5\times  \,13

24000=2^6\times  \,3\times  \,5^3

Exercise 11:

Find the number you are looking for.

search-number

The solutions are 101; 113; 137 and 149.

Exercise 12:

Use prime factor decompositions to make these fractions irreducible.

504=2^3\times  \,3^2\times  \,7;13500=2^2\times  \,3^3\times  \,5^3\\4400=2^4\times  \,5^2\times  \,11;11466=2\times  \,3^2\times  \,7^2\times  \,13

Make the following fractions irreducible: \frac{504}{4400};\frac{504}{11466};\frac{13500}{504}.

\frac{504}{4400}=\frac{63}{550};\frac{504}{11466}=\frac{4}{91};\frac{13500}{504}=\frac{375}{14};

Exercise 13:

Make the following fractions irreducible: \frac{8800}{1638};\frac{64}{4400};\frac{1260}{1638};\frac{1638}{810}.

\frac{8800}{1638}=\frac{4400}{819};\frac{64}{4400}=\frac{4}{275};\frac{1260}{1638}=\frac{10}{13};\frac{1638}{810}=\frac{91}{45};

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

\frac{60775}{114400}=\,\frac{60775:3575}{114400:3575}=\frac{17}{32} .

Exercise 17:

Let the numbers A =\,\frac{117}{63} and B = – \,\frac{8}{7}.

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

\frac{117}{63}=\,\frac{117:9}{63:9}=\frac{13}{7}

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

A-B=\frac{13}{7}-(-\frac{8}{7})=\frac{13}{7}+\frac{8}{7}=\frac{21}{7}=3

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 \,\frac{520}{336} = \,\frac{65}{42}.

GCD( 520 ; 336 ) = 8

\frac{520}{336}=\frac{520:8}{336:8}=\frac{65}{42}

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

\frac{408}{578}=\frac{408:34}{578:34}=\frac{12}{17}

Show that PGCD( 2499 ; 1911 ) = 147

\frac{2499}{1911}=\frac{2499:147}{1911:147}=\frac{17}{13}

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 \,\frac{682}{352}

clearly indicating the method used.

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

\,\frac{682}{352}=\frac{682:22}{352:22}=\frac{31}{16}

Exercise 24:

Calculate and give the result as an irreducible fraction:

\,A=\frac{5}{4}+\frac{11}{4}\times  \,\frac{20}{33}=\frac{5}{4}+\frac{11}{4}\times  \,\frac{4\times  5}{11\times  \,3}\\=\frac{5}{4}+\frac{5}{3}=\frac{5\times  \,3}{4\,\times  \,3}+\frac{5\,\times  \,4}{3\,\times  \,4}=\frac{15+20}{12} .
\,\fbox{A=\frac{35}{12}}

\,B=\frac{\frac{5}{2}}{\frac{7}{4}+\,\frac{9}{2}}=\frac{\frac{5}{2}}{\frac{7}{4}+\,\frac{18}{4}}=\frac{\frac{5}{2}}{\frac{25}{4}}=\frac{5}{2}\times  \,\frac{4}{25}=\frac{5\times  2\times  2}{2\times  5\times  5}
\,B=\fbox{\frac{2}{5}}
Exercise 25:

Calculate and give the result in scientific notation:

C=15\times  (10^7)^2\times  \,3\,\times  \,10^{-5}=15\times  \,3\times  \,10^{7\times  \,2-5}=45\times  \,10^9.
\fbox{C=4,5\times  \,10^{10}}

Exercise 26:

1.

Calculate A and give the result as a fraction.

\,A=\frac{13}{7}-\frac{2}{7}\times  \,\frac{15}{12}=\frac{78}{42}-\frac{2}{7}\times  \,\frac{15}{2\times  6}=\frac{78}{42}-\frac{15}{7\times  6}=\frac{78}{42}-\frac{15}{42}
\fbox{A=\frac{63}{42}}

2. Write B in the form \,b\sqrt{3} where b is an integer .

\,B=7\sqrt{75}-5\sqrt{27}+4\sqrt{48}=7\sqrt{25\times  3}-5\sqrt{9\times  3}+4\sqrt{16\times  3}

\,B=7\times  5\sqrt{3}-5\times  3\sqrt{3}+4\times  4\sqrt{3}=(35-15+16)\sqrt{3}
\fbox{B=36\sqrt{3}

3.

Calculate C and give the scientific form of the result.
\,C=\frac{0,23\times  \,10^3-1,7\times  \,10^2}{0,5\times  \,10^{-1}}=\frac{(23-17)\times  10^1}{5\times  \,10^{-2}}=\frac{6}{5}\times  10^3
\fbox{C=1,2\times  10^3

Exercise 27:

1. Determine the PGCD of 288 and 224.

Let’s use Euclid’s algorithm.

288=1\times  \,224+64

224=3\times  \,64+32

64=2\times  \,32+0

The PGCD being the last non-zero remainder, we deduce that pgcd(288,224)=32

2 . Write the fraction \frac{224}{288} in irreducible form.

\frac{224}{288}=\frac{224:32}{288:32}=\frac{7}{9} 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.

Photo exhibition

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

Fraction

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

Fractions

Exercise 35:

Calculations with fractions.

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.

Bracelet

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 .

Artist painter

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.

excel spreadsheet

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.

Raspberry 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.

flowers

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 \frac{3120}{2760} irreducible.

\frac{3120}{2760}=\frac{3120:120}{2760:120}=\frac{26}{23}

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.

Answer key to math exercises on arithmetic and prime factor decomposition in 3rd grade.

After consulting the answers to these math exercises on arithmetic and prime factor decomposition, you can return to the exercises in third grade.

The exercises in the third grade.

Scratch: programming exercises in 5th grade.

Exercises with Scratch in order to work on the algorithm and programming part for the students of fifth grade in cycle 4.
Assimilation of the different commands and bricks and understanding of algorithms.

Exercise 1:

Scratch Program Answer Key

Where is the cat when you click on the block?

I click on flag 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 flag is not inserted in the program.

Exercise 2:

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

cat on the move on scratch

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:

four moving programs with scratch.

Program D has just been executed.

Exercise 4:

exercise-4-scratch

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

Exercise 5:

Which of these three programs has just been run?

exercise-5-1-scratch exercise-5-2-scratch

exercise-5-scratch

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?

exercise-6-1-scratch exercise-6-scratch

Program C is suitable.

Exercise 7:

When the program starts, what will the lion do?

exercise-7-1-scratch exercise-7-scratch
The lion will retrieve the green marble and return to its original position.

Exercise 8:

Which of these three programs has just been executed?

exercise-8-1-scratch exercise-8-scratch

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?

exercise-10-1-scratch exercise-10-scratch

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-11-scratch

Answers to the exercises on Scratch and programming in 5th grade.

After having consulted the answers to these exercises on Scratch and algorithms in 5th grade, you can return to the exercises in 5th grade

Fifth grade exercises.

Scratch : corrected algorithms and programming exercises in 4th grade in PDF.

Answers to the programming exercises with scratch in the fourth grade of cycle 4.

Exercise 1:

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

scratch-ex1

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?

scratch-ex2

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?

scratch-ex3

The program executes plot 3.

Exercise 4:

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

scratch-ex4

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?

scratch-ex5

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?

scratch-ex6

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?scratch-ex7

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?

scratch-ex8

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

Exercise 9:

Which instruction calculates 3 + 4×5 – 2?

scratch-ex9

This is instruction 4.

Exercise 10:

Is it possible that the cat announces 750?

What is the maximum number that can be obtained?

scratch-ex10

This program will provide as maximum result :

10\times  ,9\times  ,8+10+9+8=720+27=747.

It is impossible for this program to provide 750.

The answers to the exercises on Scratch in 4th grade.

After having consulted the answers to these exercises on Scratch in 8th grade, you can return to the exercises in 8th grade.

Eighth grade exercises.

Scratch : correction of the exercises in 3rd grade of programming.

The answers to the exercises on scratch in cycle 4 in the 3rd grade. Know how to create a program and set up an algorithm to answer a given problem.

Exercise 1

Associate each program with the corresponding output.ex1-scratch

Program 1: plot 2.

Program 2: plot 3.

Program 3: plot 1.

Exercise 2

ex2-scratch

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?

ex3-scratch

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?

ex4-scratch

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?

ex5-scratch

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?
ex6-scratch

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?

ex7-scratch

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?

ex8-scratch

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?

ex9-scratch

This program will calculate \sqrt{12^2-15^2}=\sqrt{144-225}=\sqrt{-81} 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?

ex10-scratch

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.

The answers to the exercises on scratch in PDF in 3ème.

After having consulted the answers to these exercises on scratch in 9th grade, you can go back to the exercises in 3rd grade.

Third grade exercises.

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