**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.

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.

## 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**.