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Mathematics 6.

Table of contents
Divisibility
The multiples and divisors of natural numbers (review)
Look at the remainder
The divisibility of the sum, the difference and the product
Rules of divisibility
Divisibility based on the sum of the digits
Additional divisibility rules
Primes and compound numbers
Writing compound numbers as the product of primes
Common factors, the greatest common factor
Common multiples, lowest common multiple
Mixed exercises
How to solve problems?
Rational numbers I.
Axial symmetry
Rational numbers II.
Proportion
Percentages
Probability, statistics
Supplementary material
The divisibility of sums, differences and products
\latex{ 5 } boys and \latex{ 3 } girls want to ride the dodgem cars. Can they sit in the dodgem cars in pairs?
Can they sit in the dodgem cars in such a way that every boy and girl has a pair of the opposite sex?
The divisibility of the sum
Example 1
There are four coins in your wallet: one \latex{ 20 }-cent coin, one \latex{ 10 }-cent coin, one \latex{ 5 }-cent coin, and one \latex{ 1 }-cent coin. The sum of which two could you exchange for \latex{ 2 }-cent coins?
Solution
and
both can be exchanged for \latex{ 2 }-cent coins individually
and
neither can be exchanged for \latex{ 2 }-cent coins individually
Their sum:
+
can be exchanged for fifteen \latex{ 2 }-cent coins
even + even = even
Their sum:
+
can be exchanged for three \latex{ 2 }-cent coins
odd + odd = even
The sum of any two even natural numbers is even.

The sum of any two odd natural numbers is even.
It is also possible that only one of the two coins can be exchanged for \latex{ 2 }-cent coins. These are the following cases:
Their sum is odd; therefore, they cannot be exchanged for \latex{ 2 }-cent coins.
odd + even = odd
The sum of an even and an odd natural number is odd.
You bought red and green apples at the market. In which case can the apples be divided equally between three children?
I.  
\latex{6}
\latex{+}
\latex{9}
\latex{=}
\latex{15}
\latex{2\times 3}
\latex{+}
\latex{3\times 3}
\latex{=}
\latex{5\times 3}
\latex{+}
\latex{=}
Both addends are divisible by \latex{ 3 }, so the sum is also divisible by \latex{ 3 }.
The sum of the remainders:
\latex{\textcolor{009fe9}{0}+\textcolor{009fe9}{0}=\textcolor{009fe9}{0}.}
If every addend is divisible by a natural number, the sum is also divisible by the given number.
II.  
\latex{6}
\latex{+}
\latex{10}
\latex{=}
\latex{16}
\latex{2\times 3}
\latex{+}
\latex{3\times 3+\textcolor{009fe9}{1}}
\latex{=}
\latex{5\times 3+\textcolor{009fe9}{1}}
\latex{+}
\latex{=}
Exactly one addend is not divisible by \latex{ 3 }; therefore, the sum is not divisible by \latex{ 3 }.
The sum of the remainders:
\latex{\textcolor{009fe9}{0} +\textcolor{009fe9}{1}=\textcolor{009fe9}{1}.}
If exactly one addend is not divisible by a natural number, then the sum is not divisible by the given number either.
III.  a)
\latex{4}
\latex{+}
\latex{7}
\latex{=}
\latex{11}
\latex{1\times 3+\textcolor{009fe9}{1}}
\latex{+}
\latex{2\times 3+\textcolor{009fe9}{1}}
\latex{=}
\latex{3\times 3+\textcolor{009fe9}{2}}
\latex{+}
\latex{=}
None of the addends nor the sum are divisible by \latex{ 3 }.
The sum of the remainders:
\latex{\textcolor{009fe9}{1} +\textcolor{009fe9}{1}=\textcolor{009fe9}{2}.}
III.  b)
\latex{5}
\latex{+}
\latex{7}
\latex{=}
\latex{12}
\latex{1\times 3+\textcolor{009fe9}{2}}
\latex{+}
\latex{2\times 3+\textcolor{009fe9}{1}}
\latex{=}
\latex{4\times 3}
\latex{+}
\latex{=}
None of the addends are divisible by \latex{ 3 }, but their sum is divisible by \latex{ 3 }.
The sum of the remainders:
\latex{\textcolor{009fe9}{2}+\textcolor{009fe9}{1}=\textcolor{009fe9}{3}.}

\latex{3 = 1\times3 + \textcolor{009fe9}{0}.}
If neither of the addends is divisible by a natural number, their sum may or may not be divisible by the given number.
The remainder of the sum can be determined on the basis of the sum of the addends' remainders.
The divisibility of the difference
Example 2
You have three blue and three red number cards.
\latex{30}
\latex{31}
\latex{32}
\latex{6}
\latex{7}
\latex{8}
  1. Group the cards according to the remainder of the numbers when divided by three.
  2. Write down all the cases when the minuend is on a blue card, and the subtrahend is on a red card. In which cases is the difference divisible by \latex{ 3 }?
Solution
a)
b)
The difference divided by \latex{3}
divisible
not divisible
\latex{30-6;}
\latex{31-7;}
\latex{32-8;}
\latex{30-8;}
\latex{30-7;}
\latex{31-8;}
\latex{31-6;}
\latex{32-7;}
\latex{32-6.}
The difference is divisible by \latex{ 3 } if the remainder of the minuend and the subtrahend is equal when divided by \latex{ 3 }.
\latex{32}
\latex{8}
A difference is divisible by a natural number if the remainder of the minuend and the subtrahend are equal when divided by the given number.
The divisibility of a product
Example 3
The surface of a rectangle with an area of \latex{4\times 9} is covered with congruent rectangles whose sides are positive integers. List all the rectangles that can be used to cover the original rectangle.
Solution 
  1. The original rectangle can be covered with \latex{ 36 } unit squares and one rectangle of \latex{4\times 9} (with itself).
Any product is divisible by \latex{1} and itself.
\latex{36\times1}
\latex{1\times36}
  1. It can obviously be covered with nine \latex{ 4 }-unit or four \latex{ 9 }-unit rectangles in the following ways:
Any product is divisible by its factors.
\latex{9\times4}
\latex{4\times9}
  1. The original rectangle could be covered with \latex{ 4 }- and \latex{ 9 }-unit rectangles; therefore, it can also be covered with rectangles that can cover these unit areas. These can be
    \latex{ 2 }- or \latex{ 3 }-unit rectangles.
Any product is divisible by the divisors of its factors.
\latex{18\times2}
\latex{12\times3}
  1. Three other solutions can be obtained by converting the terms \latex{ 4 } and \latex{ 9 } to multiplications: \latex{4=2\times2} and \latex{9=3\times3}.
\latex{6\times6}
\latex{2\times18}
\latex{(2\times3) \times (2\times 3)}
You have listed all the factor pairs of \latex{4\times9=36}, so you have found all the possible solutions.
\latex{3\times12}
\latex{36}
\latex{1\times36}
\latex{2\times18}
\latex{3\times12}
\latex{4\times9}
\latex{6\times6}
Example 4
Are the following divisible?
  1. \latex{2\times7} divisible by \latex{2}
  2. \latex{8\times9} divisible by \latex{3} and \latex{4}
  3. \latex{8\times25} divisible by \latex{ 10 }?
Solution
A natural number \latex{ b } is divisible by natural number \latex{ a } if there is a natural number k\latex{ }, which multiplied by \latex{ a } gives \latex{ b }.
Based on this:
  1. \latex{2\times7} is divisible by \latex{2} because \latex{2\times7=7\times2}
  2. Write down the terms as the product of factors
\latex{8\times 9=8\times (3\times 3)=8\times 3\times 3=(8\times 3)\times 3=24\times 3.}
\latex{8\times9} is divisible by \latex{ 3 } because it can be expressed as \latex{24\times3}.
\latex{8\times 9=(4\times 2)\times 9=4\times 2\times 9=4\times (2\times 9)=4\times 18=18\times 4.}
\latex{8\times9}  is divisible by \latex{ 4 } because it can be expressed as \latex{18\times4}.
  1. Write down the terms as the product of factors.
\latex{8\times 25=4\times 2\times 5\times 5=4\times 5\times 2\times 5=20\times 10.}
\latex{8\times25} is divisible by \latex{ 10 } because it can be expressed as \latex{20\times10}.
The factors of multiplications can be arranged in several ways.
Any product is divisible by its factors and the divisors of the factors.
Exercises
{{exercise_number}}. A group of \latex{ 17 } children and \latex{ 8 } adults go to a bobsleigh track. Can they sit in pairs in the bobsleighs so that everyone has a partner?
{{exercise_number}}. Zoe kept her CDs on three shelves. In her new CD holder, she can arrange the CDs in four columns. Can she place her CDs in the holder in such a way that there are the same number of CDs in each column if she had
  1. \latex{ 44 }, \latex{ 60 }, and \latex{ 75 };
  2. \latex{ 38 }, \latex{ 45 }, and \latex{ 51 } CDs on the shelves of her old CD rack?
{{exercise_number}}. Six children are playing with beads at a table. The beads are sorted into five containers according to their colours. There are \latex{ 60 } white beads in the first container, \latex{ 38 } red ones in the second, \latex{ 43 } black ones in the third, \latex{ 71 } green ones in the fourth, and \latex{ 40 } blue ones in the fifth. Can they divide the beads equally among themselves? Justify your answer.
{{exercise_number}}. There are \latex{ 12 }, \latex{ 14 }, and \latex{ 9 } boys in three \latex{ 6th }-grade classes. Can the boys form a basketball team of \latex{ 5 } players in such a way that every boy can play and each boy is a member of only one team?
{{exercise_number}}. Four crates contain \latex{ 19 kg }, \latex{ 25 kg }, \latex{ 28 kg }, and \latex{ 30 kg } of potatoes, respectively. Can the potatoes be packed in bags of \latex{ 3 kg }?
{{exercise_number}}. At the Tricky store, the cash register is accurate on Mondays, Tuesdays, and Wednesdays. However, it makes small mistakes on Thursdays, Fridays, and Saturdays. Determine on which days the following receipts were issued with as little counting as possible.
a)
b)
c)
11 Mice Str., Cathill
ALBERT's SHOP
THANK YOU FOR SHOPPING!
RECEIPT
157.00
494.00
mineral water
cheese sticks
TOTAL:
09/08/06 17:04
bread
milk
25.00
83.00
259.00
a)
b)
c)
11 Mice Str., Cathill
ALBERT's SHOP
THANK YOU FOR SHOPPING!
RECEIPT
144.00
524.00
yoghurt
peach
TOTAL:
12/08/06 16:30
sugar
roll
168.00
112.00
67.00
ALBERT's SHOP
11 Mice Str., Cathill
RECEIPT
apple juice
chocolate
flour
potatoes
155.00
235.00
85.00
180.00
TOTAL:
659.00
10/08/06 15:27
THANK YOU FOR SHOPPING!
{{exercise_number}}. There are two switches at the two ends of a hallway for the same lamp. The lamp was turned on in the morning. During the day, the position of one of the switches was changed five times, while the other ten times. Was the lamp turned on or off at the end of the day?
Try to solve exercises \latex{8-12} without performing any calculations.
{{exercise_number}}.  Which of the following sums is divisible by \latex{ 7 }? Justify your answer.
  1. \latex{35 + 18 + 10}
  1. \latex{21 + 63 + 42 + 14}
  1. \latex{17 + 21 + 7 + 49}
{{exercise_number}}. Which of the following sums is divisible by \latex{ 4 }? Justify your answer.
  1. \latex{16 + 13 + 17}
  1. \latex{18 + 25 + 29}
  1. \latex{20 + 28 + 32 + 36}
  1. \latex{23 + 7 + 11 + 15}
{{exercise_number}}. Determine the remainder when dividing the addends by \latex{ 5 }, and then that of the sums when divided by \latex{ 5 }.
  1. \latex{26 + 34 + 12}
  1. \latex{13 + 27 + 18 + 4}
  1. \latex{19 + 42 + 53}
{{exercise_number}}. Determine the remainder when dividing the addends by \latex{ 9 }, and then that of the sums when divided by \latex{ 9 }.
  1. \latex{53 + 42 + 68 + 13}
  1. \latex{19 + 24 + 27 + 30}
  1. \latex{32 + 17 + 51 + 43}
{{exercise_number}}. Write numbers in the boxes so that the sum is divisible by
  1. \latex{5}          \latex{15 + 25 + 70 +\square}; 
  1. \latex{9}        \latex{10+23+\square};
  1. \latex{13}        \latex{17+\square+27+5};
  1. \latex{10}       \latex{20+\square+32} .
How many solutions are there in each case? What properties do these numbers have?
{{exercise_number}}. Decide whether the following statements are true or false.
  1. If none of three numbers are divisible by \latex{11}, then their sum is not divisible by \latex{11}.
  2. If one of three numbers is divisible by \latex{ 11 }, while the other two are not, their sum is also divisible by \latex{ 11 }.
  3. If one out of two numbers is divisible by \latex{7} and the other one is not, their sum is not divisible by \latex{7}.
  4. If neither of two numbers is divisible by \latex{7}, then their sum is divisible by \latex{7}.
  5. If only one of three numbers is not divisible by \latex{ 4 }, then their sum is divisible by \latex{ 4 }.
{{exercise_number}}. Without performing the subtractions, determine which differences are divisible by \latex{ 2 }.
  1. \latex{56 - 38}
  1. \latex{49- 35}
  1. \latex{73- 44}
  1. \latex{197- 109}
  1. \latex{216- 193}
{{exercise_number}}. Without performing the subtractions, determine which differences are divisible by \latex{ 5 }.
  1. \latex{417 - 305}
  1. \latex{526- 96}
  1. \latex{226- 111}
  1. \latex{345- 295}
  1. \latex{319- 248}
  1. \latex{228- 143}
  1. \latex{87- 49}
{{exercise_number}}. You had €\latex{ 20.5 } and spent €\latex{ 12.15 }.
  1. Can you exchange the remaining amount for \latex{ 2 }-cent coins?
  2. Can you exchange the remaining amount for \latex{ 5 }-cent coins?
{{exercise_number}}. Choose two, three or four of the following fruits so that you can pay only with \latex{ 5 }-cent coins.
\latex{21} ¢
\latex{62} ¢
\latex{50} ¢
€\latex{2}
\latex{45} ¢
\latex{22} ¢
{{exercise_number}}. Starting from \latex{ 2 }, the green numbers on the number line form a sequence that increases by \latex{ 3 }. What are the remainders of the green numbers when divided by \latex{ 3 }?
\latex{0}
\latex{3}
\latex{6}
\latex{9}
\latex{12}
\latex{15}
\latex{18}
\latex{+3}
\latex{+3}
\latex{+3}
\latex{+3}
\latex{+3}
\latex{2}
\latex{5}
\latex{8}
\latex{11}
\latex{14}
\latex{17}
{{exercise_number}}. Write down the two missing terms of each sequence. What are the remainders of the terms when divided by \latex{ 5 }?
  1. \latex{...; 28; 23; 18; 13; ...}
  1. \latex{...; 16; 21; 26; 31; ...}
{{exercise_number}}. Write down the two missing terms of each sequence. What are the remainders of the terms when divided by \latex{ 7 }?
  1. \latex{...; 36; 29; 22; 15; ...}
  1. \latex{...; 20; 27; 34; 41; ...}
{{exercise_number}}. A number is \latex{15} times \latex{7}.
  1. Is this number divisible by \latex{15}?
  2. Is this number divisible by \latex{5}?
  3. Is this number divisible by \latex{3}?
  4. Is this number divisible by \latex{21}?
{{exercise_number}}. Without performing the multiplication, decide whether the product of \latex{10\times21} is divisible by
  1. \latex{2};
  1. \latex{3};
  1. \latex{4};
  1. \latex{5};
  1. \latex{6};
  1. \latex{7}.
{{exercise_number}}. Substitute the symbols with numbers so that the products are
  1. multiples of \latex{6};
  2. multiples of \latex{10}.
  1. \latex{\square\times2\times3}
  1. \latex{2\times\square\times\triangle\times5}
  1. \latex{5\times\triangle\times9\times\square}
{{exercise_number}}. Write a number in the \latex{ △ } so that the product is a multiple of both \latex{ 9 } and \latex{ 10 }.
  1. \latex{3\times\triangle\times5}
  1. \latex{2\times\triangle\times3}
What is the smallest positive number that can be written in the \latex{ △ }?
{{exercise_number}}. Choose three of the following multiplications so that their sum is divisible by
  1. \latex{2};
  1. \latex{3};
  1. \latex{5};
  1. \latex{7.}
  1.  \latex{\fcolorbox{f6b900}{fef6e3}{ $2\times5$ }}
  1. \latex{\fcolorbox{f6b900}{fef6e3}{ $2\times7$ }}
  1. \latex{\fcolorbox{f6b900}{fef6e3}{ $3\times5$ }}
  1. \latex{\fcolorbox{f6b900}{fef6e3}{ $3\times7$ }}
  1. \latex{\fcolorbox{f6b900}{fef6e3}{ $5\times7$ }}
  1. \latex{\fcolorbox{f6b900}{fef6e3}{ $2\times11$ }}
{{exercise_number}}. Choose two of the following multiplications so that their difference is divisible by
  1. \latex{3};
  1. \latex{5};
  1. \latex{7};
  1. \latex{11.}
  1. \latex{\fcolorbox{f6b900}{fef6e3}{ $2\times11$ }}
  1. \latex{\fcolorbox{f6b900}{fef6e3}{ $3\times5$ }}
  1. \latex{\fcolorbox{f6b900}{fef6e3}{ $7\times11$ }}
  1. \latex{\fcolorbox{f6b900}{fef6e3}{ $3\times7$ }}
  1. \latex{\fcolorbox{f6b900}{fef6e3}{ $5\times7$ }}
  1. \latex{\fcolorbox{f6b900}{fef6e3}{ $11\times13$ }}
{{exercise_number}}. Answer without performing the operations. Justify your answer in each case. Is
  1. \latex{2\times 5+3\times 4}  divisible by \latex{2};
  1. \latex{3\times 7+2\times 15} divisible by \latex{3};
  1. \latex{3\times 7+9\times 5} divisible by \latex{9};
  1. \latex{3\times 10+2\times 9} divisible by \latex{6};
  1. \latex{7\times 10-3\times 7} divisible by \latex{7};
  1. \latex{7\times 8-4\times 11} divisible by \latex{4}?
Quiz
Matt and Sam played with nine numbered cards, marked from \latex{ 1 } to \latex{ 9 }. Matt hid one of the cards and told Sam to divide the cards into three groups so that the sum of the numbers on the cards was the same in each group. Sam solved the problem quickly and even divided the cards into four groups, with the sum of the numbers on the cards being equal in each group.
  1. Which card did Matt hide?
  2. How did Sam divide the cards into three groups?
  3. How did Sam divide the cards into four groups?

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