eTwinning Project

1. Numbers - solutions of the problems proposed
by Serbia, Polonia, Turkey
​"​Jean Monnet​"​ High School, Bucharest, Romania
Math&Mysteries e-Twinning Project
2016-2017

2. TEAM RIDDLES Image
TEAM who wants to solve it
AND SOLUTIONS
TEAM
SERBIA
RIDDLE 9
Nikola have been
wandering all
day: which is
greater and for
how much, the
sum of all even or
the sum of all odd
numbers in the
first hundred
numbers?
Team 7
Solution:
S​2​ = 2+4+6+.....+100
S​1​ = 1+3+5+.....+99
S​2​ –S​1​ = (2–1) + (4–3)
+…..+ (100–99)
S​2​ – S​1​ = 1+1+1+.....+1
S​2​ – S​1​ = 1 50×
S​2​ – S​1​ = 50
Answer:
So, S​2​ is greater with 50

3. TEAM
SERBIA
RIDDLE 10
To number all the
pages of the book,
we require 492
digits.
How many pages
that book
contain?
TEAM 8 and 5
From page 1 to 9 we
have 9 digits
From page 10 to 99 we
have (99 10+1) 2 digits– ×
= 90 2=180 digits×
From page 100 to 999 we
have (999–100+1) 3×
digits = 2700 digits.
This means that the book
has less than 999 pages
492 – 9 – 180 = 303 digits
remaining from pages 99
to end which represent
303:3=101 pages more.
The book will have
99+101=200 pages

4. TEAM
SERBIA
RIDDLE 11
Two trains cross
same number of
kilometers,
although one
traveled 6 days
more than the
other. If first train
crosses 36 km per
day and the other
60 km per day, how
many days did each
traveled?
TEAM 1
Solution:
d = how many days the
second train traveled.
36(d+6) = 60d
36d +36 x 6 = 60d
24d = 36x6
d = 36x6:24
d = 36x6:6:4
d = 9
Answer:
First train traveled 15
days.
Second train traveled 9
days.

5. TEAM
SERBIA
RIDDLE 12
Place the numbers
1, 2, 3, 4, 5, 6, 7, 8
and 9 in the
circles, so that
every side of
triangle have the
sum 17.
TEAM 8
9,8,7 should be on the middle
points in different lines as every
2 of them could not be in the
same line with other 2 numbers
(sum will be above 17 otherwise)
1,2,3 (smallest numbers) should
be on the corners as they count
in 2 lines each
On the same line with 9 we
could have 2 combinations
(9,1,2,5 or 9,1,3,4)
On the same line with 8 we
could have 3 combinations
(8,1,2,6 or 8,1,3,5 or 8,2,3,4)
On the same line with 7 we
could have 2 combinations
(7,1,3,6 or 7,2,3,5)
If we choose 1,5,9,2 as base line
then 3 is on top. We choose
3,4,8,2 on the right line and
3,7,6,1 on the left line.

6. TEAM
SERBIA
RIDDLE 13
A pillar is 15m tall.
A snail at the
bottom climbsup
3m each day and
slips back 2m each
night.
How many days
will it take the snail
to reach the top of
the pillar?
TEAM 2
The snail reaches the top
of the pillar in the
thirteenth days, because it
climbs up 12 m in 12 days
and, in the thirteenth:
12+3-2= ​15​-2
If he climbs up 3 m each
day and slips back 2 m
each night, that means he
goes up 1 m each 24
hours. So, he goes up 12 m
in 12 days and in the
thirteenth he will go up
the last 3 m, to the top of
the pillar.
Answer: 13 days

7. TEAM
SERBIA
RIDDLE 14
Fifteen people
enter a chess
tournament in
which each person
must play every
other person
exactly once.
Determine the total
number of games
that will be played.
TEAM 6
The first person plays 14
games with the others 14
players (once with every
other person), the second
person plays 13 games
with the others 13 players
and so on...until the last
person(the 14th) who
plays with the last player.
Total number of games
can be determinated using
the summ of Gauss:
S= n(n+1):2
S= 1+2+3+......+14
S=(14 15):2= 210:2=105·
Answer: 105 grames

8. TEAM
SERBIA
RIDDLE 16
At some business
meeting each
person shakes
hands exactly once
with everyone else.
If 10 handshakes
took place, how
many people were
present?
TEAM 6
We have x people who
shake hands with the
others x–1 people and
we’ll have x(x–1):2
handshakes.
The first person, shake
hands with (x–1) people,
the second person shake
hands with (x–2) people
(without the first one and
himself), the third person
shake hands with
(x–3)people and so on …
Using the sum of Gauss:
S=n(n+1):2
x(x–1):2=10 x(x–1)=20⇒
x(x–1)=5 4 x=5⇒ · ⇒
Answer: 5 people were
present

9. TEAM
POLAND
RIDDLE 17
I bought plates on
the sale. When I
came home , I saw
that 2/3 is chipped,
1/2 is cracked, 1/4
is chipped and
cracked.
How many plates I
bought if 2 plates
are whole?
TEAM 4
x=nr. of plates
⅔ by x is 8/12 by x;
½ by x is 6/12 by x;
¼ by x is 3/12 by x;
--------------------------
8/12 + 6/12 - 3/12+a=12/12
by x
11/12+a=12/12 by x ⇒
a=1/12 by x
a=2 (nr. of whole plates)
2=1/12 by x ⇒ x= 2 x
12=24 bought plates

10. TEAM
POLAND
RIDDLE 18
In a huge chest
there are 9 big
chests. In each big
chest are 9 medium
chests. In each
medium chest are
9 small chests.
How many chests
are there in total?
TEAM 3
We know that 9 small
chest are in every medium
chest, then we have:
9x9=81 small chest
We know that there are
also 9 medium chest:
81+9=90 small and
medium chest
There is as well a huge
chest, so in total there are:
90+1=91(small, medium
and huge chest).

11. TEAM
TURKEY
RIDDLES 22
A lot of the water
lilies growing on a
lake.
They double their
surface every day
and in 20 days
covering the entire
lake.
How long lilies
cover half of the
lake?
TEAM 7
In the 19th because when
they double their surface
another time they will
cover the entire lake.
TEAM 3
If in every day it doubles
and in the last day the
lake is covered, then the
half is in the 19th day:
day 1: a
day 2: 2a
…..
day 19: 2​18​
a
day 20: 2​19​
a=2 2​18​
a·

12. TEAM
TURKEY
RIDDLES 23
The sorcerer Gandoulf
proposes to his servant
Blue Dwarf to increase
his fortune. A pot of
50 gold pieces is in
front of him.
Gandoulf says:
"If you answer my
question correctly, you
take 5 pieces of the
pot.
If you're wrong, you
put two in the pot. "
After 21 questions, the
pot contains 50 gold
coins again.
How many good
answers did Blue
Dwarf give? Justify
your answer.
TEAM 5
x -number of correct
answers
y -number of wrong
answers
50 – 5x + 2y =50 5x = 2y⇒
x + y = 21 x = 21– y ,⇒
5 ( 21 – y ) = 2y ⇒
105 – 5y = 2y 7y = 105⇒
y = 15 x = 21 – 15= 6⇒ ⇒
6 correct answers
15 wrong answers
6 5 gold = 30 gold×
50–30=20 gold
20 gold stay in the pot
21 – 6 = 15 (quetions/wrong)
15 questions gold = 30 gold2×
20 + 30 = 50 gold
Remaining in the pot 50 gold.