Page 148 - NUMINO Challenge_D2
P. 148
Problem solving Therefore, Aladdin needs to try at least
1 Think about the worst case. To open the first car, 10 9 4 3 52 times before he opens all
Daniel must try 10 times. To open the second the treasure boxes.
car, he needs to try nine times. Using the same
method, he can open the tenth car after one try. 3 The pairs of numbers that add up to 30 are
Therefore, Daniel needs to try at least 10 9
8 2 1 55 times. (1,29), (3, 27), (5, 25), (7,23), (9, 21), (11, 19), and
(13, 17) and the remaining number 15 cannot be
2 The number of remaining votes is paired. The worst case when picking two
numbers that add up to 30 is to pick 15 first.
(Total Number of Votes) (Number of Checked Next, pick a number from each of the seven
Votes) (Number of Invalid Votes). Therefore pairs of numbers that add up to 30, and pick
55 14 17 11 2 11. To have Mike selected one more number from the remaining seven
as class president, the worst case is that the numbers. Therefore, you need to pick at least
remaining 11 votes are shared between Tom, 1 7 1 9 numbers.
who is in first place, and Mike, who is in third
place. If Mike wants to have more votes than 4 There are 100 27 17 10 6 40 votes left.
Tom, then he should first have 17 votes, which is
the same number of votes Tom has. Therefore, If John wants to be selected, the worst case is
Mike should first have six votes from the that he shares the remaining 40 votes with Amy,
remaining 11 votes. Next, if Mike gets three who has the most votes therefore so far. First,
votes or more from the remaining five votes, John must have the same number of votes as
then he'll be selected. Therefore, Mike should Amy, which is 27 votes, therefore he should get
get at least 6 3 9 votes. 27 10 17 votes. Then, from the remaining
40 17 23 votes, John must get at least 12
votes in order to win. Therefore, John must get
at least 17 12 29 votes.
Creative Thinking p.40~p.41 5 Crossing the River p.42~p.43
1 If you put one marble at a time in the bag, then Example A, A 10, 20 45
each bag will have one marble, therefore you'll Example white white white, white
need 9 marbles this time. If you repeat the same
steps, then each bag will have 2 marbles, Type 5-1 Least Number of Time Needed p.44~p.45
therefore 18 marbles will be in the bags. If you
put the last marble in one of the 9 bags, then one 1 Since one person must come back, then the two
bag will have 3 marbles. Therefore, there will be people who take the shortest time should leave
at least 3 marbles in one bag. first. Therefore, A, who takes one minute, and B,
who takes two minutes, should leave first.
2 The worst case when Aladdin tries to open the
2 B, 2 2 C, D B, 2
first treasure box is that he opens it after ten
tries. The worst case when he tries to open the 3 2 1 21 2 2 1 11 2 2 44 minutes
second treasure box is that he opens it after
nine tries. Using the same method, the worst
case when he tries to open the eighth treasure
box is that he opens it after three tries using the
remaining three keys.
Answer Key
1 Think about the worst case. To open the first car, 10 9 4 3 52 times before he opens all
Daniel must try 10 times. To open the second the treasure boxes.
car, he needs to try nine times. Using the same
method, he can open the tenth car after one try. 3 The pairs of numbers that add up to 30 are
Therefore, Daniel needs to try at least 10 9
8 2 1 55 times. (1,29), (3, 27), (5, 25), (7,23), (9, 21), (11, 19), and
(13, 17) and the remaining number 15 cannot be
2 The number of remaining votes is paired. The worst case when picking two
numbers that add up to 30 is to pick 15 first.
(Total Number of Votes) (Number of Checked Next, pick a number from each of the seven
Votes) (Number of Invalid Votes). Therefore pairs of numbers that add up to 30, and pick
55 14 17 11 2 11. To have Mike selected one more number from the remaining seven
as class president, the worst case is that the numbers. Therefore, you need to pick at least
remaining 11 votes are shared between Tom, 1 7 1 9 numbers.
who is in first place, and Mike, who is in third
place. If Mike wants to have more votes than 4 There are 100 27 17 10 6 40 votes left.
Tom, then he should first have 17 votes, which is
the same number of votes Tom has. Therefore, If John wants to be selected, the worst case is
Mike should first have six votes from the that he shares the remaining 40 votes with Amy,
remaining 11 votes. Next, if Mike gets three who has the most votes therefore so far. First,
votes or more from the remaining five votes, John must have the same number of votes as
then he'll be selected. Therefore, Mike should Amy, which is 27 votes, therefore he should get
get at least 6 3 9 votes. 27 10 17 votes. Then, from the remaining
40 17 23 votes, John must get at least 12
votes in order to win. Therefore, John must get
at least 17 12 29 votes.
Creative Thinking p.40~p.41 5 Crossing the River p.42~p.43
1 If you put one marble at a time in the bag, then Example A, A 10, 20 45
each bag will have one marble, therefore you'll Example white white white, white
need 9 marbles this time. If you repeat the same
steps, then each bag will have 2 marbles, Type 5-1 Least Number of Time Needed p.44~p.45
therefore 18 marbles will be in the bags. If you
put the last marble in one of the 9 bags, then one 1 Since one person must come back, then the two
bag will have 3 marbles. Therefore, there will be people who take the shortest time should leave
at least 3 marbles in one bag. first. Therefore, A, who takes one minute, and B,
who takes two minutes, should leave first.
2 The worst case when Aladdin tries to open the
2 B, 2 2 C, D B, 2
first treasure box is that he opens it after ten
tries. The worst case when he tries to open the 3 2 1 21 2 2 1 11 2 2 44 minutes
second treasure box is that he opens it after
nine tries. Using the same method, the worst
case when he tries to open the eighth treasure
box is that he opens it after three tries using the
remaining three keys.
Answer Key
