Math Challenge Club Week 3 Worksheet
1. Anna, Betty, Clara, and Diana each play different sport: karate, soccer, volleyball, or judo. Anna
does not like sports played with balls. The Judo player Betty often attends a soccer match with
Clara to watch their friend play. Who plays which sports?
Anna : Karate
Betty : Judo
Clara : Volleyball
Diana : Soccer
1. We know from the information that Betty is the Judo player.
2. Since Anna doesn’t like sports with balls, the only choices for her are Judo and Karate. But we
already know that Betty is the Judo player. So Anna has to be the Karate Kid.
3. Betty attends soccer game with Clara to watch their friend play. This means that Clara cannot
be the soccer player. So she is the volleyball player.
4. By process of elimination, Diana is the soccer player.
2. There are four buttons in a row as shown below. Two of them show happy faces, and two of them
show sad faces. If we press on a face, its expression turns to the opposite (e.g. a happy face turns
into a sad face after the touch). In addition to this, the adjacent buttons also change their
expressions. What is the least number of times you need to press a button in order to get all happy
faces? 3
ABAB ⇨ BABB ⇨ ABBB ⇨ AAAA
3. Jim needs exactly 5 quarts of water. He goes out to the river with the only two buckets he has: a 7quart bucket and a 3-quart bucket. The buckets are not see-through and they have no level
indicator markings. How does Jim get exactly 5 quarts of water?
1. Fill 7-qt bucket and pour into 3-qt bucket. Empty out 3-qt bucket.
2. Pour again into 3-qt bucket and empty it out again.
3. Now pour the remaining 1 qt of water into the 3-qt bucket.
4. Fill the 7-qt bucket and pour into the 3-qt bucket (which already has 1 qt of water inside). This
means 2 qt of water will be poured from the 7-qt bucket leaving behind 5 quarts of water in the
7-qt bucket.
5. Empty out the 3-qt bucket, now you are left with 5 qt of water in the 7-qt bucket.
Alternate solution:
1. Fill the 3-qt bucket, and pour the water into the 7-qt bucket.
2. Fill the 3-qt bucket again, and pour the water into the 7-qt bucket.
Math Challenge Club Week 3 Worksheet
3. Now there is 6 qts of water in the 7-qt bucket. Fill the 3-qt bucket and fill up the 7-qt bucket.
4. 2 qts of water will be left in the 3-qt bucket. Empty the 7-qt bucket and transfer the 2 qts of
water into the 7-qt bucket.
5. Fill up the 3-qt bucket and pour it into the 7-qt bucket which is already holding 2 qts of water to
make 5 qts.
4. On one side of a river there are 3 heroes and 3 villains. They have a boat they can use to cross
the river, but the boat can only hold 2 people and there needs to be one person inside to row it.
If there are ever more villains than heroes on one side of the river, the villains kill the heroes.
How do you get everyone across safely?
3H 3V
>1H 1V
H<
1V
3H 2V
>2V
V<
2V
3H 1V
>2H
1H 1V<
2H 2V
>2H
V<
3H
3V
>2V
V<
3H 1V
2V
>2V
1H 1V
3H 3V
5. The beaver says: “If there is a vowel on one side of a card, then there is always an even
number on the opposite side of the same card.” You only see one side of each card so you do
not know if the beaver is telling the truth. Which cards must be flipped over to determine if the
beaver is telling the truth? E & 7

E is an obvious choice since once you flip it over, if it is an
even number, then the beaver is telling the truth, but if it’s
not, then he is lying.

Some of you might be tempted to flip over 2 thinking that the flip side should be a vowel. But
that is not the case, because the statement states that “if there is a vowel on one side of a
card, then there is always an even number on the opposite side of the same card.”, this
statement does not stay true when you flip the statement around, “if there is an even
number, the opposite side will be a vowel”, that is not true at all! If there’s a consonant on
the back, the statement is still true. It didn’t say anything about what’s behind consonants!
Math Challenge Club Week 3 Worksheet

A way to understand this is let’s say I have the following statement, “Everyone in Pippi
Longstocking’s family has red hair.”, But when we flip it, it becomes “Everyone that has red
hair is in Pippi Longstocking’s family”, which we know is not true at all!

So 7 needs to be flipped to see if there is a vowel at the opposite side, if there is, then the
beaver is lying, if it’s not a vowel, then the beaver is telling the truth.
6. Move 2 matchsticks so there are exactly 4 squares.
7. What’s the next number in the “Pyramid Sequence”?
13112221
*This is not a math question, but a reading question. Each subsequent
1
row is the “write out” of the previous. For example:
11
21

1st row is 1, which can be read as One 1, hence 2nd row is 11

2nd row can be read as two 1, hence 3rd row is 21

3rd row can be read as one 2, one 1, so the 4th row is 1211
111221

4th row can be read as one 1, one 2, two 1, so the 5th row is 111221
312211
1211
and so on.
8. A: “B always lies”
B: “I sometimes lie and sometimes tell the truth”
C: “I always tell the truth.”
Of A, B and C, one of them always tells the truth, one always lies, and one sometimes lies and
sometimes tells the truth. You know for a fact that this time the one that sometimes lies and
sometimes tells the truth is lying. Which one of them is which?
Liar =

B
Truth-teller = A
Half & Half = C
1st fact we know is that the “sometimes lies and sometimes tells the truth” person lied this
time, which means out of the 3 people, 2 people lied and 1 person told the truth.
Math Challenge Club Week 3 Worksheet

The 1st person to look at is B. If B were the “sometimes lies and sometimes tells the truth”
person, then he would be telling the truth. But we already know that the half and half person
lied. So B is not the “sometimes lies and sometimes tells the truth” person and he is lying.
And B obviously isn’t the truth-teller, so B is the liar.

Now that we proved B is the liar, then A is the truth-teller (since the sometimes-truth person
is lying)

So C has to be the “sometimes lies and sometimes tells the truth” who is lying when C said
“I always tell the truth.”
9. There were 2013 inhabitants on an island. Some of them were knights and the others were liars.
The knights always tell the truth and the liars always lie. Every day, one of the inhabitants said:
"After my departure the number of knights on the island will equal the number of liars" and then left
the island. After 2013 days, there is nobody on the island. How many liars were there initially?
1006 liars
*key fact here is that everyone left the island, which means that knights have to be one more than
liars in order for that to happen. Let’s look at the different scenarios.
Let’s simplify the question to only 7 people on island.
Scenario 1 : way more liars than knights

If 5 liars and 2 knights are left on the island, the first 2 liars have no problem leaving the island,
but when it comes down to 3 liars and 2 knights, neither knight nor liar can leave the island
because if one of the 3 liars left the island, leaving behind 2 liars and 2 knights, he will be
speaking the truth, so that cannot happen, but if it were one of the knights to leave, there
would be 3 liars and 1 knight left when he leave, so he won’t be speaking the truth, so he can’t
leave either. This proves that there needs to be more knights than liars in order for everyone to
leave.
Scenario 2 : way more knights than liars

If 5 knights and 2 liars, 2 liars can leave, but the knights can never leave. This proves that
knights can only be 1 more than liars.
Scenario 3 : 1 more knight than liars
Math Challenge Club Week 3 Worksheet

If 4 knights and 3 liars, the 1st person to leave would have to be the knight, he would be
speaking the truth since 3 knights and 3 liars are left behind. 2 nd person to leave have to be
the liar, since he would be lying because once he left, only 3 knights and 2 liars are left. 3 rd
person to leave would be the knight again…so knights and liars need to leave alternately in
order to keep the statement true and false alternately also, so that everyone can leave the
island.
10. In my phone book, the telephone number of my friend Ryan has six digits, but it must be a sevendigit number as are all phone numbers in his area. I have no idea what digit I forgot to write down,
or its position in the number. How many numbers do I have to dial to be certain that I will call
Ryan’s number? (Note that in Ryan’s area phone numbers may start with any digit, including 0). 64
Let’s say the 6 digit number is 123456, a digit can be inserted at 7 different spots.
A1B2C3D4E5F6G
The digit 0-9 can be inserted into spaces A-G, there are 10 digits you can put into each A-G slot.
However, we need to be careful of some double counting here. We are building different phone
numbers, let’s take a look:

If we put 1 into A we get the number 1123456. However we also get this number when we
put 1 into B!

Similarly, if we put 2 into B it has the same result as putting 2 into C! This is true for each
slot.

Therefore, to build our different phone numbers, we can put 10 digits in A, then 9 digits in B
(we avoid the digit 1, since A=1 counted that phone number already), then 9 digits in C (no
2), then 9 digits in D (no 3), and so on. So our answer is 10+9 digits x 6 slots=64 phone
numbers.