Question

Question of logical deduction #3

Answer 1

Ms Ho gives 3 cards marked with different integers to 3 students A, B and C. They can only read their own card.

Ms Ho: The number on student A's card is a 2-digit number. The number on student B's card is a single digit number. The number on student C's card is a 2-digit number which is less than 60. It is also equal to the number on student A's card multiplying number on student B's card.
A: I can't guess the numbers they have.
B: I can't guess the numbers they have too.
C: I can't guess the numbers they have neither
A: Now, I know what the numbers on student B's and student C's cards are

What are the numbers student A, B and C have?

Answer 2

Is the question right this time?

Answer 3

I think so...
A: __ <-number of 2 digits
B: _
C: __
C<60 and C =AxB

Answer 4

The key seems to be the fact that B and C couldn't guess. That means that there are at least two factorizations of the number C has that have a single digit, multiplied by double digit number. The possibilities I've come up with so far, are
30=2x15=3x10
40=2x20=4x10
42=2x21=3x14
48=2x24=4x12
50=2x25=5x10
56=2x26=4x13
60=2x30=3x20=6x10

Since A couldn't guess the first time, that means the two digit number must appear at least twice on this list. That excludes 42, 48, and 56. Leaving us with the following possibilities:
30=2x15=3x10
40=2x20=4x10
50=5x10
60=3x20=6x10

However, with the remaining ones, A has enough information! Thus, A has 15, B has 2, and C has 30.

Answer 5

Nevermind. I forgot 60=4x15

Answer 6

Wait! The order... C is the one who speaks first. B is the next. Here's the corrected one:

Ms Ho: The number on student A's card is a 2-digit number. The number on student B's card is a single digit number. The number on student C's card is a 2-digit number which is less than 60. It is also equal to the number on student A's card multiplying number on student B's card.
A: I can't guess the numbers they have.
C: I can't guess the numbers they have too.
B: I can't guess the numbers they have neither
A: Now, I know what the numbers on student B's and student C's cards are

I'm sorry!!!! extremely sorry!!!

Answer 7

And before B there seems to be another sentence:
After listening to C, A asked B: Can you guess C and my numbers?
Sorry, the question I have is a little messy and I have to translate it.. so there are so many mistakes !!!

Answer 8

im still working it out but just want to stop and tell callisto its alright, got enuf clues to solve the question i think.

Answer 9

I'm not sure. But the order of speech and every information is important in this type of question! I think...

Answer 10

Once again:

Ms Ho: The number on student A's card is a 2-digit number. The number on student B's card is a single digit number. The number on student C's card is a 2-digit number which is less than 60. It is also equal to the number on student A's card multiplying number on student B's card.
A: I can't guess the numbers they have.
C: I can't guess the numbers they have too.
After listening to C, A asked B: Can you guess C's number and my numbers?
B: I can't guess the numbers they have neither
A: Now, I know what the numbers on student B's and student C's cards are
B: I know what A and C have.

What are the numbers student A, B and C have?

Answer 11

I'm thinking that if the omitted sentences were not important, my teacher would not add them to the question

Answer 12

B: I can't guess the numbers they have neither

This means what? B can guess the numbers? There is a double negative here, can't and neither.

Answer 13

Sorry.. B can't guess it (my poor English )

Answer 14

I've hit a brick wall for now. I also forgot that B could have 1, inducing yet another factorization.

Answer 15

B could NOT have 1.... If B has 1, AxB =C , A x1 = C , C=A
But the question stated that the 3 cards contain DIFFERENT integers!

Answer 16

derp.

Answer 17

So after C speaks, we have possibilities
30=2x15=3x10
40=2x20=4x10
42=2x21=3x14
48=2x24=4x12
50=2x25=5x10
56=2x26=4x13
60=2x30=3x20=4x15=6x10

Since A couldn't guess the first time, we have
30=2x15=3x10
40=2x20=4x10
50=5x10
60=3x20=4x15=6x10

Now, since B couldn't guess, that narrows it down even further to
30=2x15=3x10
40=2x20=4x10
60=3x20=4x15

But now A has enough information. And after that, B has enough information. What am I missing? There's one insight I need, and it'll be done, but I don't have it.

Answer 18

You missed the answer?!
(Sorry.. Perhaps I should have told you all that I have the numerical answers to this question and the answers should be correct)

Answer 19

So the solution is not on that last list? Now I'm confused.

Answer 20

The solution I have is not on the list... I'm sorry!

Answer 21

Argh wait i got a mistake, correcting. so far im liking this question.

Answer 22

I missed
36=2x18=3x12
48=3x16
52=2x26=3x18
56=2x28=4x14

If we include these, we also get
30=2x15=3x10
36=2x18=3x12
40=2x20=4x10
42=3x14
48=4x12
50=5x10
52=3x18
56=4x14
60=3x20=4x15=6x10

Answer 23

I see the solution there!!!!

Answer 24

I'm still stuck there for now though, and I must sleep. So good night, and good luck.

Answer 25

Good night and thanks!!~

Answer 26

Given that C<60, so the last one can be ruled out

Answer 27

Since student A cannot be a single digit number, you can rule out B having 6 or 7 or 8 or 9 since (multiplying by 10 will give 60,70,80,90)

B can only have 2 - 5


When A said he couldnt guess, his number cannot be anything above 28 since having 29 he will know B is 2 and C is 58.

Then C knowing that A=10-28 and B=2-5, and knowing his own number but still cannot guess, goes to show that his number is not a prime number and in the range of 30-54 and must have both factors of 2 and 3.
A further limited to 15-28 (cause 30/2 = 15)
And B becomes further limited to 2-3 (since 4*15 or 5*15 is not allowed)
Then A also cannot have anything above 19 (become 15-18) as he would have guessed that B is 2.
Then C also cannot have anything above 36. (become 30-36) if not he would have know B is 3.
Then A asked B if he could guess, B said no.
That goes to show B is 2.

From here on, my brain shut down and i couldnt continue.
I just narrowed down to, 2* 15 =30 and 2 *18-36

All the different variations screwed up my brain cells.

Those 3 children A and B and C must have been geniuses.

Answer 28

Before the speech of A, we can know what 10≦ A <30 (since the smallest possible number of B is 2)
After the speech of A, we know that 10≦A<20 (if A>20, she/he knows B gets 2)

Answer 29

I'm gonna start over real quick.
After A speaks the first time, A is limited to \(10\leq A<20\)
After C speaks, we have possibilities
30=2x15=3x10
36=2x18=3x12
40=2x20=4x10
42=2x21=3x14
48=2x24=3x16=4x12
50=2x25=5x10
52=2x26=4x13
54=2x27=3x18
56=2x28=4x14
But \(A<20\) so that makes
30=2x15=3x10
36=2x18=3x12
40=4x10
42=3x14
48=3x16=4x12
50=5x10
52=4x14
54=3x18
56=4x14
But it still needs to have two factorizations, so that means it's one of
30=2x15=3x10
36=2x18=3x12
48=3x16=4x12
Since B doesn't know yet, we can throw out 48=4x12, Leaving us with
30=2x15=3x10
36=2x18=3x12
48=3x16
Now A knows what the others have since there are 5 unique possibilities for A, and A knows what s/he has. But then B also knows. And that's where I hit yet another wall :(

Answer 30

From what I've written on my paper, I ruled out 48=3x16
And I've written A must have the number 12, B must NOT have 4, so, B must have 3
But I don't understand why :S

Answer 31

I think the thing I was missing was the second thing A said. A said "NOW I know what the numbers on student B's and student C's cards are." This implies that before B spoke, A was still unable to guess. Hence, using the possibilities
30=2x15=3x10
36=2x18=3x12
48=3x16=4x12
A's number must be repeated. Thus, A has 12. But since B couldn't guess, we throw out 48=4x12. From this, we have that A=12, B=3, C=36.

Answer 32

That's quite true....
Thanks!!!! You're a genius!!!

Answer 33

You're welcome :) That was quite difficult.