The riddles about three children
$begingroup$
We have three children A , B , C
Child A says: " I ate sweets more than B , and B ate sweets more than C "
Child B says: " C ate sweets more than me , also C ate sweets more than A "
Child C says: " B ate sweets more than me , also I and A ate sweets equal amount "
Suppose between the two children who ate less sweets say more right.
Which of the following sentences can not be correct?
1- C ate sweets more than A and B.
2- B ate sweets more than A and C.
3- A ate sweets more than B .
4- B ate sweets more than A.
5- B ate sweets more than C.
6- C ate sweets more than B.
7- C ate sweets more than A.
8- A ate sweets more than C.
9- All children ate sweets equal amount .
puzzle
asked Dec 23 '18 at 9:19
dolfi224dolfi224
33
$endgroup$
add a comment |
$begingroup$
We have three children A , B , C
Child A says: " I ate sweets more than B , and B ate sweets more than C "
Child B says: " C ate sweets more than me , also C ate sweets more than A "
Child C says: " B ate sweets more than me , also I and A ate sweets equal amount "
Suppose between the two children who ate less sweets say more right.
Which of the following sentences can not be correct?
1- C ate sweets more than A and B.
2- B ate sweets more than A and C.
3- A ate sweets more than B .
4- B ate sweets more than A.
5- B ate sweets more than C.
6- C ate sweets more than B.
7- C ate sweets more than A.
8- A ate sweets more than C.
9- All children ate sweets equal amount .
puzzle
asked Dec 23 '18 at 9:19
dolfi224dolfi224
33
$endgroup$
1
$begingroup$
I think that there is a whole stack exchange dedicated to puzzles of this sort.
$endgroup$
– badjohn
Dec 23 '18 at 9:29
$begingroup$
Also, this is a very finite Problem. There are only $13$ possible situations you might want to play through … Nevertheless, as a first Observation note that at least one of A$'s claims must be false.
$endgroup$
– Hagen von Eitzen
Dec 23 '18 at 9:34
$begingroup$
Is a riddle like that? Can you introduce a sample?
$endgroup$
– dolfi224
Dec 23 '18 at 16:57
add a comment |
$begingroup$
We have three children A , B , C
Child A says: " I ate sweets more than B , and B ate sweets more than C "
Child B says: " C ate sweets more than me , also C ate sweets more than A "
Child C says: " B ate sweets more than me , also I and A ate sweets equal amount "
Suppose between the two children who ate less sweets say more right.
Which of the following sentences can not be correct?
1- C ate sweets more than A and B.
2- B ate sweets more than A and C.
3- A ate sweets more than B .
4- B ate sweets more than A.
5- B ate sweets more than C.
6- C ate sweets more than B.
7- C ate sweets more than A.
8- A ate sweets more than C.
9- All children ate sweets equal amount .
puzzle
asked Dec 23 '18 at 9:19
dolfi224dolfi224
33
$endgroup$
We have three children A , B , C
Child A says: " I ate sweets more than B , and B ate sweets more than C "
Child B says: " C ate sweets more than me , also C ate sweets more than A "
Child C says: " B ate sweets more than me , also I and A ate sweets equal amount "
Suppose between the two children who ate less sweets say more right.
Which of the following sentences can not be correct?
1- C ate sweets more than A and B.
2- B ate sweets more than A and C.
3- A ate sweets more than B .
4- B ate sweets more than A.
5- B ate sweets more than C.
6- C ate sweets more than B.
7- C ate sweets more than A.
8- A ate sweets more than C.
9- All children ate sweets equal amount .
puzzle
puzzle
asked Dec 23 '18 at 9:19
dolfi224dolfi224
33
asked Dec 23 '18 at 9:19
dolfi224dolfi224
33
edited Dec 23 '18 at 9:25
dolfi224
asked Dec 23 '18 at 9:19
dolfi224dolfi224
33
asked Dec 23 '18 at 9:19
dolfi224dolfi224
33
asked Dec 23 '18 at 9:19
dolfi224dolfi224
33
33
1
$begingroup$
I think that there is a whole stack exchange dedicated to puzzles of this sort.
$endgroup$
– badjohn
Dec 23 '18 at 9:29
$begingroup$
Also, this is a very finite Problem. There are only $13$ possible situations you might want to play through … Nevertheless, as a first Observation note that at least one of A$'s claims must be false.
$endgroup$
– Hagen von Eitzen
Dec 23 '18 at 9:34
$begingroup$
Is a riddle like that? Can you introduce a sample?
$endgroup$
– dolfi224
Dec 23 '18 at 16:57
add a comment |
1
$begingroup$
I think that there is a whole stack exchange dedicated to puzzles of this sort.
$endgroup$
– badjohn
Dec 23 '18 at 9:29
$begingroup$
Also, this is a very finite Problem. There are only $13$ possible situations you might want to play through … Nevertheless, as a first Observation note that at least one of A$'s claims must be false.
$endgroup$
– Hagen von Eitzen
Dec 23 '18 at 9:34
$begingroup$
Is a riddle like that? Can you introduce a sample?
$endgroup$
– dolfi224
Dec 23 '18 at 16:57
1
1
$begingroup$
I think that there is a whole stack exchange dedicated to puzzles of this sort.
$endgroup$
– badjohn
Dec 23 '18 at 9:29
$begingroup$
I think that there is a whole stack exchange dedicated to puzzles of this sort.
$endgroup$
– badjohn
Dec 23 '18 at 9:29
$begingroup$
Also, this is a very finite Problem. There are only $13$ possible situations you might want to play through … Nevertheless, as a first Observation note that at least one of A$'s claims must be false.
$endgroup$
– Hagen von Eitzen
Dec 23 '18 at 9:34
$begingroup$
Also, this is a very finite Problem. There are only $13$ possible situations you might want to play through … Nevertheless, as a first Observation note that at least one of A$'s claims must be false.
$endgroup$
– Hagen von Eitzen
Dec 23 '18 at 9:34
$begingroup$
Is a riddle like that? Can you introduce a sample?
$endgroup$
– dolfi224
Dec 23 '18 at 16:57
$begingroup$
Is a riddle like that? Can you introduce a sample?
$endgroup$
– dolfi224
Dec 23 '18 at 16:57
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Child A says: " I ate sweets more than B , and B ate sweets more than C "
Child B says: " C ate sweets more than me , also C ate sweets more than A "
Child C says: " B ate sweets more than me , also I and A ate sweets equal amount "
Suppose between the two children who ate less sweets say more right.
Two children who ate less sweets ?
A: ate more than B
B: Ate more than C
B: c ate more than B
C: ate more than A
B: ate more than C
C: ate same as A
Which children ate less sweets ?
Here the number represents how many times the letter ate less
‘A’: 3
‘B’: 3
‘C’: 3
As you can see, All letters ate equally less sweets.
Therefore, the answer is 9 - All children ate sweets equal amount.
answered Dec 23 '18 at 9:47
Dan Dan
93431133
$endgroup$
$begingroup$
I did not understand. if answer is the sentences that can not be correct? and All letters ate equally less sweets ,why the answer is 9 also A is 2 . why do you say 3?
$endgroup$
– dolfi224
Dec 23 '18 at 16:53
add a comment |
$begingroup$
What each child said:
- Child A. A > B and B > C. A ate the most
- Child B. C > B and C > A. C ate the most
- Child C. B > C and A = C. B ate the most
There are rules that can be inferred from the statements. Any of the following sentences that breaks one of these rules cannot be correct:
(1) These 2 children must say at least 1 thing right (1 or 2 right things). This comes from “Suppose between the two children who ate less sweets say more right”.
(2) At most one child says 2 right things. This is the case because every child says at least one statement that contradicts at least one statement from every other child
(3) The child who eats the most cannot say anything right. Whilst “Suppose between the two children who ate less sweets say more right” implies that the child who eats the most could say one thing right, this would mean that the other two children must say two right things, which is impossible.
Going through the 9 sentences:
1- C ate sweets more than A and B. C > A and C > B
Child A: says either 0 or 1 right things. B > C is wrong, but we don’t know if A > B
Child B: says 2 things right
Child C: says 2 wrong things
If A > B, then the rules are followed
2- B ate sweets more than A and C. B > A and B > C
Child A: says 1 thing right (B > C)
Child B: says 0 or 1 things right. C > B is wrong. We don’t know if C > A
Child C: says 1 or 2 things right. B > C is right. We don’t know if A = C
If A ≥ C, then Child B says 0 things right, and other children say at least 1 thing right. So the rules are followed.
3- A ate sweets more than B. A > B
Child A: says at least one thing right. To avoid contradicting rule (3), then C > A > B. This makes Child C the child who ate the most and have 2 incorrect statements, whilst Childs A and B have one correct statement. The rules are followed
4- B ate sweets more than A.
This can follow the same pattern as in sentence 2 to follow the rules.
5- B ate sweets more than C.
This can follow the same pattern as in sentence 2 to follow the rules.
6- C ate sweets more than B.
This can follow the same pattern as in sentence 1 to follow the rules.
7- C ate sweets more than A.
This can follow the same pattern as in sentence 1 to follow the rules.
8- A ate sweets more than C.
If A > C, and also B > A > C, then this can follow one of the patterns mentioned in sentence 2 to follow the rules.
9- All children ate sweets equal amount. A = B = C
In this case, both child A and B said 0 correct things. This contradicts rule (1) and hence must always be false.
Hence, by going through all the available options, we can conclude the only sentence that cannot ever be correct is sentence 9.
answered Dec 24 '18 at 13:54
MBorgMBorg
1751114
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Child A says: " I ate sweets more than B , and B ate sweets more than C "
Child B says: " C ate sweets more than me , also C ate sweets more than A "
Child C says: " B ate sweets more than me , also I and A ate sweets equal amount "
Suppose between the two children who ate less sweets say more right.
Two children who ate less sweets ?
A: ate more than B
B: Ate more than C
B: c ate more than B
C: ate more than A
B: ate more than C
C: ate same as A
Which children ate less sweets ?
Here the number represents how many times the letter ate less
‘A’: 3
‘B’: 3
‘C’: 3
As you can see, All letters ate equally less sweets.
Therefore, the answer is 9 - All children ate sweets equal amount.
answered Dec 23 '18 at 9:47
Dan Dan
93431133
$endgroup$
$begingroup$
I did not understand. if answer is the sentences that can not be correct? and All letters ate equally less sweets ,why the answer is 9 also A is 2 . why do you say 3?
$endgroup$
– dolfi224
Dec 23 '18 at 16:53
add a comment |
$begingroup$
Child A says: " I ate sweets more than B , and B ate sweets more than C "
Child B says: " C ate sweets more than me , also C ate sweets more than A "
Child C says: " B ate sweets more than me , also I and A ate sweets equal amount "
Suppose between the two children who ate less sweets say more right.
Two children who ate less sweets ?
A: ate more than B
B: Ate more than C
B: c ate more than B
C: ate more than A
B: ate more than C
C: ate same as A
Which children ate less sweets ?
Here the number represents how many times the letter ate less
‘A’: 3
‘B’: 3
‘C’: 3
As you can see, All letters ate equally less sweets.
Therefore, the answer is 9 - All children ate sweets equal amount.
answered Dec 23 '18 at 9:47
Dan Dan
93431133
$endgroup$
$begingroup$
I did not understand. if answer is the sentences that can not be correct? and All letters ate equally less sweets ,why the answer is 9 also A is 2 . why do you say 3?
$endgroup$
– dolfi224
Dec 23 '18 at 16:53
add a comment |
$begingroup$
Child A says: " I ate sweets more than B , and B ate sweets more than C "
Child B says: " C ate sweets more than me , also C ate sweets more than A "
Child C says: " B ate sweets more than me , also I and A ate sweets equal amount "
Suppose between the two children who ate less sweets say more right.
Two children who ate less sweets ?
A: ate more than B
B: Ate more than C
B: c ate more than B
C: ate more than A
B: ate more than C
C: ate same as A
Which children ate less sweets ?
Here the number represents how many times the letter ate less
‘A’: 3
‘B’: 3
‘C’: 3
As you can see, All letters ate equally less sweets.
Therefore, the answer is 9 - All children ate sweets equal amount.
answered Dec 23 '18 at 9:47
Dan Dan
93431133
$endgroup$
Child A says: " I ate sweets more than B , and B ate sweets more than C "
Child B says: " C ate sweets more than me , also C ate sweets more than A "
Child C says: " B ate sweets more than me , also I and A ate sweets equal amount "
Suppose between the two children who ate less sweets say more right.
Two children who ate less sweets ?
A: ate more than B
B: Ate more than C
B: c ate more than B
C: ate more than A
B: ate more than C
C: ate same as A
Which children ate less sweets ?
Here the number represents how many times the letter ate less
‘A’: 3
‘B’: 3
‘C’: 3
As you can see, All letters ate equally less sweets.
Therefore, the answer is 9 - All children ate sweets equal amount.
answered Dec 23 '18 at 9:47
Dan Dan
93431133
answered Dec 23 '18 at 9:47
Dan Dan
93431133
answered Dec 23 '18 at 9:47
Dan Dan
93431133
answered Dec 23 '18 at 9:47
Dan Dan
93431133
93431133
$begingroup$
I did not understand. if answer is the sentences that can not be correct? and All letters ate equally less sweets ,why the answer is 9 also A is 2 . why do you say 3?
$endgroup$
– dolfi224
Dec 23 '18 at 16:53
add a comment |
$begingroup$
I did not understand. if answer is the sentences that can not be correct? and All letters ate equally less sweets ,why the answer is 9 also A is 2 . why do you say 3?
$endgroup$
– dolfi224
Dec 23 '18 at 16:53
$begingroup$
I did not understand. if answer is the sentences that can not be correct? and All letters ate equally less sweets ,why the answer is 9 also A is 2 . why do you say 3?
$endgroup$
– dolfi224
Dec 23 '18 at 16:53
$begingroup$
I did not understand. if answer is the sentences that can not be correct? and All letters ate equally less sweets ,why the answer is 9 also A is 2 . why do you say 3?
$endgroup$
– dolfi224
Dec 23 '18 at 16:53
add a comment |
$begingroup$
What each child said:
- Child A. A > B and B > C. A ate the most
- Child B. C > B and C > A. C ate the most
- Child C. B > C and A = C. B ate the most
There are rules that can be inferred from the statements. Any of the following sentences that breaks one of these rules cannot be correct:
(1) These 2 children must say at least 1 thing right (1 or 2 right things). This comes from “Suppose between the two children who ate less sweets say more right”.
(2) At most one child says 2 right things. This is the case because every child says at least one statement that contradicts at least one statement from every other child
(3) The child who eats the most cannot say anything right. Whilst “Suppose between the two children who ate less sweets say more right” implies that the child who eats the most could say one thing right, this would mean that the other two children must say two right things, which is impossible.
Going through the 9 sentences:
1- C ate sweets more than A and B. C > A and C > B
Child A: says either 0 or 1 right things. B > C is wrong, but we don’t know if A > B
Child B: says 2 things right
Child C: says 2 wrong things
If A > B, then the rules are followed
2- B ate sweets more than A and C. B > A and B > C
Child A: says 1 thing right (B > C)
Child B: says 0 or 1 things right. C > B is wrong. We don’t know if C > A
Child C: says 1 or 2 things right. B > C is right. We don’t know if A = C
If A ≥ C, then Child B says 0 things right, and other children say at least 1 thing right. So the rules are followed.
3- A ate sweets more than B. A > B
Child A: says at least one thing right. To avoid contradicting rule (3), then C > A > B. This makes Child C the child who ate the most and have 2 incorrect statements, whilst Childs A and B have one correct statement. The rules are followed
4- B ate sweets more than A.
This can follow the same pattern as in sentence 2 to follow the rules.
5- B ate sweets more than C.
This can follow the same pattern as in sentence 2 to follow the rules.
6- C ate sweets more than B.
This can follow the same pattern as in sentence 1 to follow the rules.
7- C ate sweets more than A.
This can follow the same pattern as in sentence 1 to follow the rules.
8- A ate sweets more than C.
If A > C, and also B > A > C, then this can follow one of the patterns mentioned in sentence 2 to follow the rules.
9- All children ate sweets equal amount. A = B = C
In this case, both child A and B said 0 correct things. This contradicts rule (1) and hence must always be false.
Hence, by going through all the available options, we can conclude the only sentence that cannot ever be correct is sentence 9.
answered Dec 24 '18 at 13:54
MBorgMBorg
1751114
$endgroup$
add a comment |
$begingroup$
What each child said:
- Child A. A > B and B > C. A ate the most
- Child B. C > B and C > A. C ate the most
- Child C. B > C and A = C. B ate the most
There are rules that can be inferred from the statements. Any of the following sentences that breaks one of these rules cannot be correct:
(1) These 2 children must say at least 1 thing right (1 or 2 right things). This comes from “Suppose between the two children who ate less sweets say more right”.
(2) At most one child says 2 right things. This is the case because every child says at least one statement that contradicts at least one statement from every other child
(3) The child who eats the most cannot say anything right. Whilst “Suppose between the two children who ate less sweets say more right” implies that the child who eats the most could say one thing right, this would mean that the other two children must say two right things, which is impossible.
Going through the 9 sentences:
1- C ate sweets more than A and B. C > A and C > B
Child A: says either 0 or 1 right things. B > C is wrong, but we don’t know if A > B
Child B: says 2 things right
Child C: says 2 wrong things
If A > B, then the rules are followed
2- B ate sweets more than A and C. B > A and B > C
Child A: says 1 thing right (B > C)
Child B: says 0 or 1 things right. C > B is wrong. We don’t know if C > A
Child C: says 1 or 2 things right. B > C is right. We don’t know if A = C
If A ≥ C, then Child B says 0 things right, and other children say at least 1 thing right. So the rules are followed.
3- A ate sweets more than B. A > B
Child A: says at least one thing right. To avoid contradicting rule (3), then C > A > B. This makes Child C the child who ate the most and have 2 incorrect statements, whilst Childs A and B have one correct statement. The rules are followed
4- B ate sweets more than A.
This can follow the same pattern as in sentence 2 to follow the rules.
5- B ate sweets more than C.
This can follow the same pattern as in sentence 2 to follow the rules.
6- C ate sweets more than B.
This can follow the same pattern as in sentence 1 to follow the rules.
7- C ate sweets more than A.
This can follow the same pattern as in sentence 1 to follow the rules.
8- A ate sweets more than C.
If A > C, and also B > A > C, then this can follow one of the patterns mentioned in sentence 2 to follow the rules.
9- All children ate sweets equal amount. A = B = C
In this case, both child A and B said 0 correct things. This contradicts rule (1) and hence must always be false.
Hence, by going through all the available options, we can conclude the only sentence that cannot ever be correct is sentence 9.
answered Dec 24 '18 at 13:54
MBorgMBorg
1751114
$endgroup$
add a comment |
$begingroup$
What each child said:
- Child A. A > B and B > C. A ate the most
- Child B. C > B and C > A. C ate the most
- Child C. B > C and A = C. B ate the most
There are rules that can be inferred from the statements. Any of the following sentences that breaks one of these rules cannot be correct:
(1) These 2 children must say at least 1 thing right (1 or 2 right things). This comes from “Suppose between the two children who ate less sweets say more right”.
(2) At most one child says 2 right things. This is the case because every child says at least one statement that contradicts at least one statement from every other child
(3) The child who eats the most cannot say anything right. Whilst “Suppose between the two children who ate less sweets say more right” implies that the child who eats the most could say one thing right, this would mean that the other two children must say two right things, which is impossible.
Going through the 9 sentences:
1- C ate sweets more than A and B. C > A and C > B
Child A: says either 0 or 1 right things. B > C is wrong, but we don’t know if A > B
Child B: says 2 things right
Child C: says 2 wrong things
If A > B, then the rules are followed
2- B ate sweets more than A and C. B > A and B > C
Child A: says 1 thing right (B > C)
Child B: says 0 or 1 things right. C > B is wrong. We don’t know if C > A
Child C: says 1 or 2 things right. B > C is right. We don’t know if A = C
If A ≥ C, then Child B says 0 things right, and other children say at least 1 thing right. So the rules are followed.
3- A ate sweets more than B. A > B
Child A: says at least one thing right. To avoid contradicting rule (3), then C > A > B. This makes Child C the child who ate the most and have 2 incorrect statements, whilst Childs A and B have one correct statement. The rules are followed
4- B ate sweets more than A.
This can follow the same pattern as in sentence 2 to follow the rules.
5- B ate sweets more than C.
This can follow the same pattern as in sentence 2 to follow the rules.
6- C ate sweets more than B.
This can follow the same pattern as in sentence 1 to follow the rules.
7- C ate sweets more than A.
This can follow the same pattern as in sentence 1 to follow the rules.
8- A ate sweets more than C.
If A > C, and also B > A > C, then this can follow one of the patterns mentioned in sentence 2 to follow the rules.
9- All children ate sweets equal amount. A = B = C
In this case, both child A and B said 0 correct things. This contradicts rule (1) and hence must always be false.
Hence, by going through all the available options, we can conclude the only sentence that cannot ever be correct is sentence 9.
answered Dec 24 '18 at 13:54
MBorgMBorg
1751114
$endgroup$
What each child said:
- Child A. A > B and B > C. A ate the most
- Child B. C > B and C > A. C ate the most
- Child C. B > C and A = C. B ate the most
There are rules that can be inferred from the statements. Any of the following sentences that breaks one of these rules cannot be correct:
(1) These 2 children must say at least 1 thing right (1 or 2 right things). This comes from “Suppose between the two children who ate less sweets say more right”.
(2) At most one child says 2 right things. This is the case because every child says at least one statement that contradicts at least one statement from every other child
(3) The child who eats the most cannot say anything right. Whilst “Suppose between the two children who ate less sweets say more right” implies that the child who eats the most could say one thing right, this would mean that the other two children must say two right things, which is impossible.
Going through the 9 sentences:
1- C ate sweets more than A and B. C > A and C > B
Child A: says either 0 or 1 right things. B > C is wrong, but we don’t know if A > B
Child B: says 2 things right
Child C: says 2 wrong things
If A > B, then the rules are followed
2- B ate sweets more than A and C. B > A and B > C
Child A: says 1 thing right (B > C)
Child B: says 0 or 1 things right. C > B is wrong. We don’t know if C > A
Child C: says 1 or 2 things right. B > C is right. We don’t know if A = C
If A ≥ C, then Child B says 0 things right, and other children say at least 1 thing right. So the rules are followed.
3- A ate sweets more than B. A > B
Child A: says at least one thing right. To avoid contradicting rule (3), then C > A > B. This makes Child C the child who ate the most and have 2 incorrect statements, whilst Childs A and B have one correct statement. The rules are followed
4- B ate sweets more than A.
This can follow the same pattern as in sentence 2 to follow the rules.
5- B ate sweets more than C.
This can follow the same pattern as in sentence 2 to follow the rules.
6- C ate sweets more than B.
This can follow the same pattern as in sentence 1 to follow the rules.
7- C ate sweets more than A.
This can follow the same pattern as in sentence 1 to follow the rules.
8- A ate sweets more than C.
If A > C, and also B > A > C, then this can follow one of the patterns mentioned in sentence 2 to follow the rules.
9- All children ate sweets equal amount. A = B = C
In this case, both child A and B said 0 correct things. This contradicts rule (1) and hence must always be false.
Hence, by going through all the available options, we can conclude the only sentence that cannot ever be correct is sentence 9.
answered Dec 24 '18 at 13:54
MBorgMBorg
1751114
answered Dec 24 '18 at 13:54
MBorgMBorg
1751114
answered Dec 24 '18 at 13:54
MBorgMBorg
1751114
answered Dec 24 '18 at 13:54
MBorgMBorg
1751114
1751114
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1
$begingroup$
I think that there is a whole stack exchange dedicated to puzzles of this sort.
$endgroup$
– badjohn
Dec 23 '18 at 9:29
$begingroup$
Also, this is a very finite Problem. There are only $13$ possible situations you might want to play through … Nevertheless, as a first Observation note that at least one of A$'s claims must be false.
$endgroup$
– Hagen von Eitzen
Dec 23 '18 at 9:34
$begingroup$
Is a riddle like that? Can you introduce a sample?
$endgroup$
– dolfi224
Dec 23 '18 at 16:57