This blog is the online extension of our intermediate classrooms. Our goal is to enhance and document our learning experience throughout the school year, and share this journey with teachers, parents and students. We welcome your constructive feedback, and we look forward to learning with you!
Grade 7/8 POTW We know the sum is 14, and that Alpha says Beta and Gamma have different numbers, Beta says he/she already knew all 3 were different, and that Gamma now knows the numbers. First, Alpha says Beta and Gamma are different. This can only happen if Alpha has an odd number as then, Beta and Gamma would have to have an odd and even number instead of both the same. Beta then says that he knew everyone had different numbers before. We can use the same information as before and know that Beta has an odd number, therefore knowing the other 2 don't have the same number. To ensure that no one has any of the same number, Beta must have the number 7 or more (0 wouldn't count as a positive integer). The numbers Beta can have are 7, 9, or 11. Now, Alpha knows all 3 numbers. The information given now would be that Beta has numbers 7, 9, or 11, Alpha has an odd number less than 7, but still odd, or 5, 3, or 1, and Gamma has an even number less than 7, or 6, 4, or 2. Now from here, Gamma knows all 3 numbers. If he had a 2, the other 2 could have a 5 and 7, a 3 or 9, or 1 or 11. If he had a 4, the other 2 could have a 1 or 9, or 3 or 7. Both have multiple possibilities. However, if he had a 6, the others can only have a 1 and 7. Therefore, Alpha has a 1, Beta has a 7, and Gamma has a 6.
Grade 7/8 POTW I will uncover the answer of this question step by step. Firstly, I know that the sum of the three numbers is 14. When Alpha said that he knew that Beta and Gamma had different numbers, it meant that his number had to be odd. Possibilites for his number are 1, 3, 5, 7, 9, 11. Then, after, Beta said that each number was different. This further proves that Alphas number has to be odd. Knowing this and following the same rule, Beta must have a number of 7 or more to ensure that none of the kids have the same number. Therefore, her number can be 7, 9, or 11. If Alphas number is odd, and Beta's number is an odd number larger than 7, Gamma's number has to be an even number less than 7. Now, I can find out who had which number by ruling out the possibilities based on Gamma's number. With a 2, the possibilities would be a 2, 3, 9 . 2, 5, 7 . 2, 1, 11. With a 4, the possibilities would be a 4, 3, 7 . 4, 1, 9. Lastly, with a 6, the possibilities would be a 6, 1, and 7 since 7 is the least possible value for Beta. Therefore, this was the right possibility. The numbers would be 7 for Beta, 1 for Alpha, and 6 for Gamma.
Grade 7/8 POTW: Info: - Alpha: “I know that Beta and Gamma have different numbers” - Beta: “I already knew that all three numbers were different” - Gamma: I now know what all three of the numbers are” - Totals 14
Let a represent Alpha’s number, b represent Beta’s number and g represent Gamma’s number.
All we know right now is that a + b + g = 14. Alpha already said that he knows that Beta and Gamma different numbers. This means that his number must make sure that Beta and Gamma’s numbers add up to an odd number. This means that Alpha must have an odd number. It must be one of the following: 1, 3, 5, 7, 9, 11, 13.
Beta must also have an odd number because if he said he knows that all the numbers are different, so he must also have an odd number. It must also be larger than 7 because then he would know for sure no one else has his number. He can have a 7,9 or 11.
For Gamma to know that they are all different, Alpha must have a small odd number such as 1, 3 or 5.
Now, we can trial and error. If Gamma had a 2 and 4, there would be multiple answers for both. Yet if Gamma had a nice card with a 6, there is only 1 and 7 that would work in this case.
Alpha has a 1, Beta obtained a 7 and Gamma grabbed a 6 (heheh, synonyms)
Alpha already knows that the other two have different numbers. Therefore, he must have an odd number, because if he had an even number, say, 2, 14 - 2 is 12, so the other two would both have to have 6. Alpha could have: 1, 3, 5, 7, 9, 11
Beta then says he already knew all of the other numbers are different. It must be larger than 7, because then he would know for sure that no one else has his number, options: 7, 9, 11.
Gamma knew the answer, and to determine whether or not they are all different, Alpha needs to have a small odd number, options 1, 3, 5.
From here, we can determine using guess and check, that Alpha obtained a 1, Gamma obtained a 6, and Beta obtained a 7.
Grade 7/8 POTW
ReplyDeleteWe know the sum is 14, and that Alpha says Beta and Gamma have different numbers, Beta says he/she already knew all 3 were different, and that Gamma now knows the numbers.
First, Alpha says Beta and Gamma are different. This can only happen if Alpha has an odd number as then, Beta and Gamma would have to have an odd and even number instead of both the same. Beta then says that he knew everyone had different numbers before. We can use the same information as before and know that Beta has an odd number, therefore knowing the other 2 don't have the same number. To ensure that no one has any of the same number, Beta must have the number 7 or more (0 wouldn't count as a positive integer). The numbers Beta can have are 7, 9, or 11. Now, Alpha knows all 3 numbers. The information given now would be that Beta has numbers 7, 9, or 11, Alpha has an odd number less than 7, but still odd, or 5, 3, or 1, and Gamma has an even number less than 7, or 6, 4, or 2. Now from here, Gamma knows all 3 numbers. If he had a 2, the other 2 could have a 5 and 7, a 3 or 9, or 1 or 11. If he had a 4, the other 2 could have a 1 or 9, or 3 or 7. Both have multiple possibilities. However, if he had a 6, the others can only have a 1 and 7.
Therefore, Alpha has a 1, Beta has a 7, and Gamma has a 6.
Grade 7/8 POTW
ReplyDeleteI will uncover the answer of this question step by step. Firstly, I know that the sum of the three numbers is 14. When Alpha said that he knew that Beta and Gamma had different numbers, it meant that his number had to be odd. Possibilites for his number are 1, 3, 5, 7, 9, 11. Then, after, Beta said that each number was different. This further proves that Alphas number has to be odd. Knowing this and following the same rule, Beta must have a number of 7 or more to ensure that none of the kids have the same number. Therefore, her number can be 7, 9, or 11. If Alphas number is odd, and Beta's number is an odd number larger than 7, Gamma's number has to be an even number less than 7. Now, I can find out who had which number by ruling out the possibilities based on Gamma's number. With a 2, the possibilities would be a 2, 3, 9 . 2, 5, 7 . 2, 1, 11. With a 4, the possibilities would be a 4, 3, 7 . 4, 1, 9. Lastly, with a 6, the possibilities would be a 6, 1, and 7 since 7 is the least possible value for Beta. Therefore, this was the right possibility. The numbers would be 7 for Beta, 1 for Alpha, and 6 for Gamma.
Grade 7/8 POTW:
ReplyDeleteInfo:
- Alpha: “I know that Beta and Gamma have different numbers”
- Beta: “I already knew that all three numbers were different”
- Gamma: I now know what all three of the numbers are”
- Totals 14
Let a represent Alpha’s number, b represent Beta’s number and g represent Gamma’s number.
All we know right now is that a + b + g = 14.
Alpha already said that he knows that Beta and Gamma different numbers. This means that his number must make sure that Beta and Gamma’s numbers add up to an odd number. This means that Alpha must have an odd number. It must be one of the following: 1, 3, 5, 7, 9, 11, 13.
Beta must also have an odd number because if he said he knows that all the numbers are different, so he must also have an odd number. It must also be larger than 7 because then he would know for sure no one else has his number. He can have a 7,9 or 11.
For Gamma to know that they are all different, Alpha must have a small odd number such as 1, 3 or 5.
Now, we can trial and error. If Gamma had a 2 and 4, there would be multiple answers for both. Yet if Gamma had a nice card with a 6, there is only 1 and 7 that would work in this case.
Alpha has a 1, Beta obtained a 7 and Gamma grabbed a 6 (heheh, synonyms)
POTW:
ReplyDeleteAlpha already knows that the other two have different numbers. Therefore, he must have an odd number, because if he had an even number, say, 2, 14 - 2 is 12, so the other two would both have to have 6. Alpha could have: 1, 3, 5, 7, 9, 11
Beta then says he already knew all of the other numbers are different. It must be larger than 7, because then he would know for sure that no one else has his number, options: 7, 9, 11.
Gamma knew the answer, and to determine whether or not they are all different, Alpha needs to have a small odd number, options 1, 3, 5.
From here, we can determine using guess and check, that Alpha obtained a 1, Gamma obtained a 6, and Beta obtained a 7.