1) for any three consecutive numbers, what can you say about odd numbers and even numbers? Explain.
2 A)Mirari conjectures that, for any three consecutive numbers, one number would be divisible by 3. Do you think Mirari is correct? Explain.
2 B) Gia claims that the sum of any theee consecutive whole numbers is divisible by 6. Is this true? Explain.
2 C) Kim claims that the product of any three consecutive whole numbers is divisible by 6. Is this true? Explain.
2 D) Does the product of any four consecutive whole numbers have any interesting properties? Explain.
3) How many consecutive numbers do you need to guarantee that one of the numbers is divisible by 5?
4) How many consecutive numbers do you need to guarantee that one of the numbers is divisible by 6?
Someone can probably check YOUR answers.
Ms. Sue, I think we know.
ok
so stop
i think mirari is correct because 33 is divisible by 3
Look at 31 or 32
1) For any three consecutive numbers, one number will always be even and the other two will be odd. This is because when you have three consecutive numbers, the middle number will always be even, while the number before and after it will be odd. For example, if we take the numbers 4, 5, and 6, we can see that 5 is odd and both 4 and 6 are even.
2 A) Mirari's conjecture that one number in any three consecutive numbers is divisible by 3 is correct. To prove this, we can start with any arbitrary number, let's say n. The three consecutive numbers would be n-1, n, and n+1. If we consider their remainders when divided by 3, we can see that there are three possibilities for the remainders: 0, 1, and 2.
- If n has a remainder of 0 when divided by 3, then n is divisible by 3.
- If n has a remainder of 1 when divided by 3, then n-1 has a remainder of 0 and is divisible by 3.
- If n has a remainder of 2 when divided by 3, then n+1 has a remainder of 0 and is divisible by 3.
Therefore, in any three consecutive numbers, we can always find one number that is divisible by 3.
2 B) Gia's claim that the sum of any three consecutive whole numbers is divisible by 6 is incorrect. Let's try an example to show this. Consider the numbers 1, 2, and 3. Their sum is 1 + 2 + 3 = 6, which is divisible by 6. However, if we shift the consecutive numbers by one, we get the numbers 2, 3, and 4. Their sum is 2 + 3 + 4 = 9, which is not divisible by 6. Therefore, Gia's claim is not true.
2 C) Kim's claim that the product of any three consecutive whole numbers is divisible by 6 is also incorrect. Let's consider the numbers 1, 2, and 3. Their product is 1 * 2 * 3 = 6, which is divisible by 6. However, if we shift the consecutive numbers by one, we get the numbers 2, 3, and 4. Their product is 2 * 3 * 4 = 24, which is not divisible by 6. Therefore, Kim's claim is not true.
2 D) The product of any four consecutive whole numbers has an interesting property. It is always divisible by 24. To understand why, we can consider the four consecutive numbers as n, n+1, n+2, and n+3. The product is (n) * (n+1) * (n+2) * (n+3). We can factor out 2 from two pairs of consecutive numbers, and we can also factor out 3 from one pair of consecutive numbers. This gives us a product of (2 * 3) * (n) * (n+1) * (n+2) * (n+3), which simplifies to 6 * (n) * (n+1) * (n+2) * (n+3). Since 6 is divisible by both 2 and 3, the product of any four consecutive whole numbers is divisible by 6.
3) To guarantee that one of the numbers is divisible by 5, you need a sequence of 6 consecutive numbers. This is because if you take any six consecutive numbers, there will always be a multiple of 5 present in the sequence. For example, consider the sequence starting from 10: 10, 11, 12, 13, 14, 15. The number 15 is divisible by 5. Therefore, with a sequence of 6 consecutive numbers, you can guarantee that one of them will be divisible by 5.
4) To guarantee that one of the numbers is divisible by 6, you need a sequence of 7 consecutive numbers. This is because if you take any seven consecutive numbers, at least one of them will be divisible by both 2 and 3, making it divisible by 6. For example, consider the sequence starting from 12: 12, 13, 14, 15, 16, 17, 18. The number 18 is divisible by 6. Therefore, with a sequence of 7 consecutive numbers, you can guarantee that one of them will be divisible by 6.