Consecutive Counting Numbers (CCNs)

2008-04-24 3:46 pm
To prove or disprove...

4. The sum of any number of CCNs is a multiple of that number. [False]
5. The sum of any odd number of CCNs is a multiple of that odd number. [True]
6. Odd multiples of 3 (except 3 itself) can always be expressed in at least two ways as sums of CCNs. [True]
7. Odd primes can be expressed in only one way as the sum of CCNs. [True]
10. The sum of a sequence of CCNs is a multiple of at least one of the numbers in the sequence. [False]

Example:
1) The sum of two consecutive counting numbers (CCNs) is always odd.
First, we will prove that the sum of 2 consecutive counting numbers (CCNs) is always odd.
Let n = 1st CCN,
Then n+1 = 2nd CCN
Then the sum of 2 CCNs is n+(n+1) = 2n+1
which is odd because 2n is always even so 1 added to an even number is always odd.
Therefore, this statement “the sum of two consecutive counting numbers (CCNs) is always odd.” is true.

2) The sum of three CCNs is a multiple of 3.
First, we will prove that the sum of 3 CCNs is a multiple of 3.
Let n-1 = 1st CCN, n>1
Then n = 2nd CCN
And n+1 = 3rd CCN
Then the sum of 3 CCNs is (n-1) + n + (n+1) = 3n
Any counting number multiplied by 3 equals a multiple of 3.

Let n = 1st CCN
Then n+1 = 2nd CCN
And n+2 = 3rd CCN
Then the sum of 3CCNs is n + (n+1) + (n+2) = 3n+3 =3(n+1)
where n+1 is a counting number.
3 times any counting number equals a multiple of 3.
Therefore, this statement “The sum of three CCNs is a multiple of 3.” is true.

參考: http://www2.edc.org/cwm/files/Chapter1.pdf

回答 (1)

2008-04-24 8:24 pm
✔ 最佳答案
4. Let n = 1st CCN
Then n+1 = 2nd CCN
Sum of the 2 CNNs = 2n + 1
which is not a multiple of 2.

5. Let n = 1st CCN
n+1 = 2nd CCN
.
.
.
(2n + 1)th CNN = n + (2n + 1) - 1 = 3n
Sum of the CNNs = n + (n + 1) + ... + 3n
= (n + 3n)(2n + 1)/2
= 2n (2n + 1)
which is always a multiple of (2n + 1)

6. Let 3(2n + 1) be an odd multiple of 3, where n>= 1
Since 3(2n + 1)
= 6n + 3
= 2n + (2n + 1) + (2n + 2)
But 6n + 3
= (n - 2) + (n - 1) + n + (n + 1) + (n + 2) + (n + 3)
So, it can always be expressed in at least two ways as sums of CCNs.



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