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There are no consectutive numbers that add to 64.
If the numbers are consecutive then they form an arithmetic progression. Let `a_1` be the first number in the progression and `a_n` be the last number where there are n numbers in the progression.
The sum is given by `S=n((a_1+a_n)/2)` . The `n^(th)` term is found by `a_n=a_1+(n-1)d` where d is the common difference between terms.
In our case the sum is 64 and d=1 as the numbers are consecutive. Substituting we get:
`64=n((a_1+a_1+(n-1))/2)` Multiply both sides by 2:
`n(2a_1+(n-1))=128` Now 128 is a power of 2 and has only even factors.
(1) Suppose n is even. Let n=2k for some integer k.
`2k(2a_1+2k-1)=128` Divide by 2k:
The left side is an integer. For the right side to be an integer, k must be even.
With k even the right side is even; with k even the left side is odd. This is a contradiction so n cannot be even.
(2) Suppose n is odd. Then one of the factors of 128 is odd since `2a_1+n-1` is an integer. But all factors of 128 are even so this is a contradiction.
Since n is not even and n is not odd, the assumption that there are consecutive numbers that add to 64 is false.
I need to correct this answer.
There is a subtle flaw in my "proof". N is even (there are 128 terms; each term from 1 to 63 has its opposite which is 126 terms, plus the zero term and 64 which makes 128.)
So my "proof" says that k must be even which is true. ``This implies that the left hand side is odd, which is also true. The problem is that the right side does not have to be even -- it can be , and is, one.
Let k=64; `2a_1+128-1=128/128 ==> a_1=-63` which gives the answer.
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