- #1
numbthenoob
- 10
- 0
Let T represent an odd target number to be factored. Let m represent any odd number > 1.
If T is congruent to zero mod m, (T-1)/2 is congruent to (m-1)/2 mod m, and vice-versa.
For example
T
35 = 0 mod 5, 17 = 2 mod 5
77 = 0 mod 7, 38 = 3 mod 7
77 = 0 mod 11, 38 = 5 mod 11
47 = 0 mod 47, 23 = 23 mod 47
81 = 0 mod 3, 40 = 1 mod 3
81 = 0 mod 9, 40 = 4 mod 9
81 = 0 mod 27, 40 = 13 mod 27
and so on...
It's easy to see why; for any odd number divisible by another odd number, the halfway point through the dividend is going to correspond to the halfway point through the divisor's middle length.
My question is, what's up with the target number's congruences for moduli other than its factors?
35 = 2 mod 3, 17 = 2 mod 3
2 is one more than (3-1)/2, so wouldn't you expect 35 to be congruent to 1 mod 3? I would, I mean, it's necessarily congruent to zero when 17 is congruent to a number that's zero more than (m-1)/2...so, what am I missing?
Any help here would be greatly appreciated. Thanks.
If T is congruent to zero mod m, (T-1)/2 is congruent to (m-1)/2 mod m, and vice-versa.
For example
T
35 = 0 mod 5, 17 = 2 mod 5
77 = 0 mod 7, 38 = 3 mod 7
77 = 0 mod 11, 38 = 5 mod 11
47 = 0 mod 47, 23 = 23 mod 47
81 = 0 mod 3, 40 = 1 mod 3
81 = 0 mod 9, 40 = 4 mod 9
81 = 0 mod 27, 40 = 13 mod 27
and so on...
It's easy to see why; for any odd number divisible by another odd number, the halfway point through the dividend is going to correspond to the halfway point through the divisor's middle length.
My question is, what's up with the target number's congruences for moduli other than its factors?
35 = 2 mod 3, 17 = 2 mod 3
2 is one more than (3-1)/2, so wouldn't you expect 35 to be congruent to 1 mod 3? I would, I mean, it's necessarily congruent to zero when 17 is congruent to a number that's zero more than (m-1)/2...so, what am I missing?
Any help here would be greatly appreciated. Thanks.