There are no 5s in the set (2,2,3), so the first number divisible by 1 through 5 has to be (2,2,3,5) = 60.Ħ is 2*3, and we already have that, so we’ve found the number divisible by 1 through 6. So 2*2*3 = 12 is the smallest number divisible by 2, 3 and 4. (2,2,3) is the smallest set that includes the sets (2), (3) and (2,2). To find the smallest number divisible by 1, 2, 3 and 4, we don’t have to multiply all of them together because there’s some redundancies. The smallest number divisible by 1, 2, and 3 is 1*2*3 = 6. So (2n)k = 1 (mod m), which means 2kn = 1 (mod m) for any integer, k.Ģφ(m) = 1 (mod m), by Euler’s Totient Theorem. Now 2n = m+1, so 2n = 1 (mod m), but no smaller (but still positive) power of 2 is equivalent to 1 (mod m).
#Aa2 mods for abs mod#
Suppose a is a number coprime to n, and b and c are two different numbers, also coprime to n, in the range and ab=ac, mod n.Ī(b-c)=nk, and since b-c is smaller than n in absolute value, a must be larger than k in absolute value.Ī and n have no factors in common, so a|k, but a can’t divide k, since a is larger than k, a contradiction.
Let A be the set of numbers coprime to n: A = congruent, mod n? Proof: take a to be any positive integer coprime to n.
Now I will present a wonderful fact about the coprimes to n - they form a group under multiplication mod n, which leads to:
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