Factors of n help
For each integer A such that A is the product of 4 distinct primes
let
1=a<b<c<d<...<p=A
where the lowercase letters are the 16 positive integer divisors of A
Prove if A<1995 then i-h cannot equal 22
I read through the solution for this question and right at the start it said
'Note that 35*57=1995=3*5*7*19.
Suppose that n<1995and i-h=22; then hi=A'
and the rest of the proof relies on that statement
How does n<1995 and i-h=22 make hi=A?
I've shown it true when the largest prime is smaller than the product of the two smaller primes but to go through the rest of the cases would takes ages and isn't very intuitive.
let
1=a<b<c<d<...<p=A
where the lowercase letters are the 16 positive integer divisors of A
Prove if A<1995 then i-h cannot equal 22
I read through the solution for this question and right at the start it said
'Note that 35*57=1995=3*5*7*19.
Suppose that n<1995and i-h=22; then hi=A'
and the rest of the proof relies on that statement
How does n<1995 and i-h=22 make hi=A?
I've shown it true when the largest prime is smaller than the product of the two smaller primes but to go through the rest of the cases would takes ages and isn't very intuitive.
(edited 7 years ago)
Original post by MathMoFarah
How does n<1995 and i-h=22 make hi=A?
How does n<1995 and i-h=22 make hi=A?
They seem to assume that h,i are , though not necessarily in that order, and I can see no justification for it either. Where are the four prime factors in ascending order.
(edited 7 years ago)
The problem is that p4 can be larger than p1p2p3, which means they aren't the primes (I assume in which case it would be p1p2p3, p4)
but the number of inequalities which mess with the order are a bit ridiculous :/
but the number of inequalities which mess with the order are a bit ridiculous :/
Original post by MathMoFarah
but the number of inequalities which mess with the order are a bit ridiculous :/
but the number of inequalities which mess with the order are a bit ridiculous :/
Agreed - which makes me think their solution is suspect.
Had to write a programme to confirm their "conjecture"
There may be something in the i-h =22, but I can't see how to utilise it for that initial claim.
Original post by ghostwalker
Agreed - which makes me think their solution is suspect.
Had to write a programme to confirm their "conjecture"
There may be something in the i-h =22, but I can't see how to utilise it for that initial claim.
Had to write a programme to confirm their "conjecture"
There may be something in the i-h =22, but I can't see how to utilise it for that initial claim.
Am I missing something?
Original post by DFranklin
Given a factor X | A, A/X is also a factor of A. Surely this implies letters equidistant from the central position will always multiply to equal A?
That's the missing piece - PRSOM.
My number theory is rustier than I imagined.
(edited 7 years ago)
Original post by ghostwalker
That's the missing piece - PRSOM.
My number theory is rustier than I imagined.
My number theory is rustier than I imagined.
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