Join TSR now and get all your revision questions answeredSign up now
    • Thread Starter
    Offline

    2
    ReputationRep:
    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.
    • Study Helper
    Online

    13
    (Original post by MathMoFarah)
    How does n<1995 and i-h=22 make hi=A?
    They seem to assume that h,i are p_1p_4,p_2p_3, though not necessarily in that order, and I can see no justification for it either. Where p_1,p_2,p_3,p_4 are the four prime factors in ascending order.
    • Thread Starter
    Offline

    2
    ReputationRep:
    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 :/
    • Study Helper
    Online

    13
    (Original post by MathMoFarah)
    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.
    Offline

    17
    ReputationRep:
    (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.
    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?

    Am I missing something?
    • Study Helper
    Online

    13
    (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.
    Offline

    17
    ReputationRep:
    (Original post by ghostwalker)
    That's the missing piece - PRSOM.

    My number theory is rustier than I imagined.
    Well, the "suppose that ..., then hi = A" line kind of leads you in the wrong direction. TBH, the fact that I skimmed the OP probably helped by not sending me up the same garden path...(!)
 
 
 
Poll
Which pet is the best?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.