12

I was doing questions from previous year's exam paper and I'm stuck on this question.

Suppose $P, Q, R$ are positive integers such that $$PQR + PQ + QR + RP + P + Q + R = 1000$$ Find $P + Q + R$.

It seems easy but I am not getting the point from where I should start. Thank you.

Rodrigo de Azevedo
  • 1
aMighty
  • 267

2 Answers2

33

Note that

$(1 + P) (1 + Q) (1 + R) = 1 + P + Q + R$ $+ PQ+ PR + QR + PQR = 1 + 1000 = 1001, \tag{1}$

by the hypothesis on $P$, $Q$, $R$; also,

$1001 = 7 \cdot 11 \cdot 13, \tag{2}$

all primes; thus we may take

$P = 6; \;\; R = 10; \;\; Q = 12, \tag{3}$

or some permutation thereof; in any event, we have

$P + Q + R = 28. \tag{4}$

Note Added in Edit, Saturday 12 August 2017 10:04 PM PST: The solution is unique up to a permutation of $(P, Q, R) = (6, 10 ,12)$ by virtue of the fact that $1001$ may be factored into three non-unit positive integers in exactly one way, $1001 = 7 \cdot 11 \cdot 13$, since $7$, $11$, and $13$ are all primes. This follows from the Fundamental Theorem of Arithmetic. These remarks added at the suggestion of user21820 made 5 May 2015. Sorry about the delayed response. End of Note.

Robert Lewis
  • 71,180
  • 3
    Beat me to it. Once you spot the factorization, the rest is trivial. – smci May 05 '15 at 09:08
  • 1
    @amighty: you're more than welcome! Blessings to you as well! And thanks for the "acceptance"! Cheers! ;-)! – Robert Lewis May 05 '15 at 09:40
  • 3
    Do add the justification why there are no other solutions. – user21820 May 05 '15 at 13:12
  • How should one go about solving this if you don't spot the factorization? – Gareth Rees May 05 '15 at 14:20
  • @GarethRees: It is possible to spot the factorization without prior knowledge - in my case, it began when I factored the first two terms into PQ(R+1), and I remarked that I could make the term (R+1) emerge elsewhere. – Jean Hominal May 05 '15 at 14:32
  • @Gareth Rees the symmetry of the expressions $P+Q+R$, $PQ+QR+RP$, $PQR$ should suggest this has something to do with the roots of a cubic polynomial. Or you could try letting $S = P+Q+R$, expanding $S^2$ and $S^3$, and constructing a cubic equation for $S$. – alephzero May 05 '15 at 14:35
  • @JeanHominal: I'm not asking, "how do I factorize a polynomial?" I'm asking, "what feature of the problem indicates that factorization is a good first step?" which alephzero has answered. – Gareth Rees May 05 '15 at 14:36
  • @user21820: I will add a few remarks along those lines after I get a little more sleep! Cheers! – Robert Lewis May 05 '15 at 15:46
  • @smci: yes, factoring $P + Q + R + PQ+ PR + QR + PQR$ is the key. Think elementary symmetric functions. That's what I did! Cheers! – Robert Lewis May 05 '15 at 16:56
  • It's funny that factorisation of LHS (seen some question before) just came into my brain just 10 sec after seeing the statement. I probably knew it but didn't try factorising the RHS. haha – Someone May 05 '15 at 17:50
  • @user21820: I finally honored my word to add what you requested. See above. Sorry it took so long. Cheers! – Robert Lewis Aug 13 '17 at 05:19
  • Not a problem at all. I completely forgot about it myself! Have a nice day! =) – user21820 Aug 13 '17 at 06:48
2

If you allow some $p,q,r$ to be $0$, you also get $$p+q+r=1000$$ $$p+q+r=88$$ $$p+q+r=100$$ $$p+q+r=148$$

for $$p=1000,q=0,r=0$$ $$p=76,q=12,r=0$$ $$p=90,q=10,r=0$$ $$p=142,q=6,r=0$$

respectively, or any permutation thereof.

Tobias Shuxue Laoshi
  • 145