Maths Olympiad Prep

Track / Stage 4 / 289 of 340 #549 of 1964

Problem 549

AMC 12 late, AIME early
Number theory Difficulty 5.0 Prove it Team selection tests for JBMO 2018 · Saudi Arabia · 2018

Distinct prime numbers pp, qq, rr satisfy the equation
2pqr+50pq=7pqr+55pr=8pqr+12qr=A 2 p q r + 50 p q = 7 p q r + 55 p r = 8 p q r + 12 q r = A
for some positive integer AA. Find AA.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Review the given condition as
pq(2r+50)=pr(7q+55)=qr(8p+12)=A. p q (2 r + 50) = p r (7 q + 55) = q r (8 p + 12) = A.
This implies that AA is a multiple of pp, qq and rr so the value K=ApqrK = \frac{A}{p q r} is an integer. Dividing through, we have
K=8+12p=7+55q=2+50r. K = 8 + \frac{12}{p} = 7 + \frac{55}{q} = 2 + \frac{50}{r}.
Hence, p12p \mid 12, q55q \mid 55, r50r \mid 50. So we have 3 cases as follow
- p=2p = 2, q=11q = 11, r=5r = 5.
- p=3p = 3, q=11q = 11, r=5r = 5.
- p=3p = 3, q=5q = 5, r=2r = 2.
We can check that only (p,q,r)=(3,11,5)(p, q, r) = (3, 11, 5) works so K=12K = 12 and the value of AA is 19801980. \square

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