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The prime numbers p, q, r satisfy the simultaneous equations
pq + pr = 80 and pq + qr = 425. Find the value of p + q + r.

Sagot :

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[tex]first\ equation :\\pq+pr=80\\pq=80-pr..........[1]\\\\secon\ equation :\\pq+qr=425..........[2][/tex]

[tex]Now\ substituting\ equation\ 1\ into\ 2:\\\\pq+qr=425\\(80-pr)+qr=425\\-pr+qr=425-80\\-pr+qr=345\\r(q-p)=345\\\\we\ have :\\ p(q+r)=80=2^{4} \times 5,\\q(p+r)=425=5^{2} \times 17\\r(q-p)=345=3 \times 5 \times 23\\\\so\ : p = 2\ or\ 5, ~q=5\ or\ 17,~ r = 3,5,or\ 23.[/tex]

[tex]From\ : p(q+r)=80, if~p=5~then\ q+r=16, which\ has\ no\ solution\\So~~p=2, then\ ~q+r=40, hence~~q=17~~and~~r=23.\\\\\therefore~~ p+q+r~~~is~~~\boxed{2+17+23=42}[/tex]
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