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Mathematics 8 Online
OpenStudy (anonymous):

solve in prime numbers\[pqr=11(p+q+r)\]

hartnn (hartnn):

either one of the p,q,r should be 11

OpenStudy (anonymous):

yep

hartnn (hartnn):

so pq=p+q+11

ganeshie8 (ganeshie8):

q=2 p=13

OpenStudy (anonymous):

why 11...... ?

OpenStudy (anonymous):

q=3 p=7

OpenStudy (anonymous):

oh got it

hartnn (hartnn):

q=(p+11)/(p-1)

OpenStudy (anonymous):

these questions all are about factoring

hartnn (hartnn):

q=1+12/(p-1)

OpenStudy (anonymous):

So, only two solutions?

OpenStudy (anonymous):

\[q=1+\frac{12}{p-1}\]

hartnn (hartnn):

so 12 is divisible by p-1

OpenStudy (anonymous):

So, p cannot be greater than 13

OpenStudy (anonymous):

yep

hartnn (hartnn):

so p-1=1,2,3,4,6,12

OpenStudy (anonymous):

So, we substitute all the prime number between 2 and 13 for p and see

OpenStudy (anonymous):

yup

OpenStudy (anonymous):

Interesting questions........ Do u have some more problems

hartnn (hartnn):

so p=3,7,11

OpenStudy (anonymous):

p is not 11

OpenStudy (anonymous):

these questions comes from my mind...let me make another one

hartnn (hartnn):

(p,q,r)=(3,7,11) in any order

OpenStudy (anonymous):

OMG....... TRUE GENIUS

OpenStudy (anonymous):

(2,11,13)

OpenStudy (anonymous):

also (2,13,11) in any order

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