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solve in prime numbers\[pqr=11(p+q+r)\]
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either one of the p,q,r should be 11
yep
so pq=p+q+11
q=2 p=13
why 11...... ?
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q=3 p=7
oh got it
q=(p+11)/(p-1)
these questions all are about factoring
q=1+12/(p-1)
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So, only two solutions?
\[q=1+\frac{12}{p-1}\]
so 12 is divisible by p-1
So, p cannot be greater than 13
yep
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so p-1=1,2,3,4,6,12
So, we substitute all the prime number between 2 and 13 for p and see
yup
Interesting questions........ Do u have some more problems
so p=3,7,11
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p is not 11
these questions comes from my mind...let me make another one
(p,q,r)=(3,7,11) in any order
OMG....... TRUE GENIUS
(2,11,13)
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also (2,13,11) in any order
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