Question: The congruence
equation ′′ ( 5a +3 )x ≡^(3a + 4) 19 ′′ is given.
Find the sum of digits of
the smallest three −digit natural number ” a ”
such that the assumed equation has
no solution in ” Z ”.
If xyz ∈R^+ , xyz=1 , prove that the following inequality holds:
(x/(2x^5 +x+4))+(y/(2y^5 +y+4))+(z/(2z^5 +z+4))≥(3/7).
Solution please with an advice to get better
at inequalities and which book would you recommend.
Thanks in advance!