superpus07 wrote:
All the terms in Set S are integers. Five terms in S are even, and four terms are multiples of 3. How many terms in S are even numbers that are not divisible by 3?
(1) The product of all the even terms in Set S is a multiple of 9.
(2) The integers in S are consecutive.
\(?\,\,\, = \,\,\,x\,\,\, = \,\,\,5 - \left( {\# \,\,{\rm{multiples}}\,\,{\rm{of}}\,\,6} \right)\)
\(\left( 1 \right)\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,S = \left\{ {\,2\,,2 \cdot 3\,,2 \cdot {3^2},\,2 \cdot {3^{3\,}},2 \cdot {3^4}\,} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 5 - 4 = 1 \hfill \cr \\
\,{\rm{Take}}\,\,S = \left\{ {\,2\,,{2^2}\,,2 \cdot {3^2},\,2 \cdot {3^{3\,}},2 \cdot {3^4},3\,} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 5 - 3 = 2 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{INSUFF}}{\rm{.}}\)
\(\left( 2 \right)\,\,\,{\rm{different}}\,\,{\rm{classes}}\,\,{\rm{of}}\,\,{\rm{remainders}}\,\,{\rm{by}}\,\,3\,\,{\rm{and}}\,\,5\,\,\left( {{\rm{with}}\,\,5\,\,{\rm{even}}\,\,{\rm{numbers}}} \right)\,\,{\rm{:}}\,\,\,0,2,4\,\,{\rm{for}}\,\,{\rm{the}}\,\,{\rm{smaller}}\,\,{\rm{even}}\,\,{\rm{in}}\,\,{\rm{S}}\)
The idea is crucial: 0, 2 and 4 represent ALL possible scenarios for the first even integer belonging to S, even when negative integers are considered.
(We have presented the 6, 8 and 10 "next group" so that the "repetition of the cyclic behavior" becomes clear!)
\(\left. {\left\{ \matrix{\\
\,{\rm{0}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}} - 1} \right),1,3,5,7\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,9} \right)\,\, \to \,\,\,\,\,0,\,3,6,9\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 5 - 2 = 3 \hfill \cr \\
\,{\rm{2}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}1} \right),3,5,7,9\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,11} \right)\,\, \to \,\,\,\,\,3,\,6,9\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,{\rm{not}}\,\,{\rm{viable}} \hfill \cr \\
\,{\rm{4}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}3} \right),5,7,9,11\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,13} \right)\,\, \to \,\,\,\,\,3,\,6,9,12\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,? = 5 - 2 = 3 \hfill \cr \\
\,{\rm{6}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}5} \right),7,9,11,13\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,15} \right)\,\, \to \,\,\,\,\,6,\,9,12,15\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,? = 5 - 2 = 3 \hfill \cr \\
\,{\rm{8}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}7} \right),9,11,13,15\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,17} \right)\,\, \to \,\,\,\,\,9,\,12,15\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,{\rm{not}}\,\,{\rm{viable}} \hfill \cr \\
{\rm{10}}\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}9} \right),11,13,15,17\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,19} \right)\,\, \to \,\,\,\,\,9,\,12,15,18\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,? = 5 - 2 = 3 \hfill \cr} \right.\,\,\,} \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 3\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}{\rm{.}}\,\)
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)