AAPL wrote:Manhattan Prep

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!)

$${\left\{ \matrix{
\,{\rm{0}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}} - 1} \right),1,3,5,7\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,9} \right)\,\, \to \,\,\,\,\,0,\,3,6,9\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 5 - 2 = 3 \hfill \cr
\,{\rm{2}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}1} \right),3,5,7,9\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,11} \right)\,\, \to \,\,\,\,\,3,\,6,9\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,{\rm{not}}\,\,{\rm{viable}} \hfill \cr
\,{\rm{4}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}3} \right),5,7,9,11\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,13} \right)\,\, \to \,\,\,\,\,3,\,6,9,12\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,? = 5 - 2 = 3 \hfill \cr
\,{\rm{6}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}5} \right),7,9,11,13\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,15} \right)\,\, \to \,\,\,\,\,6,\,9,12,15\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,? = 5 - 2 = 3 \hfill \cr
\,{\rm{8}}\,\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly}}\,{\rm{ }}7} \right),9,11,13,15\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,17} \right)\,\, \to \,\,\,\,\,9,\,12,15\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,{\rm{not}}\,\,{\rm{viable}} \hfill \cr
{\rm{10}}\,\, \to \,\,\,{\rm{odds}}:\,\,\left( {{\rm{possibly }}\, 9} \right),11,13,15,17\,\left( {{\rm{and}}\,\,{\rm{possibly}}\,\,19} \right)\,\, \to \,\,\,\,\,9,\,12,15,18\,\,{\rm{multiples}}\,\,{\rm{of}}\,\,3\,\,\,\,\,\, \Rightarrow \,\,\,? = 5 - 2 = 3 \hfill \cr} \right.\,\,\,}$$
$$ \Rightarrow \,\,\,\,\,\,? = 3\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Statement 1
The product of all even terms in set S as a multiple of 9
The smallest even multiple of 9 which should be a product of 5 even terms could be either
$$2\cdot2\cdot2\cdot2\cdot2\cdot9\ or\ 2\cdot2\cdot4\cdot4\cdot4\cdot9$$
There are so many terms that will satisfy the criteria information provided in statement is not enough to provide the answers;
statement 1 is INSUFFICIENT.

Statement 2
The integers in 5 are consecutive. For consecutive numbers in range 1 to 11 we are going to have 1,2,3,4,5,6,7,8,9,10.
From the set of numbers, we will have 5 even numbers but only three terms are multiple of 3 and 3 is part of only 1 even number
which is 6 we cannot consider it.
For consecutive numbers in range 3 to 13 this will give us 3,4,5,6,7,8,9,10,11,12 and we will have 5 even terms, 4 multiples of 3
even numbers 6 and 12 are divisible by 3.
Number of terms in S not divisible by
$$3=5-2$$ $$\le3\ terms\ $$
$$answer\ is\ option\ B$$