Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

The terms of a sequence a1, a2, a3, ... are defined by an=an-2+3. Is 411 a term of the sequence?

1) a1=111
2) a2=112

Very nice problem, Max. Congrats!
$$S\,\,{\rm{sequence:}}\,\,\left\{ \matrix{
{{\rm{a}}_{\rm{1}}},{a_2}, \ldots \hfill \cr
{a_n} = {a_{n - 2}} + 3\,\,,\,\,{\rm{for}}\,\,{\rm{all}}\,\,n \ge 3 \hfill \cr} \right.\,\,\,\,\,\left( * \right)$$
$$411\,\,\mathop \in \limits^? \,\,\,S$$
$$\left( 1 \right)\,\,{a_1} = 111\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,{a_3} = 111 + 3\,\,\,\, \Rightarrow \,\,\,{a_5} = 111 + 3 + 3 = 111 + 2 \cdot 3\,\,\, \Rightarrow \,\,\,\,{a_7} = 111 + 3 \cdot 3\,\,\,\, \Rightarrow \,\,\, \ldots $$
$${\text{Hence}}\,\,411 = 111 + 100 \cdot 3\,\, \in \,\,\,S\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle $$
$$\left( 2 \right)\,\,{a_2} = 112\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,{a_4} = 112 + 3\,\,\,\, \Rightarrow \,\,\,{a_6} = 112 + 2 \cdot 3\,\,\, \Rightarrow \,\,\,\, \ldots $$
$$411 \ne 112 + k \cdot 3\,,\,\,{\rm{for}}\,\,{\rm{all}}\,\,k\,\, \ge 1\,\,{\mathop{\rm int}} \,\,\,\,\left\{ \matrix{
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,{{\rm{a}}_{\rm{1}}}{\rm{ = 111}}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr
\,Take\,\,{a_1} = 112\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.$$
Obs.: note that defining a1 and a2 , the sequence S is uniquely defined.


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

Regards,
Fabio.

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The formula an=an-2+3 tells us that alternate terms have the same remainder when they are divided by 3.
Since a1 = 111 = 3*37 is a multiple of three, all multiples of three greater than 111 can be obtained as odd-numbered terms. Therefore, 411= 3*137 is one of the odd-numbered terms, and 411 is in the sequence.
Condition 1) is sufficient.

a2 = 112 = 3*37 + 1 and all even-numbered terms have a remainder of 1 when they are divided by 3. Since 411 = 3*137, it is not an even-numbered term. Since we don't know any of the odd-numbered terms, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A