Problem Solving

BTGmoderatorDC wrote:If 2 and 17 are factors of positive integer n, then which of the following must also divide into n?

I. 34
II. 68
III. 136

A. I only
B. II only
C. I and II only
D. I and III only
E. None
Source: Manhattan Prep

$$n \ge 1\,\,{\mathop{\rm int}} $$
$$\left\{ \matrix{
{n \over 2} = {\mathop{\rm int}} \hfill \cr
{n \over {17}} = {\mathop{\rm int}} \hfill \cr} \right.\,\,\,\,\,\,\,$$
$$?\,\,\,:\,\,\,{n \over {\,{\rm{I}}\,{\rm{,}}\,{\rm{II}}\,{\rm{,}}\,{\rm{III}}\,}}\,\,\,\mathop = \limits^? \,\,\,{\mathop{\rm int}} $$

$${\rm{I}}.\,\,\,\,{n \over {34}}\,\, = \,\,{n \over {2 \cdot 17}}\,\,\mathop = \limits^{GCF\,\left( {2,17} \right)\,\, = \,\,1} \,\,\,\,{\mathop{\rm int}} \,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,$$
$${\rm{II}}.\,\,\,\,{n \over {68}}\,\,\,\mathop = \limits^? \,\,\,{\mathop{\rm int}} \,\,\,\,,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\left[ {\,{\rm{Take}}\,\,n = 34\,} \right]$$
$${\rm{III}}.\,\,\,\,{n \over {136}}\,\,\,\mathop = \limits^? \,\,\,{\mathop{\rm int}} \,\,\,\,,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\left[ {\,{\rm{Take}}\,\,n = 34\,} \right]$$
$$\left( {\,\,\left\langle {{\rm{NO}}} \right\rangle = \,\,{\rm{not}}\,\,{\rm{necessarily}}\,} \right)$$

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

Regards,
Fabio.

BTGmoderatorDC wrote:If 2 and 17 are factors of positive integer n, then which of the following must also divide into n?

I. 34
II. 68
III. 136

A. I only
B. II only
C. I and II only
D. I and III only
E. None

2 and 17 have no factors in common.
Since n must be divisible by both 2 and 17 -- and 2*17 = 34 -- n must be a MULTIPLE OF 34.
Thus, n must be divisible by I.
Eliminate any answer choice that does not include I.
Eliminate B and E.
If n=34, then n is not divisible by II or III.
Eliminate any remaining answer choice that include II or III.
Eliminate C and D.

The correct answer is A.