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

What is the units digit of a positive integer n?

1) n is a common multiple of 13 and 14.
2) n is a common multiple of 13 and 15.

$$n \ge 1\,\,{\mathop{\rm int}} $$
$$?\,\,\,:\,\,n\,\,{\rm{units}}\,\,{\rm{digit}}$$
$$\left( 1 \right)\,\,n\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,LCM\left( {13,14} \right) = 13 \cdot 14\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,n = 13 \cdot 14\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,{\rm{ 2}}\,\, \hfill \cr
\,{\rm{Take}}\,\,n = 13 \cdot 14 \cdot 2\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,{\rm{ 4}} \hfill \cr} \right.$$
$$\left( 2 \right)\,\,n\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,LCM\left( {13,15} \right) = 13 \cdot 15\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,n = 13 \cdot 15\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,{\rm{ 5}}\,\, \hfill \cr
\,{\rm{Take}}\,\,n = 13 \cdot 15 \cdot 2\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,{\rm{ 0}} \hfill \cr} \right.$$
$$\left( {1 + 2} \right)\,\,n\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,13 \cdot 14 \cdot 15 = 13 \cdot 7 \cdot \underline {2 \cdot 5} \cdot 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,n\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,10\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{?}}\,\,{\rm{ = }}\,{\rm{ 0}}\,\,\,\,\,$$

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

Regards,
Fabio.

n is a positive integer, hence $$n\ >0$$

Statement 1
n is a common multiple of 13 and 14
multiple of 13 = 1 and 13
multiple of 14 = 1, 2,7, and 14
common multiple =1
n = 1

Statement 2
n is a common multiple of 13 and 15
multiple of 13 = 1, 13
multiple of 15 = 1, 3, 5 and 15
common multiple = 1
n=1

Statement 1 and 2 are SUFFICIENT.
$$answer\ is\ option\ D$$

n is a positive integer, hence $$n\ >0$$

Statement 1
n is a common multiple of 13 and 14
multiple of 13 = 1 and 13
multiple of 14 = 1, 2,7, and 14
common multiple =1
n = 1

Statement 2
n is a common multiple of 13 and 15
multiple of 13 = 1, 13
multiple of 15 = 1, 3, 5 and 15
common multiple = 1
n=1

Statement 1 and 2 are SUFFICIENT
$$answer\ is\ option\ D$$

=>

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.

Since we have 1 variable (n) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
Since n is a common multiple of 13 and 14, n is a multiple of lcm(13,14) = 182. Thus units digits of multiples of n are 0, 2, 4, 6, 8 since the units digit of 182 is 2.
Condition 1) is not sufficient.

Condition 2)
Since n is a common multiple of 13 and 15, n is a multiple of lcm(13,15) = 195. Thus units digits of multiples of n are 0, 5 since the units digit of 182 is 2.
Condition 2) is not sufficient.

Conditions 1) & 2)
The common units digit of multiples of 182 and 195 is 0 only.
Both conditions together are sufficient.

Answer: C

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.