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

The four-digit number 486X, where X is the units digit of 486X, is a multiple of 36. What is the value of X?

A. 0
B. 2
C. 4
D. 6
E. 8

$$? = X$$
$${{\left\langle {486X} \right\rangle } \over {{2^2} \cdot {3^2}}} = {\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,\,\left\langle {486X} \right\rangle \,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,4\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {6X} \right\rangle \,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,4\,\,\,\,\, \Rightarrow \,\,\,\,X \in \left\{ {0,4,8} \right\}\,\,\,\left( * \right) \hfill \cr
\,\,\left\langle {486X} \right\rangle \,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,9\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {18 + X\,\,,\,\,{\rm{hence}}} \right)\,\,X\,\,{\rm{is}}\,\,{\rm{a}}\,\,{\rm{multiple}}\,\,{\rm{of}}\,\,9\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,? = X = 0 \hfill \cr} \right.$$

The correct answer is therefore (A).


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

Regards,
Fabio.

=>

Since X is a one-digit integer, 0 ≤ X ≤ 9.
36 = 4*9, so 486X is a multiple of both 4 and 9.
Since 486X is a multiple of 4, 6*10 + X must be a multiple of 4.
Thus, X is one of values 0, 4 and 8.
In addition, since 486X is a multiple of 9, the sum of all of its digits, 4 + 8 + 6 + X = X + 18 is a multiple of 9.
Thus, X must be 0 or 9.

0 is the only one of these digits for which 486X is a multiple of 36.

Therefore, the answer is A.
Answer: A