The four-digit number 486X, where X is the units digit of 486X, is a m

So,here is my approach ..

I would divide 486 by 36 and see what is the remainder.

If you divide 486/36 the remainder is 18..Now what the next digit of quotient should come
after 18 and that is x such that 18x/36 is divisible...clearly x is zero..so 180/36=5

hence OA is A

Press Kudos if it helps!!

\(? = X\)

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


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

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