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

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The four-digit number 486X, where X is the units digit of 486X, is a m  [#permalink]

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New post 31 Oct 2018, 23:37
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[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 \(X\)?

\(A. 0\)
\(B. 2\)
\(C. 4\)
\(D. 6\)
\(E. 8\)

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Re: The four-digit number 486X, where X is the units digit of 486X, is a m  [#permalink]

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New post 01 Nov 2018, 00:18
MathRevolution 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 \(X\)?

\(A. 0\)
\(B. 2\)
\(C. 4\)
\(D. 6\)
\(E. 8\)



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

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Re: The four-digit number 486X, where X is the units digit of 486X, is a m  [#permalink]

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New post 01 Nov 2018, 00:32
2
Divisibility of 4 : Last 2 digits should be divisible by 4
Divisibility of 9 : Sum of digits should be divisible by 9

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Re: The four-digit number 486X, where X is the units digit of 486X, is a m  [#permalink]

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New post 01 Nov 2018, 05:42
MathRevolution 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 \(X\)?

\(A. 0\)
\(B. 2\)
\(C. 4\)
\(D. 6\)
\(E. 8\)

\(? = X\)

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


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

Regards,
Fabio.
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Re: The four-digit number 486X, where X is the units digit of 486X, is a m  [#permalink]

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New post 02 Nov 2018, 16:10
1
MathRevolution 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 \(X\)?

\(A. 0\)
\(B. 2\)
\(C. 4\)
\(D. 6\)
\(E. 8\)


Since 36 = 4 x 9, we see that 486X must be divisible by 4 and 9. From the divisibility rules, we see that the sum of the digits must be a multiple of 9, and the number formed by the last two digits, 6X, must be a multiple of 4.

Since 4 + 8 + 6 = 18, so X can be 0 or 9 if 486X is divisible by 9, or it can be 0, 4 or 8 if 486X is divisible by 4. However, since 486X is divisible by both 9 and 4, X must be 0.

Answer: A
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Re: The four-digit number 486X, where X is the units digit of 486X, is a m  [#permalink]

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New post 04 Nov 2018, 19:13
=>

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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Re: The four-digit number 486X, where X is the units digit of 486X, is a m &nbs [#permalink] 04 Nov 2018, 19:13
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