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11 Oct 2018, 13:52
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Difficulty:
95%
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Question Stats:
39% (01:53) correct
61%
(01:54)
wrong
based on 120
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\(j\) and \(k\) are positive integers, and \(n = 10^j + k\). Is \(n\) divisible by 15?
(1) j and k are each divisible by 3
(2) j and k are each divisible by 5
*kudos for all correct solutions
(1) j and k are each divisible by 3
(2) j and k are each divisible by 5
*kudos for all correct solutions
11 Oct 2018, 15:29
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GMATPrepNow
10 to the power of any positive integer of j will have a units of 0.
To be divisible by 15 it must be divisible by both 3 and 5.
If j = 3 then 1,000 and if k = 3
1,003 not divisble by 15.
J/3 = integer.
J = 3 * integer. The same also applies to K
For n to be divisible by 5 the units digit must be 0 or 5
k must be 15,30,45 etc.. however this will not satisfy divisibility by 3.
The 1,000 + any number divisible by 3 will not be divisible by 3. Because the sum of the digits will be 1 + a number divisible by 3
Sufficient to answer that is it not divisible by 15
Statement 2)
From the analysis above we know that the units digit of 10^j will be positive.
If k = 5
Then 100,005 is divisible by 15.
K = 10
100,010 is not divisible by 15.
Insufficient.
Answer choice A
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11 Oct 2018, 19:19
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GMATPrepNow
\(n = {\rm{1}}{{\rm{0}}^{\rm{j}}} + k\,\,\,\,;\,\,\,\,\,j,k\,\,\, \ge 1\,\,\,\,{\rm{ints}}\,\,\,\left( * \right)\)
\({n \over {3 \cdot 5}}\,\,\,\mathop = \limits^? \,\,\,{\mathop{\rm int}}\)
\(\left( 1 \right)\,\, \cap \,\,\,\left( * \right)\,\,\, \Rightarrow \,\,\,\left\{ \matrix{\\
j = 3M,\,\,M \ge 1\,\,{\mathop{\rm int}} \hfill \cr \\
k = 3L,\,\,L \ge 1\,\,{\mathop{\rm int}} \hfill \cr} \right.\)
\(n\,\,\, = \underbrace {{{\left( {{{10}^{M\, \ge \,1}}} \right)}^3}}_{\left\langle {100 \ldots 0} \right\rangle \,\,{\rm{not}}\,\,{\rm{div}}\,\,{\rm{by}}\,\,3} + \underbrace {3L}_{{\rm{div}}\,\,{\rm{by}}\,\,3}\,\, = \,\,{\rm{not}}\,\,{\rm{div}}\,\,{\rm{by}}\,\,\,3\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle\)
\(\left( 2 \right)\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {j,k} \right) = \left( {5,5} \right)\,\,\,\,\,\, \Rightarrow \,\,\,n = 100005\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\left( {\sum\nolimits_{{\rm{digits}}} {\,\,{\rm{and}}\,\,{\rm{final}}\,\,{\rm{digit}}} } \right) \hfill \cr \\
\,{\rm{Take}}\,\,\left( {j,k} \right) = \left( {5,10} \right)\,\,\,\,\,\, \Rightarrow \,\,\,n = 100010\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\left( {\sum\nolimits_{{\rm{digits}}} {} } \right) \hfill \cr} \right.\)
The correct answer is therefore (A).
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
