February 23, 2019 February 23, 2019 07:00 AM PST 09:00 AM PST Learn reading strategies that can help even nonvoracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT February 24, 2019 February 24, 2019 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score.
Author 
Message 
TAGS:

Hide Tags

CEO
Joined: 11 Sep 2015
Posts: 3447
Location: Canada

j and k are positive integers, and n = 10^j
[#permalink]
Show Tags
11 Oct 2018, 12:52
Question Stats:
38% (01:50) correct 63% (01:39) wrong based on 53 sessions
HideShow timer Statistics
\(j\) 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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Test confidently with gmatprepnow.com



Director
Joined: 19 Oct 2013
Posts: 508
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)

Re: j and k are positive integers, and n = 10^j
[#permalink]
Show Tags
11 Oct 2018, 14:29
GMATPrepNow wrote: \(j\) 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 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 Posted from my mobile device



GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 772

Re: j and k are positive integers, and n = 10^j
[#permalink]
Show Tags
11 Oct 2018, 18:19
GMATPrepNow wrote: \(j\) 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
\(n = {\rm{1}}{{\rm{0}}^{\rm{j}}} + k\,\,\,\,;\,\,\,\,\,j,k\,\,\, \ge 1\,\,\,\,{\rm{ints}}\,\,\,\left( * \right)\) \({n \over {3 \cdot 5}}\,\,\,\mathop = \limits^? \,\,\,{\mathop{\rm int}}\) \(\left( 1 \right)\,\, \cap \,\,\,\left( * \right)\,\,\, \Rightarrow \,\,\,\left\{ \matrix{ j = 3M,\,\,M \ge 1\,\,{\mathop{\rm int}} \hfill \cr k = 3L,\,\,L \ge 1\,\,{\mathop{\rm int}} \hfill \cr} \right.\) \(n\,\,\, = \underbrace {{{\left( {{{10}^{M\, \ge \,1}}} \right)}^3}}_{\left\langle {100 \ldots 0} \right\rangle \,\,{\rm{not}}\,\,{\rm{div}}\,\,{\rm{by}}\,\,3} + \underbrace {3L}_{{\rm{div}}\,\,{\rm{by}}\,\,3}\,\, = \,\,{\rm{not}}\,\,{\rm{div}}\,\,{\rm{by}}\,\,\,3\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle\) \(\left( 2 \right)\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {j,k} \right) = \left( {5,5} \right)\,\,\,\,\,\, \Rightarrow \,\,\,n = 100005\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\left( {\sum\nolimits_{{\rm{digits}}} {\,\,{\rm{and}}\,\,{\rm{final}}\,\,{\rm{digit}}} } \right) \hfill \cr \,{\rm{Take}}\,\,\left( {j,k} \right) = \left( {5,10} \right)\,\,\,\,\,\, \Rightarrow \,\,\,n = 100010\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\left( {\sum\nolimits_{{\rm{digits}}} {} } \right) \hfill \cr} \right.\) The correct answer is therefore (A). This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our highlevel "quant" preparation starts here: https://gmath.net



SVP
Joined: 26 Mar 2013
Posts: 2068

j and k are positive integers, and n = 10^j
[#permalink]
Show Tags
12 Oct 2018, 02:49
GMATPrepNow wrote: \(j\) 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 j & k are positive integers, So zero is NOT valid in any case here. To make n divisible by 15, n must have BOTH unit digit either 0 or 5, and sum of the digits divisible by 3. For example: 105, 10050, 135...etc (1) j and k are each divisible by 3We can use plugin or use reasoning. I would choose the latter. \(10^j\) have always sum of 1. Therefore, if 1 is added to any number divisible by the 3, will make the total not divisible by 3 and hence not divisible by 15 Sufficient (2) j and k are each divisible by 5Let j=2 k =5..........n=105.........Answer is Yes Let j=2 k =25..........n=105.........Answer is No Insufficient Answer: A



CEO
Joined: 11 Sep 2015
Posts: 3447
Location: Canada

j and k are positive integers, and n = 10^j
[#permalink]
Show Tags
12 Oct 2018, 05:22
GMATPrepNow wrote: \(j\) 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 Given: n = 10^j + k Target question: Is n divisible by 15?Key concepts:  If n is divisible by 15, then n must be divisible by 3 AND by 5  If n is divisible by 3, then the sum of n's digits must be divisible by 3 (for example, we know that 747 is divisible by 3, because 7+4+7 = 18, and 18 is divisible by 3) Statement 1: j and k are each divisible by 3 First recognize that 10^j = 1 followed by j zeros. For example, 10^15 = 1,000,000,000,000,000 (1 followed by 15 zeros) Next, recognize that, if k is divisible by 3, then the sum of n's digits must be divisible by 3So, if n = 10^j + k, then the sum of n's digits will be 1 greater than some multiple of 3. [since we're adding 1 and several zeros to a number that is divisible by 3]For example, if j = 6 and k = 24, then n = 10^j + k = 10^6 + 24 = 1,000,024. In this case, the sum of n's digits = 1+0+0+0+0+2+4 = 7, which is 1 greater than a multiple of 3. Likewise, if j = 9 and k = 75, then n = 10^j + k = 10^9 + 75 = 1,000, 000,075. In this case, the sum of n's digits = 13, which is 1 greater than a multiple of 3. And, if j = 15 and k = 99, then n = 10^15 + 99 = 1,000, 000,000,000,099. In this case, the sum of n's digits = 19, which is 1 greater than a multiple of 3. And so on. Since the sum of n's digits will always be 1 greater than some multiple of 3, we can be certain that n is NOT divisible by 3 So, by the above property, n is NOT divisible by 15Since we can answer the target question with certainty, statement 1 is SUFFICIENT Statement 2: j and k are each divisible by 5 There are several values of j and k that satisfy statement 2. Here are two: Case a: j = 5 and k = 5. In this case, n = 10^j + k = 10^5 + 5 = 100,005. Since 100,005 is divisible by 3 and by 5, the answer to the target question is YES, n IS divisible by 15Case b: j = 5 and k = 10. In this case, n = 10^j + k = 10^5 + 10 = 100,010. Since 100,010 is NOT divisible by 3, the answer to the target question is NO, n is NOT divisible by 15Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT Answer: A Cheers, Brent
_________________
Test confidently with gmatprepnow.com




j and k are positive integers, and n = 10^j
[#permalink]
12 Oct 2018, 05:22






