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# j and k are positive integers, and n = 10^j

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Joined: 12 Sep 2015
Posts: 4007
j and k are positive integers, and n = 10^j  [#permalink]

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11 Oct 2018, 13:52
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00:00

Difficulty:

95% (hard)

Question Stats:

40% (01:58) correct 60% (01:43) wrong based on 58 sessions

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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

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Re: j and k are positive integers, and n = 10^j  [#permalink]

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11 Oct 2018, 15:29
1
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.

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Joined: 12 Oct 2010
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Re: j and k are positive integers, and n = 10^j  [#permalink]

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11 Oct 2018, 19:19
1
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.
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j and k are positive integers, and n = 10^j  [#permalink]

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12 Oct 2018, 03:49
1
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 3

We can use plug-in 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 5

Let j=2 k =5..........n=105.........Answer is Yes

Let j=2 k =25..........n=105.........Answer is No

Insufficient

GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4007
j and k are positive integers, and n = 10^j  [#permalink]

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12 Oct 2018, 06:22
Top Contributor
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 3
So, 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 15
Since 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 15
Case 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 15
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Cheers,
Brent
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j and k are positive integers, and n = 10^j   [#permalink] 12 Oct 2018, 06:22
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