GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Nov 2019, 15:06

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

If x, y, and k are positive integers, is k < 10 ?

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Intern
Intern
avatar
G
Joined: 18 Nov 2013
Posts: 46
Reviews Badge
If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post Updated on: 14 Oct 2018, 18:10
26
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

23% (02:29) correct 77% (02:19) wrong based on 253 sessions

HideShow timer Statistics

If x, y, and k are positive integers, is k < 10 ?


(1) \(45! = x(10^k)\)

(2) \(y=\sqrt[3]{1.25*10^k}\)

Originally posted by harish1986 on 08 Oct 2018, 08:53.
Last edited by chetan2u on 14 Oct 2018, 18:10, edited 3 times in total.
Renamed the topic, corrected the OA
Most Helpful Expert Reply
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 8180
Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 08 Oct 2018, 09:05
4
2
If x, y, and k are positive integers, is k < 10 ?

1. \(45! = x(10^k)\)
Max possible value of k is when x does not have any 10s in it..
So number of 10s in 45! is number of 5s = \(\frac{45}{5}+\frac{45}{25}=9+1=10\)
So if x does not have any 5s then k is 10, otherwise it will be <10
Insufficient

2. y=\(3\sqrt{1.25(10^k)}\)
I believe it is the cube root ..
So \(y^3=1.25*10^k=5^3*10^{k-2}\)
So minimum value of k is 2 or k can be 5,8,11 and so on
Insufficient

Combined k cannot be 10 as per statement II and \(k\leq{10}\)
So k<10
Sufficient

C

Editing the OA as it cannot be A
_________________
General Discussion
Manager
Manager
User avatar
S
Joined: 10 Sep 2015
Posts: 70
Location: India
Concentration: Finance, Human Resources
GMAT 1: 640 Q47 V31
GMAT 2: 660 Q47 V35
GMAT 3: 700 Q49 V36
GPA: 4
Reviews Badge
Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 14 Oct 2018, 11:57
Hi chetan2u,
I took the statement 2 as 3*(1.25(10k))^-2
and hence i got E. Please confirm if my understanding is incorrect or question is written in a bad format.
GMATH Teacher
User avatar
P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 14 Oct 2018, 14:02
harish1986 wrote:
If x, y, and k are positive integers, is k < 10 ?

(1) \(45! = x(10^k)\)

(2) \(y\) is the cubic root of \(1.25*(10^k)\)

\(x,y,k\,\, \ge 1\,\,\,{\rm{ints}}\,\,\,\left( * \right)\)

\(k\,\,\mathop < \limits^? \,\,10\)

\(\left( 1 \right)\,\,45! = x \cdot {10^k}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,k\,\,\, \le \,\,\,\left\lfloor {{{45} \over 5}} \right\rfloor + \left\lfloor {{{45} \over {25}}} \right\rfloor \, = 10\,\,\,\,\,\left( {**} \right)\,\,\)

\(\left( {**} \right)\) See my explanation (and notation) here:
https://gmatclub.com/forum/if-n-is-the- ... 75460.html

\(\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {k{\kern 1pt} \,;\,x} \right) = \left( {10\,;\,\,{{45!} \over {{{10}^{10}}}}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\left( {x = {{45!} \over {{{10}^{10}}}}\,\, \ge 1\,\,{\mathop{\rm int}} \,\,\,{\rm{by}}\,\,\left( {**} \right)} \right)\,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {k\,;\,x} \right) = \left( {1\,;\,\,{{45!} \over {10}}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\, \hfill \cr} \right.\)


\(\left( 2 \right)\,\,\,y = \root {3\,} \of {{5 \over 4}\left( {{{10}^k}} \right)} \,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,{5 \over 4}\left( {{2^k} \cdot {5^k}} \right) = {2^{k - 2}} \cdot {5^{k + 1}}\,\,\,\,{\rm{positive}}\,\,{\rm{perfect}}\,\,{\rm{cube}}\,\,\,\,\,\)

\(\Rightarrow \,\,\,\,\,\,\left\{ \matrix{
\,k - 2 = {\rm{mult}}\,\,{\rm{of}}\,\,{\rm{3}} \hfill \cr
k + 1 = \,\,{\rm{mult}}\,\,{\rm{of}}\,\,3 \hfill \cr} \right.\,\,\,\,\,\,\,\,\left( {k \ge 2} \right)\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,k \ge 2\,\,\,\,\,{\rm{divided}}\,\,{\rm{by}}\,\,3\,\,\,{\rm{has}}\,\,{\rm{remainder}}\,\,2\,\)

\(\left\{ \matrix{
\,{\rm{Take}}\,\,k = 2\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,k = 2 + 3 \cdot 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\, \hfill \cr} \right.\)


\(\left( {1 + 2} \right)\,\,\,\,\,\left\{ \matrix{
\,k \le 10\,\,\,\,{\rm{by}}\,\,\,\,\left( 1 \right) \cap \left( {**} \right) \hfill \cr
\,k \ne 10\,\,\,{\rm{by}}\,\,\left( 2 \right) \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)


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

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
GMATH Teacher
User avatar
P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 14 Oct 2018, 14:18
P.S.: \(y = 3 \cdot \,\,\root {} \of {{5 \over 4}\left( {{{10}^k}} \right)}\) cannot be an integer, for any given positive integer \(k\).

Reason:

\(y = \,\,\sqrt {\,9 \cdot {5 \over 4}\left( {{2^k} \cdot {5^k}} \right)} \,\,\,\, = \,\,\,\,\sqrt {\,{2^{k - 2}} \cdot {3^2} \cdot {5^{k + 1}}} \,\,\,\,\,\,\left( {k \ge 1\,\,\,{\mathop{\rm int}} } \right)\,\,\,\)

\({2^{k - 2}} \cdot {3^2} \cdot {5^{k + 1}}\,\,\,{\rm{perfect}}\,\,{\rm{square}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\{ \matrix{
\,k - 2\,\,\, \ge 0\,\,\,{\rm{even}} \hfill \cr
\,k + 1\,\,\, \ge 0\,\,\,{\rm{even}} \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{\rm{impossible}}!\,\)

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
SVP
SVP
User avatar
V
Joined: 26 Mar 2013
Posts: 2341
Reviews Badge CAT Tests
Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 14 Oct 2018, 15:17
chetan2u wrote:
If x, y, and k are positive integers, is k < 10 ?

1. \(45! = x(10^k)\)
Max possible value of k is when x does not have any 10s in it..
So number of 10s in 45! is number of 5s = \(\frac{45}{5}+\frac{45}{25}=9+1=10\)
So if x does not have any 5s then k is 10, otherwise it will be <10
Insufficient

2. y=\(3\sqrt{1.25(10^k)}\)
I believe it is the cube root ..
So \(y^3=1.25*10^k=5^3*10^{k-2}\)
So minimum value of k is 2 or k can be 5,8,11 and so on
Insufficient

Combined k cannot be 10 as per statement II and \(k\leq{10}\)
So k<10
Sufficient

C

Editing the OA as it cannot be A



Can you please re-format statement 2? I does not show cubic root.

Thanks
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 8180
Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 14 Oct 2018, 18:11
asthagupta wrote:
Hi chetan2u,
I took the statement 2 as 3*(1.25(10k))^-2
and hence i got E. Please confirm if my understanding is incorrect or question is written in a bad format.



Yes, you are correct.
It is bad formatting, which I have corrected now.
_________________
Manager
Manager
avatar
B
Joined: 04 Oct 2017
Posts: 73
Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 14 Oct 2018, 19:14
[quote="chetan2u"]If x, y, and k are positive integers, is k < 10 ?

1. \(45! = x(10^k)\)
Max possible value of k is when x does not have any 10s in it..
So number of 10s in 45! is number of 5s = \(\frac{45}{5}+\frac{45}{25}=9+1=10\)
So if x does not have any 5s then k is 10, otherwise it will be <10
Insufficient

2. y=\(3\sqrt{1.25(10^k)}\)
I believe it is the cube root ..
So \(y^3=1.25*10^k=5^3*10^{k-2}\)
So minimum value of k is 2 or k can be 5,8,11 and so on
Insufficient

Combined k cannot be 10 as per statement II and [m]k\leq{10}[

Hi,

Please explain Number of 10s in 45! is number of 5s. I didnot understand this. Am lost with factorials. I understood the second statement.

Thanks.
Senior Manager
Senior Manager
User avatar
G
Joined: 21 Jun 2017
Posts: 307
Location: India
Concentration: Finance, Economics
Schools: IIM
GPA: 3
WE: Corporate Finance (Commercial Banking)
CAT Tests
If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 14 Oct 2018, 21:24
1
Hi Kezia9

I'll try to explain your query.

In order to find number of zeros or 5 in a complex no , following has to be done.

1. list down numbers ending with 5 and 0
With our case in hand , the list will be
5 10 15 20 25 30 35 40 and 45
2. each of these will have one 5 in them.
3. watchout for perfect squares ...25 will have two.

so 1+1+1+1+2+1+1+1+1= 10 5's


PS: Always remember in order to find zeros or number of 5's you don't need to calculate 2's, because they are plenty. Every alternative number will be even hence, we will just find out the limiting 5.

Posted from my mobile device
_________________
Even if it takes me 30 attempts, I am determined enough to score 740+ in my 31st attempt. This is it, this is what I have been waiting for, now is the time to get up and fight, for my life is 100% my responsibility.

Dil ye Ziddi hai !!!

GMAT 1 - 620 .... Disappointed for 6 months. Im back Im back. Bhai dera tera COMEBACK !!!


My explanation may not be clear to you, but my requirement of Kudo to survive in my struggle against GMAT is. So spread love , spread Kudo, look at my profile. I'm so benevolent
GMATH Teacher
User avatar
P
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 15 Oct 2018, 06:45
1
Kezia9 wrote:
Please explain Number of 10s in 45! is number of 5s. I didnot understand this. Am lost with factorials.
Thanks.


Hi, Kezia9!

I have explained this (step-by-step) here:
https://gmatclub.com/forum/if-n-is-the- ... 75460.html

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Manager
Manager
avatar
B
Joined: 04 Oct 2017
Posts: 73
Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 15 Oct 2018, 09:36
ShankSouljaBoi wrote:
Hi Kezia9

I'll try to explain your query.

In order to find number of zeros or 5 in a complex no , following has to be done.

1. list down numbers ending with 5 and 0
With our case in hand , the list will be
5 10 15 20 25 30 35 40 and 45
2. each of these will have one 5 in them.
3. watchout for perfect squares ...25 will have two.

so 1+1+1+1+2+1+1+1+1= 10 5's


PS: Always remember in order to find zeros or number of 5's you don't need to calculate 2's, because they are plenty. Every alternative number will be even hence, we will just find out the limiting 5.

Posted from my mobile device


Appreciate your efforts to clarify my doubts. THANK YOU.
Manager
Manager
avatar
B
Joined: 04 Oct 2017
Posts: 73
Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 15 Oct 2018, 09:37
1
fskilnik wrote:
Kezia9 wrote:
Please explain Number of 10s in 45! is number of 5s. I didnot understand this. Am lost with factorials.
Thanks.


Hi, Kezia9!

I have explained this (step-by-step) here:
https://gmatclub.com/forum/if-n-is-the- ... 75460.html

Regards,
Fabio.



Appreciate your efforts to clarify my doubts. THANK YOU.
Manager
Manager
avatar
B
Joined: 04 Jun 2010
Posts: 87
Location: India
GMAT 1: 660 Q49 V31
GPA: 3.22
CAT Tests
Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 26 Oct 2019, 07:57
chetan2u wrote:
If x, y, and k are positive integers, is k < 10 ?

1. \(45! = x(10^k)\)
Max possible value of k is when x does not have any 10s in it..
So number of 10s in 45! is number of 5s = \(\frac{45}{5}+\frac{45}{25}=9+1=10\)
So if x does not have any 5s then k is 10, otherwise it will be <10
Insufficient

2. y=\(3\sqrt{1.25(10^k)}\)
I believe it is the cube root ..
So \(y^3=1.25*10^k=5^3*10^{k-2}\)
So minimum value of k is 2 or k can be 5,8,11 and so on
Insufficient

Combined k cannot be 10 as per statement II and \(k\leq{10}\)
So k<10
Sufficient

C

Editing the OA as it cannot be A



Sir,

should'nt x*10^k imply that 10 is at its highest power?? .........i would write 200 as 2*10^2 and not 20*10^1............or maybe I can.... :|
Intern
Intern
avatar
B
Joined: 08 May 2019
Posts: 17
Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 27 Oct 2019, 01:33
chetan2u wrote:
If x, y, and k are positive integers, is k < 10 ?

1. \(45! = x(10^k)\)
Max possible value of k is when x does not have any 10s in it..
So number of 10s in 45! is number of 5s = \(\frac{45}{5}+\frac{45}{25}=9+1=10\)
So if x does not have any 5s then k is 10, otherwise it will be <10
Insufficient

2. y=\(3\sqrt{1.25(10^k)}\)
I believe it is the cube root ..
So \(y^3=1.25*10^k=5^3*10^{k-2}\)
So minimum value of k is 2 or k can be 5,8,11 and so on
Insufficient

Combined k cannot be 10 as per statement II and \(k\leq{10}\)
So k<10
Sufficient

C

Editing the OA as it cannot be A


So if x does not have any 5s then k is 10, otherwise it will be <10

Regarding above statement in your answer, I think there is no 5 remaining from which we can carry 0's , hence statement 1 is sufficient.
Kindly correct me what i am missing.

10 can be the max zeros here.
Director
Director
avatar
P
Joined: 24 Nov 2016
Posts: 778
Location: United States
If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 06 Nov 2019, 08:10
harish1986 wrote:
If x, y, and k are positive integers, is k < 10 ?

(1) \(45! = x(10^k)\)
(2) \(y=\sqrt[3]{1.25*10^k}\)


(x,y,k) = positive integers > 0

(1) \(45! = x(10^k)\) insufic.

45! can take a max. of 45/5=9 + 45/25=1, 9+1=10 trailing zeros: 0<k≤10

(2) \(y=\sqrt[3]{1.25*10^k}\) insufic.

\(y^3=1.25*10^k…y^3=125*10^{-2}*10^k…y^3=125*10^{k-2}=integer\)
\(y=\sqrt[3]{125*10^{k-2}}=integer…\sqrt[3]{5^3*10^{k-2}}=integer…(k-2)/3=integer…\)
\(k=m(3)+2…k=[2,5,8,11,14…]\)

(1 & 2) sufic.

\(0<k≤10\)
\(k=m(3)+2…k=[2,5,8,11,14…]\)
\(combined:k=[2,5,8]<10\)

Ans. (C)
Intern
Intern
avatar
B
Joined: 03 Mar 2019
Posts: 22
GMAT 1: 640 Q45 V32
Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

Show Tags

New post 06 Nov 2019, 18:04
Really tricky and quality question. I got trapped on option A because I just applied common sense when seeing the factorial.
GMAT Club Bot
Re: If x, y, and k are positive integers, is k < 10 ?   [#permalink] 06 Nov 2019, 18:04
Display posts from previous: Sort by

If x, y, and k are positive integers, is k < 10 ?

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne