If x, y, and k are positive integers, is k

Question

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

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

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

chetan2u · Accepted Answer

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

asthagupta · Answer

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.

fskilnik · Answer

x,y,k\,\, \ge 1\,\,\,{ m{ints}}\,\,\,\left( * ight) k\,\,\mathop < \limits^? \,\,10 \left( 1 ight)\,\,45! = x \cdot {10^k}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,\,\,\,k\,\,\, \le \,\,\,\left\lfloor {{{45} \over 5}} ight floor + \left\lfloor {{{45} \over {25}}} ight floor \, = 10\,\,\,\,\,\left( {**} ight)\,\, \left( {**} ight) See my explanation (and notation) here: https://gmatclub.com/forum/if-n-is-the- ... 75460.html \,\left\{ \matrix{ \,{ m{Take}}\,\,\left( {k{\kern 1pt} \,;\,x} ight) = \left( {10\,;\,\,{{45!} \over {{{10}^{10}}}}} ight)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{ m{NO}}} ight angle \,\,\,\,\,\,\,\,\,\,\,\left( {x = {{45!} \over {{{10}^{10}}}}\,\, \ge 1\,\,{\mathop{ m int}} \,\,\,{ m{by}}\,\,\left( {**} ight)} ight)\,\, \hfill \cr \,{ m{Take}}\,\,\left( {k\,;\,x} ight) = \left( {1\,;\,\,{{45!} \over {10}}} ight)\,\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{ m{YES}}} ight angle \,\,\,\,\,\, \hfill \cr} ight. \left( 2 ight)\,\,\,y = oot {3\,} \of {{5 \over 4}\left( {{{10}^k}} ight)} \,\,\,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,\,\,\,{5 \over 4}\left( {{2^k} \cdot {5^k}} ight) = {2^{k - 2}} \cdot {5^{k + 1}}\,\,\,\,{ m{positive}}\,\,{ m{perfect}}\,\,{ m{cube}}\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \matrix{ \,k - 2 = { m{mult}}\,\,{ m{of}}\,\,{ m{3}} \hfill \cr k + 1 = \,\,{ m{mult}}\,\,{ m{of}}\,\,3 \hfill \cr} ight.\,\,\,\,\,\,\,\,\left( {k \ge 2} ight)\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,k \ge 2\,\,\,\,\,{ m{divided}}\,\,{ m{by}}\,\,3\,\,\,{ m{has}}\,\,{ m{remainder}}\,\,2\, \left\{ \matrix{ \,{ m{Take}}\,\,k = 2\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\left\langle {{ m{YES}}} ight angle \,\, \hfill \cr \,{ m{Take}}\,\,k = 2 + 3 \cdot 3\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{ m{NO}}} ight angle \,\,\,\,\,\, \hfill \cr} ight. \left( {1 + 2} ight)\,\,\,\,\,\left\{ \matrix{ \,k \le 10\,\,\,\,{ m{by}}\,\,\,\,\left( 1 ight) \cap \left( {**} ight) \hfill \cr \,k e 10\,\,\,{ m{by}}\,\,\left( 2 ight) \hfill \cr} ight.\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\langle {{ m{YES}}} ight angle This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.

fskilnik · Answer

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.

Mo2men · Answer

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

Thanks

chetan2u · Answer

Yes, you are correct.
It is bad formatting, which I have corrected now.

Kezia9 · Answer

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}[

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.

ShankSouljaBoi · Answer

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

fskilnik · Answer

Hi, Kezia9 ! I have explained this (step-by-step) here: https://gmatclub.com/forum/if-n-is-the- ... 75460.html Regards, Fabio.

Kezia9 · Answer

Appreciate your efforts to clarify my doubts. THANK YOU.

Kezia9 · Answer

Appreciate your efforts to clarify my doubts. THANK YOU.

avikroy · Answer

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....

Harsh2111s · Answer

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.

exc4libur · Answer

(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 {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 {125*10^{k-2}}=integer…\sqrt {5^3*10^{k-2}}=integer…(k-2)/3=integer… 
 k=m(3)+2…k=

(1 & 2)    sufic.

0<k≤10 
 k=m(3)+2…k=  
 combined:k= <10

Ans. (C)

churuand · Answer

Really tricky and quality question. I got trapped on option A because I just applied common sense when seeing the factorial.

sthahvi · Answer

hello  chetan2u  
please explain the first step, how did you derive that, I tried to understand from the link mentioned by some user below but unable to

chetan2u · Answer

45! =1*2*3...44*45
When we multiply 2 with a 5, we get a zero in the end. 
45! Will have more numbers of 2 as compared to numbers of 5. So, number of 5 will give you number of zeroes in the end. 
Now, # of 5s in 45! = 1*..*5*...*10*....*15*....*20*.....*25*....*30*.....*35*....*40*.....45
So one 5 from each of 5,10,15,20,30,35,40 and 45, and two 5s from 25. Total =10.

The max value of k is 10, when x is not a multiple of 5. If x is a multiple of 5, one 5 will move from k to x, and k will become less than 10.

Bambi2021 · Answer

(1) Two variables and only one equation, not possible.

(2) y^3 = 1,25*10^k, and this is on the same form as in (1), which together with (1) gives us:

y^3= 45!
x=1,25

45! = 1,25*10^k

and we can solve for k.

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

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