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

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If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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Updated on: 14 Oct 2018, 18:10
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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
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Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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
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Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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.
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Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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.
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Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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.
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Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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
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Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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.
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Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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.
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If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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
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If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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.
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Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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.
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Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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.
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Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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....
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Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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.
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If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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)
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Re: If x, y, and k are positive integers, is k < 10 ?  [#permalink]

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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.
Re: If x, y, and k are positive integers, is k < 10 ?   [#permalink] 06 Nov 2019, 18:04
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