Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 44654

If a and b are positive integers, what is the greatest possible value [#permalink]
Show Tags
27 Jul 2017, 01:36
1
This post received KUDOS
Expert's post
12
This post was BOOKMARKED
Question Stats:
46% (02:08) correct 54% (01:45) wrong based on 204 sessions
HideShow timer Statistics



Manager
Joined: 02 Nov 2015
Posts: 172

Re: If a and b are positive integers, what is the greatest possible value [#permalink]
Show Tags
27 Jul 2017, 03:00
Imo B. Solving we get 2^(2*a*b) X 5^(3*a*b) < 2^100. To get maximum value of A we need to minimise B. Thus the least value that can b assigned to B is 1. So after putting different values I found that the maximum value of A to be 5 because if we put A as 10 then the left hand side becomes greater than right hand side. So I will go for B. # Bunuel if I have missed something, kindly correct me. Thanks in advance. Sent from my Lenovo TAB S850LC using GMAT Club Forum mobile app



BSchool Forum Moderator
Joined: 26 Feb 2016
Posts: 2442
Location: India
GPA: 3.12

If a and b are positive integers, what is the greatest possible value [#permalink]
Show Tags
27 Jul 2017, 03:29
5
This post received KUDOS
3
This post was BOOKMARKED
Given data : \(500^{ab}<2^{100}\) We have been asked to find out the greatest possible value for a. 500 when prime factorized gives \(2^2 * 5^3\) Therefore, \(500^{ab} = 2^{2ab} * 5^{3ab}\) For the maximum possible value of a, we need to have the lowest value for b. The lowest positive value for b = 1. The expression now becomes \(2^{2a} * 5^{3a} < 2^{100}\) Now going by answer options, If we go by answer option E, a=13 : \(2^{26} * 5^{39} < 2^{100} => 5^{39} < \frac{2^{100}}{2^{26}}\) \(5^{39} > 2^{74} => (5^2)^{19} *5 > (2^4)^{18} *4\) (Extrapolating 5^2 > 2^4) For answer option D, a=11 : \(2^{22} * 5^{33} < 2^{100} => 5^{33} < \frac{2^{100}}{2^{22}}\) \(5^{33} < 2^{78} => (5^2)^{16} * 5 < (2^4)^{19} * 4\) (Extrapolating 5^2 > 2^4) Hence, a=11 is the highest value of a possible such that the expression holds true(Option D)
_________________
Stay hungry, Stay foolish
20172018 MBA Deadlines
Class of 2020: Rotman Thread  Schulich Thread Class of 2019: Sauder Thread



Manager
Joined: 02 Nov 2015
Posts: 172

Re: If a and b are positive integers, what is the greatest possible value [#permalink]
Show Tags
27 Jul 2017, 03:41
1
This post received KUDOS
pushpitkc wrote: Given data : \(500^{ab}<2^{100}\) We have been asked to find out the greatest possible value for a. 500 when prime factorized gives \(2^2 * 5^3\) Therefore, \(500^{ab} = 2^{2ab} * 5^{3ab}\)
For the maximum possible value of a, we need to have the lowest value for b. The lowest positive value for b = 1. The expression now becomes \(2^{2a} * 5^{3a} < 2^{100}\)
Now going by answer options, If we go by answer option E, a=13 : \(2^{26} * 5^{39} < 2^{100} => 5^{39} < \frac{2^{100}}{2^{26}}\) \(5^{39} > 2^{74} => (5^2)^{19} *5 > (2^4)^{18} *4\) (Extrapolating 5^2 > 2^4)
For answer option D, a=11 : \(2^{22} * 5^{33} < 2^{100} => 5^{33} < \frac{2^{100}}{2^{22}}\) \(5^{33} < 2^{88} => (5^2)^{16} * 5 < (2^4)^{22}\) (Extrapolating 5^2 > 2^4)
Hence, a=11 is the highest value of a possible such that the expression holds true(Option D) Kudos !! Great solution. Now I understand my mistake. Thanks for such good illustration. Sent from my Lenovo TAB S850LC using GMAT Club Forum mobile app



Director
Joined: 18 Aug 2016
Posts: 631

Re: If a and b are positive integers, what is the greatest possible value [#permalink]
Show Tags
27 Jul 2017, 05:01
5
This post received KUDOS
2
This post was BOOKMARKED
Bunuel wrote: \(500^{ab}<2^{100}\)
If a and b are positive integers, what is the greatest possible value of a?
(A) 2 (B) 5 (C) 10 (D) 11 (E) 13 500^ab < 512^100/9 since ab = ~11.11 and minimum value of b will be 1 a can take highest value of 11 D
_________________
We must try to achieve the best within us
Thanks Luckisnoexcuse



Manager
Joined: 19 Aug 2016
Posts: 73

Re: If a and b are positive integers, what is the greatest possible value [#permalink]
Show Tags
20 Oct 2017, 00:26
pushpitkc wrote: Given data : \(500^{ab}<2^{100}\) We have been asked to find out the greatest possible value for a. 500 when prime factorized gives \(2^2 * 5^3\) Therefore, \(500^{ab} = 2^{2ab} * 5^{3ab}\)
For the maximum possible value of a, we need to have the lowest value for b. The lowest positive value for b = 1. The expression now becomes \(2^{2a} * 5^{3a} < 2^{100}\)
Now going by answer options, If we go by answer option E, a=13 : \(2^{26} * 5^{39} < 2^{100} => 5^{39} < \frac{2^{100}}{2^{26}}\) \(5^{39} > 2^{74} => (5^2)^{19} *5 > (2^4)^{18} *4\) (Extrapolating 5^2 > 2^4)
For answer option D, a=11 : \(2^{22} * 5^{33} < 2^{100} => 5^{33} < \frac{2^{100}}{2^{22}}\) \(5^{33} < 2^{78} => (5^2)^{16} * 5 < (2^4)^{19} * 4\) (Extrapolating 5^2 > 2^4)
Hence, a=11 is the highest value of a possible such that the expression holds true(Option D) what do u mean by extrapolating? pls explain..thanks



Manager
Joined: 19 Aug 2016
Posts: 73

Re: If a and b are positive integers, what is the greatest possible value [#permalink]
Show Tags
20 Oct 2017, 00:28
Luckisnoexcuse wrote: Bunuel wrote: \(500^{ab}<2^{100}\)
If a and b are positive integers, what is the greatest possible value of a?
(A) 2 (B) 5 (C) 10 (D) 11 (E) 13 500^ab < 512^100/9 since ab = ~11.11 and minimum value of b will be 1 a can take highest value of 11 D Howdid u get 512^100/9? Pls explain thanks



Director
Joined: 18 Aug 2016
Posts: 631

Re: If a and b are positive integers, what is the greatest possible value [#permalink]
Show Tags
20 Oct 2017, 01:05
zanaik89 wrote: Luckisnoexcuse wrote: Bunuel wrote: \(500^{ab}<2^{100}\)
If a and b are positive integers, what is the greatest possible value of a?
(A) 2 (B) 5 (C) 10 (D) 11 (E) 13 500^ab < 512^100/9 since ab = ~11.11 and minimum value of b will be 1 a can take highest value of 11 D Howdid u get 512^100/9? Pls explain thanks 2^9 = 512
_________________
We must try to achieve the best within us
Thanks Luckisnoexcuse



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 2464
Location: United States (CA)

Re: If a and b are positive integers, what is the greatest possible value [#permalink]
Show Tags
24 Oct 2017, 06:39
Bunuel wrote: \(500^{ab}<2^{100}\)
If a and b are positive integers, what is the greatest possible value of a?
(A) 2 (B) 5 (C) 10 (D) 11 (E) 13 Since we have to maximize the value of a, we must minimize the value of b. Since b is a positive integer, the smallest value of b is 1. Thus, letting b = 1, we have: 500^a < 2^100 Since 2^9 = 512, which is slightly larger than 500, we can say: 500^a < (2^9)^(100/9) 500^a < 512^(100/9) Since 500 < 512, we see that if a ≤ 100/9 = 11.11, then 500^a < 2^100. Therefore, so far we can say the maximum value of a is 11. However, can a be 13 and 500^a < 2^100 still be true? Let’s prove (or disprove) it: If a = 13, then 500^13 = (2^2 x 5^3)^13 = 2^26 x 5^39, so the question becomes: Is 2^26 x 5^39 < 2^100? Is 5^39 < 2^74? The answer is no, since 5^39 > 4^39 = (2^2)^39 = 2^78. Since 5^39 is greater than 2^78, so it must be greater (not less) than 2^74. So, we can see that a can’t be 13. Thus, the largest value of a must be 11. Answer: D
_________________
Scott WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Intern
Joined: 15 Jan 2016
Posts: 29

Re: If a and b are positive integers, what is the greatest possible value [#permalink]
Show Tags
24 Oct 2017, 16:09
500^a < (2^9)^(100/9)
500^a < 512^(100/9)
I get how you got to 512, but can you please explain why you divided 100 by 9 specifically? I’m a bit confused on why used 9..
Thanks again
Posted from my mobile device



BSchool Forum Moderator
Joined: 26 Feb 2016
Posts: 2442
Location: India
GPA: 3.12

If a and b are positive integers, what is the greatest possible value [#permalink]
Show Tags
08 Nov 2017, 12:25
1
This post was BOOKMARKED
syedazeem3 wrote: 500^a < (2^9)^(100/9)
500^a < 512^(100/9)
I get how you got to 512, but can you please explain why you divided 100 by 9 specifically? I’m a bit confused on why used 9..
Thanks again
Posted from my mobile device syedazeem3This is a very important formula that deals with exponents: \((a^m)^n = a^{m*n}\) Going by that, \(2^{100} = (2^9)^{\frac{100}{9}}\) because \(100 = 9*\frac{100}{9}\) Apologies for the delay in clearing your confusion. Hope that helps you!
_________________
Stay hungry, Stay foolish
20172018 MBA Deadlines
Class of 2020: Rotman Thread  Schulich Thread Class of 2019: Sauder Thread



Manager
Joined: 18 May 2016
Posts: 159
Location: India
WE: Marketing (Other)

Re: If a and b are positive integers, what is the greatest possible value [#permalink]
Show Tags
08 Nov 2017, 13:33
Here is how I approach this, experts please let me know if this isn't the right approach:
500^ab = (5^3 * 2^2)^ab = 5^3ab * 2^2ab ~ (2^2)^3ab * 2^2ab  Taking 5 ~ 4 = 2^2 = 2^6ab *2^2ab = 2^8ab
Now, 2^8ab < 2^100
that is : 8ab < 100 or, ab < 12.5
taking b =1 as its least value, greatest value of a will be one nearest to 12.5 ie 11, since we have approximated 5 to 4 and hence a little loss of value of a gets compensated.
And, eliminating the options will also give us 11 as the greatest possible value of a. Option D
Hope this helps!



Manager
Joined: 24 Jun 2017
Posts: 122

Re: If a and b are positive integers, what is the greatest possible value [#permalink]
Show Tags
09 Nov 2017, 19:01
500^ab<2^100 (2^2ab)*(5^3ab)<2^100 taking natural log ln((2^2ab)*(5^3ab))<ln(2^100) 2ab*ln(2) + 3ab*ln(5)<100*ln(2) > this is a standard log property, actually 2 of them, log (5*5) can be written as log5 +log5 and log(5^5) can be written as 5*log5 (5 times log of 5)
ln 2 ~ 0.7 ln 5 ~ 1.6 b=1 (when a is maximum)
2a*0.7+3a*1.6<70 6.2a<70 only one answer matched 11




Re: If a and b are positive integers, what is the greatest possible value
[#permalink]
09 Nov 2017, 19:01






