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If a and b are positive integers, what is the greatest possible value

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If a and b are positive integers, what is the greatest possible value [#permalink]

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New post 27 Jul 2017, 01:36
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Re: If a and b are positive integers, what is the greatest possible value [#permalink]

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New post 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.

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If a and b are positive integers, what is the greatest possible value [#permalink]

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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)
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Re: If a and b are positive integers, what is the greatest possible value [#permalink]

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New post 27 Jul 2017, 03:41
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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.

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

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

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New post 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

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

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New post 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

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

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New post 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
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Re: If a and b are positive integers, what is the greatest possible value [#permalink]

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New post 24 Oct 2017, 06:39
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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
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Re: If a and b are positive integers, what is the greatest possible value [#permalink]

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New post 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

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If a and b are positive integers, what is the greatest possible value [#permalink]

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New post 08 Nov 2017, 12:25
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

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syedazeem3

This 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!
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Re: If a and b are positive integers, what is the greatest possible value [#permalink]

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New post 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!

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

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New post 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

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