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

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27 Jul 2017, 02:00

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.

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

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27 Jul 2017, 02:29

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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)

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

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27 Jul 2017, 02:41

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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)

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

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19 Oct 2017, 23:26

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)

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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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

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

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09 Nov 2017, 18:01

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