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

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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

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It’s easy to establish this inequality equation:

2^100=2*512^11 > 500^11

Thus, 11 is a possible answer.

But, we still need to rule out E=13.

Firstly, we have:

500^2=(5^6)*(2^4) > (4^6)*(2^4)=2^16>2*(1.024)^11=2*(512/500)^11

Then we have:

500^2>2*(512/500)^11

=> (500^11)*(500^2) > (500^11)*(2*(512/500)^11)

=>500^13>2*512^11=2^100

So, 13 is not possible, and we can rule out E.

Finally, D is our answer.

Be Cheers!

I thought of the aforementioned problem by following way………

(500)ab ˂ 2100

→ (125 × 4)ab ˂ 2100

→ (125 × 22)ab ˂ 2100

Here assume 125 = 27- (27 = 128 and 27- means less than 128 but close to 125)

→ (27- × 22)ab ˂ 2100

→ (29-)ab ˂ 2100

As a & b are both integer value then for greatest possible value of a; the integer value of b needs to be minimum. So, b=1

→ (29-)a ˂ 2100

Hence, (9-)a = 99 (to make the inequality true)

→ a = 11+ (here +ve sign means slightly larger)

Answer is : D

Answer : (D) 11

Explanation:

Taking log on both sides we get

=> a*b*log(500) a*b a*b < 11.15 (using simple log tables of 2 &5)

As a and b are both positive and we need to find the max. value of 'a', we use the assumption that b=1.

Hence, a=11

Another simpler approach involving a bit of estimation also gives the correct answer and is really quick in application but not recommended as it may give incorrect results in some case. But it is still useful to give some idea of the answer.

Apporach 2:

We know 2^9 =512

So we can say,

500^1<2^9

And by same reasoning

500^(1*ab) < 2^(9*11.11)

Hence we can compare the exponents to get

a*b < 11.11

Again taking b=1 to get the maximum value of a we have : a=11.

Some text seems to be missing in my reply. I am re-typing the portion requiring correction

Explanation:

Taking log on both sides we get

=> ab*log(500) ab ab < 11.15 (using simple log tables of 2 &5)

As a and b are both positive and we need to find the max. value of 'a', we use the assumption that b=1.

Hence, a=11

Hi, I tried this question, wasn’t very confident with it but somehow I managed to form an inequality equation out of it. This equation is 5^(3ab)<2^(100-2ab). Now at this point, I have no idea what to do but I progressed by thinking that this inequality means that 100-2ab should obviously be much larger than 3ab. And then solving for this inequality I reached a<100/5b and hence a<20/b which means b should be least to make a larger and that can be by setting b=1, hence a should be less than 20, since the greater possible value in these options which is also less than 20 is 13. So I think 13 is the answer, ie E.

I think the answer will be D =11.

if we present 2^100 as (2^9)11= 512^11.

500^ab, and ab we can present as 11*1.

so we have 500^1*11<512^11

Ans: D

The answer is A.

To maximize the value of a, we should minimize the value of b, setting it equal to 1. So 500^a < 2^100.

We can factor the left side a bit: (5^3 * 2^2)^a 5^(3a) * 2^(2a) < 2^100.

Backsolving and starting with the greatest answer choice, (E) (a=13) doesn't work:

5^39 * 2^26 < 2^100

5^39 * 2^26 < 2^74 * 2^26

5^39 2^2, so 5^39 > (2^2)^39 –> 5^39 > 2^78 > 2^74. So the inequality is contradictory.

The next greatest answer choice, (D) (a=11), does work:

5^33 * 2^22 < 2^100

5^33 * 2^22 < 2^78 * 2^22

5^33 < 2^78

Looking for a way to compare the two terms, we find that 5^3 < 2^7.

So (5^3)^11 < (2^7)^11

5^33 < 2^77 < 2^78

The inequality holds, so the correct answer choice is D.

Oops, sorry for all the typos—I was eager to get my answer in!

The first line should obviously read “The answer is D” (not A) to agree with the work and the last line.

There should be a right-facing arrow after the first term of the work in the line beginning “We can factor…”

In the line that begins “5^39 2^2, so…” it should read “5 > 2^2, so…”

I think that’s it. I wish this site had an Edit feature for comments. Thanks, Chris, for your sinister twisters!