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Re: Try this one - 700 Level, Number Properties [#permalink]
23 Oct 2010, 14:31

6

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Expert's post

4

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The answer is indeed D (4 solutions). Good work everyone!

Now for the explanation. I tend to get a little verbose... Bear with me.

Given \(a^b = b^a\) and a and b are distinct integers. First thing that comes to mind is that if we didn't need distinct integers then the answer would have simply been infinite since \(1^1 = 1^1, 2^2 = 2^2, 3^3 = 3^3\) and so on... Next, integers include positive and negative numbers. If a result is true for positive a and b, it will also be true for negative a and b and vice versa. The reason for this is that both a and b will be either even or both will be odd because \((Even)^{Odd}\)cannot be equal to \((Odd)^{Even}\) Also, it is not possible that a is positive while b is negative or vice versa because then one side of the equation will have negative power and the other side will have positive power.

So basically, I need to consider positive integers (I can mirror it on to the negative integers subsequently). Also, I will consider only numbers where a < b because the equation is symmetrical in a and b. So if I get a solution of two distinct such integers (e.g. 2 and 4), it will give me two solutions since a can take 2 or 4 which implies that b will take 4 or 2.

Let me take a look at 0. It cannot be 'a' since it will lead to \(0^b = b^0\), not possible. Next, a cannot be 1 either since it will lead to \(1^b = b^1\), not possible. Let us consider a = 2. \(2^3 < 3^2\); \(2^4 = 4^2\)(Got my first solution); \(2^5 > 5^2\); \(2^6 > 6^2\) and the difference keeps on widening. This is where pattern recognition comes in the picture. The gap will keep widening. Now I will consider a = 3. \(3^4 > 4^3\) (first term itself is greater); \(3^5 > 5^3\) and the gap keeps widening. I can try a couple more values but the pattern should be clear by now. \(4^5 > 5^4, 5^6 > 6^5\) and so on... and as the values keep increasing, the difference in the two terms will keep increasing...

Note: Generally, out of \(a^b\) and \(b^a\), the term where the base is smaller will be the bigger term (I am considering only positive integers here.). In very few cases will it be smaller or equal (only in case of a = 1, \(2^3\) and \(2^4\)).

So I have four solutions (2, 4), (4, 2), (-2, -4) and (-4, -2). This question is pattern recognition based.

Now, we know that if the question did not have the word 'distinct', the answer would have been different, but what if the question did not have the word 'integer'? Would it make a difference? - Something to think about...

Re: Try this one - 700 Level, Number Properties [#permalink]
23 Oct 2010, 15:54

Any suggestions on how to be sure that only (2,4) (4,2) and ofcourse the negatives of these solution. I came up with this solution in around 2 mins but spent the next 2 mins confirming that there cannot be other solutions like (3,9) with where a is a square of b or (3,27). _________________

Please give me kudos, if you like the above post. Thanks.

Re: Try this one - 700 Level, Number Properties [#permalink]
24 Oct 2010, 15:34

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Expert's post

scheol79 - Yes, pattern recognition is a beneficial skill to have on GMAT. It could save you precious time. nravi549 & devashish - This question tests your logic and pattern recognition skills. Perhaps tests your exposure to number properties. But still, if all you curious people out there are wondering whether we can prove it mathematically, we sure can! But I must warn you, it involves Math beyond the scope of GMAT and hence is irrelevant. Nevertheless, wait for a few mins, I will post it! _________________

\((a+1)^a = a^a + a^a + a(a-1)/2. a^{a-2} + a(a-1)(a-2)/6. a^{a-3} + ...+1\) Note that here the right hand side of the equation has (a+1) terms. The last term of 1 is extra.

Now, if we compare \(a^a + a^a + a^a ......a times\) and \(a^a + a^a + a(a-1)/2. a^{a-2} + a(a-1)(a-2)/6. a^{a-3} + ...+1\) term by term, first two terms are the same but every subsequent term of the second expression is less than the corresponding term of the first expression.

Then why doesn't it work for 2? That is because the comparison in case of 2 looks like this: \(2^2 + 2^2\)is compared with \(2^2 + 2^2 + 1\) The first two terms, as we said before, are anyway equal but the second expression has an extra term of 1. Hence the second expression is greater.

In case of 3 and greater integers, \(3^3 + 3^3 + 3^3\) is compared with \(3^3 + 3^3 + 3.2/2.3^1 + 1\) The extra term of 1 cannot make up for the deficit of the third term. Hence, as the numbers keep increasing, the gap will keep getting wider! Note: This Math is beyond the scope of GMAT. _________________

Re: Try this one - 700 Level, Number Properties [#permalink]
24 Oct 2010, 15:56

6

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I think I can prove this mathematically without using binomial theorem or any higher level math. Let mw know what you think :

Mathematical Proof

\(a^b=b^a\) without loss of generality, I assume a>b. therefore, \(a=b^{a/b}\)

Also it is easy to see any solution (a,b) has to be either both positive integers or both negative (Otherwise one side will become fractional with abs value less than 1, and the other side will have an abs value>1)

Assume for the moment, a,b>0 Since a and b are integers, a>b, (a/b)>1. For \(b^{a/b}\) to be an integer, a/b must be an integer. Thus in all cases, a has to be a multiple of b whenever this equation has a solution. Let a=kb, where k is an integer, such that k cannot be 1, since in that case a=b which is not allowed Hence, \(kb=b^k\) \(b^k-bk=0\) \(b(b^{k-1}-k)=0\) Either b=0 OR \(b=k^{\frac{1}{k-1}}\) b=0 implies a=kb=0. Hence it is an invalid solution.

Thus only solutions are given by \(b=k^{\frac{1}{k-1}}\) Clearly for k=2, this is an integer \(b=2^1=2\) & \(a=bk=4\) For any higher k, this is not an integer : k=3, b=3^(1/2) k=4, b=4^(1/3) k=5, b=5^(1/4) .. and so on ... which always gives irrational values of b which are not permitted

Hence only solution (positive) is b=2,a=4

Now notice that if (a,b) is a solution then (-a,-b) is also always a solution as long as both a and b are either odd or both are even which is the case here.

Hence b=-2,a=-4 is also a solution

By symmetry the pairs b=4,a=2 & b=-4,a=-2 are also solutions

Hence overall there are four pairs of solutions. And we have proved there can be no more feasible solutions _________________

Re: Try this one - 700 Level, Number Properties [#permalink]
24 Oct 2010, 17:51

Expert's post

That is an absolutely solid proof, in my opinion. Kudos for the very good work shrouded1. Anyone else would like to take a shot at proving it mathematically in a different way? Try it!

Note: The discussion of the proof here is for intellectual stimulation only. Please do not get lost in the mathematics if it doesn't interest you. The takeaway from the question is pattern recognition. _________________

If a and b are distinct integers and a^b = b^a [#permalink]
26 Apr 2013, 22:03

1

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Question: If a and b are distinct integers and a^b = b^a, how many solutions does the ordered pair (a, b) have?

(A) None (B) 1 (C) 2 (D) 4 (E) Infinite

Hi Experts, I'm stumped by this question, and have seen quite a few questions that usually test understanding of exponential questions. From the first impression of this question, I found that this question will test positive,negative, 0, even, odd every possibility. Can someone please post a detailed solution of this.

Re: If a and b are distinct integers and a^b = b^a [#permalink]
27 Apr 2013, 00:31

1

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karishmatandon wrote:

i could only get one solution - 2^4 + 4^2

Someone please explain the solution to this..

We are looking for values of a, b such that \(a^b=b^a\), those values have to be different (a=1, b=1 will not count for example)

One combination as you say is (2,4), but because the order does matter this values give us 2 pairs (2,4) (4,2) \(2^4=4^2\) or \(16=16\)

The other pair of values that respect that condition is -2,-4 so other 2 possible solutions here : (-2,-4) and (-4,-2) \((-2)^-^4=(-4)^-^2\) or \(\frac{1}{16}=\frac{1}{16}\)

So 4 possible solutions: D

Hope it's clear, let me know _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: If a and b are distinct integers and a^b = b^a [#permalink]
27 Apr 2013, 00:36

Zarrolou wrote:

karishmatandon wrote:

i could only get one solution - 2^4 + 4^2

Someone please explain the solution to this..

We are looking for values of a, b such that \(a^b=b^a\), those values have to be different (a=1, b=1 will not count for example)

One combination as you say is (2,4), but because the order does matter this values give us 2 pairs (2,4) (4,2) \(2^4=4^2\) or \(16=16\)

The other pair of values that respect that condition is -2,-4 so other 2 possible solutions here : (-2,-4) and (-4,-2) \((-2)^-^4=(-4)^-^2\) or \(\frac{1}{16}=\frac{1}{16}\)

So 4 possible solutions: D

Hope it's clear, let me know

Yeah..it is clear.. Thnks _________________

Consider giving +1 Kudo when my post helps you. Also, Good Questions deserve Kudos..!

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