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# If a and b are distinct integers and a^b = b^a, how many

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If a and b are distinct integers and a^b = b^a, how many [#permalink]  23 Oct 2010, 05:56
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A 700-level question for the GMAT committed souls studying over the weekend.

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

Hint: Use logic, not Math. Look for patterns.
[Reveal] Spoiler: OA

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews  Kaplan GMAT Prep Discount Codes Knewton GMAT Discount Codes Manhattan GMAT Discount Codes Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6062 Location: Pune, India Followers: 1597 Kudos [?]: 8953 [6] , given: 195 Re: Try this one - 700 Level, Number Properties [#permalink] 23 Oct 2010, 14:31 6 This post received KUDOS Expert's post 4 This post was BOOKMARKED 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... (A lot verbose, actually!) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Try this one - 700 Level, Number Properties [#permalink]  24 Oct 2010, 15:56
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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
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Re: Try this one - 700 Level, Number Properties [#permalink]  23 Oct 2010, 07:14
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My take is:D
(a,b) pairs possible are: (2,4) (4,2) (-2,-4), (-4,-2)

What is the mathematical way rather than number substituation?

Cheers!
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Re: Try this one - 700 Level, Number Properties [#permalink]  23 Oct 2010, 07:16
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D: 4

(2,4), (4,2), (-2,-4) and (-4,-2)
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Re: Try this one - 700 Level, Number Properties [#permalink]  23 Oct 2010, 07:22
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Oh yes !! I didn't take -ve values.

The answer should be 4 -D
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Re: Try this one - 700 Level, Number Properties [#permalink]  24 Oct 2010, 15:34
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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!
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Last edited by VeritasPrepKarishma on 24 Oct 2010, 15:54, edited 1 time in total. Senior Manager Joined: 07 Sep 2010 Posts: 335 Followers: 4 Kudos [?]: 352 [1] , given: 136 If a and b are distinct integers and a^b = b^a [#permalink] 26 Apr 2013, 22:03 1 This post received KUDOS 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. Thanks _________________ +1 Kudos me, Help me unlocking GMAT Club Tests VP Status: Far, far away! Joined: 02 Sep 2012 Posts: 1123 Location: Italy Concentration: Finance, Entrepreneurship GPA: 3.8 Followers: 157 Kudos [?]: 1571 [1] , given: 219 Re: If a and b are distinct integers and a^b = b^a [#permalink] 27 Apr 2013, 00:31 1 This post received KUDOS 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. Kant , Critique of Pure Reason Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b] Manager Joined: 07 Oct 2010 Posts: 180 Followers: 5 Kudos [?]: 130 [0], given: 10 Re: Try this one - 700 Level, Number Properties [#permalink] 23 Oct 2010, 07:02 My answer is C There are only two pairs possible i.e. 2^4 = 4^2 and -2^-4 = -4^-2 you will get it with trial and error method. Current Student Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2797 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Followers: 204 Kudos [?]: 1285 [0], given: 235 Re: Try this one - 700 Level, Number Properties [#permalink] 23 Oct 2010, 07:07 VeritasPrepKarishma wrote: A 700-level question for the GMAT committed souls studying over the weekend. Q. 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 Hint: Use logic, not Math. Look for patterns. (2,4) and (4,2) a=1 b=1 and a=2 b=2 also has the solution. But we need distinct. Now when we will increase a>4, b^a increase more rapidly than a^b. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : 670-to-710-a-long-journey-without-destination-still-happy-141642.html Senior Manager Joined: 05 Jul 2010 Posts: 359 Followers: 15 Kudos [?]: 46 [0], given: 17 Re: Try this one - 700 Level, Number Properties [#permalink] 23 Oct 2010, 07:43 [strike]Answer is E. Infinite 2^2^2=2^4=4^2 2^2^2^2=2^8=8^2 2^2^2^2^2=4^8=8^4=2^16=16^2 ...... so-on Ofcourse also negatives.[/strike] I was crazy when I wrote that :p My bad. D is correct. Posted from my mobile device Last edited by abhicoolmax on 23 Oct 2010, 12:05, edited 2 times in total. Manager Joined: 11 Jul 2010 Posts: 227 Followers: 1 Kudos [?]: 62 [0], given: 20 Re: Try this one - 700 Level, Number Properties [#permalink] 23 Oct 2010, 08:45 2 to power 8 is not equal to 8 to the power 2... its D.... not infinite Manager Joined: 16 Jun 2010 Posts: 188 Followers: 2 Kudos [?]: 50 [0], given: 5 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. Current Student Joined: 15 Jul 2010 Posts: 256 GMAT 1: 750 Q49 V42 Followers: 8 Kudos [?]: 122 [0], given: 65 Re: Try this one - 700 Level, Number Properties [#permalink] 23 Oct 2010, 20:33 D An interesting question that made me think a little. I found 4 right away, but spent another 2 minutes trying other numbers. I should have recognized the pattern developing, but it somehow never crossed my mind. _________________ Consider KUDOS if my post was helpful. My Debrief: 750-q49v42-105591.html#p825487 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6062 Location: Pune, India Followers: 1597 Kudos [?]: 8953 [0], given: 195 Re: Try this one - 700 Level, Number Properties [#permalink] 24 Oct 2010, 15:53 Expert's post 1 This post was BOOKMARKED We are considering only positive integers where a < b. It we prove that $$a^{a+1} > (a+1)^a$$, I think the rest will follow. LHS:$$a^{a+1} = a^a.a^1 = a^a + a^a + a^a ......a times$$ Here the right hand side of the equation has a terms. RHS: Using Binomial, $$(a+1)^a = a^a + aC1.a^{a-1} + aC2.a^{a-2} + aC3.a^{a-3} + ... +1$$ $$(a+1)^a = a^a + a^a + a(a-1)/2. a^{a-2} + a(a-1)(a-2)/6. a^{a-3} + ...+1$$ 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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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.
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Re: If a and b are distinct integers and a^b = b^a [#permalink]  27 Apr 2013, 00:26
i could only get one solution - 2^4 equals 4^2

Someone please explain the solution to this..
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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..
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Re: If a and b are distinct integers and a^b = b^a [#permalink]  27 Apr 2013, 02:26
Hi Zarrolou,

Would you mind giving a shot on this one with an algebraic/graphical way..

Regards,
H
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Re: If a and b are distinct integers and a^b = b^a   [#permalink] 27 Apr 2013, 02:26

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