Quarter Wit, Quarter Wisdom: Pattern Recognition

Quarter Wit, Quarter WisdomIf you are hoping for a 700+ in GMAT, you need to develop the ability to recognize patterns. GMAT does not test advanced concepts but you can certainly get advanced questions on simple concepts. For such questions, the ability to quickly observe patterns can come in quite handy. We will discuss a complicated question today which can be easily solved by observing the pattern.

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

Solution: Given a^b = b^a (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. Note that ‘a’ and ‘b’ must have the same sign. It is not possible that ‘a’ is a positive integer while ‘b’ is a negative integer or vice versa because then the side of the equation that has negative power will get flipped and will not be an integer (except if the base is 1 but in that case one side will be positive and the other will be negative)

So basically, we need to consider only positive integers (we can mirror them on to the negative side subsequently). Also, we need to consider only numbers where a < b because the equation is symmetrical in a and b. So if we get a solution of two distinct integers satisfying the equation (e.g. a = 2 and b = 4), it will give us one more solution a = 4 and b = 2.

Let us take a look at 0 separately. ‘a’ cannot be 0 since it will lead to 0^b = b^0. The left hand side is 0 and the right hand side is 1. This will not hold for any value of b.

Next, let’s consider a = 1; 1^b = b^1; again, this is not possible because left hand side is 1 while the right hand side cannot be 1 (recall that ‘b’ should not be equal to ‘a’)

Let us consider a = 2 now. We need to look for values where 2^b = b^2. Let’s put in some values of b now.

2^3 < 3^2;

2^4 = 4^2 (our first solution);

2^5 > 5^2;

2^6 > 6^2

and the difference between the left hand side and the right hand side will keep widening.

This is where pattern recognition comes in the picture – observe that the gap will keep widening.

Another approach is to look at the graphs of b^2 and 2^b.

b^2 is a quadratic so its graph will be a parabola with its vertex on (0, 0)

2^b is an exponential function. It intersects the x axis at 1.

When will these two be equal?

 

They intersect at two points x = 2 and x = 4 (the diagram is not to scale). After that the exponential function rises much faster than the quadratic function. So after intersecting at (2, 4), they will never intersect again. We ignore the point (2, 2) since a and b should not be equal.

Now consider a = 3.

3^4 > 4^3; 3^5 > 5^3 and the gap keeps widening (already, the left hand side is greater than the right hand side)

Here, the exponential function is already greater than the quadratic. So going further to the right, they will never intersect.

The pattern should be clear by now. 4^5 > 5^4, 5^6 > 6^5 and so on…

As the value of ‘a’ keeps increasing, the difference in the two terms will keep increasing.

So we have four solutions (2, 4), (4, 2), (-2, -4) and (-4, -2).

Answer (D)

Note: Generally, out of a^b and b^a, the term where the base is smaller will be the bigger term (discussing only positive integers). In very few cases will it be smaller or equal.

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…

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!

2 Responses

  1. gp says:

    Hello Karishma, For the question you’ve asked, I get answer – None. Please advice if I did it right. Thank you.

    If a and b are distinct ‘non-integers’ and a^b = b^a, how many solutions does the ordered pair (a, b) have?

    Both a and b less than 1:
    Say a = 1/5 = 0.2

    0.2^ 0.3 < 0.3^0.2
    0.2^0.4 < 0.4^0.2
    —-
    —-
    0.2^0.9 < 0.9^0.2

    Again checking for a = 0.3

    0.3^0.4 < 0.4^0.3
    0.3^0.5 < 0.5^0.3
    —-
    —-
    0.3^0.9 0.1^1.9
    —-
    —-
    1.9^0.5 > 0.5^1.9
    —-
    —-
    1.9^0.9 > 0.9^1.9

    • gp says:

      I am sorry the last part is unclear.

      For cases greater then 1: Say a = 1.9

      1.9^0.1 > 0.1^1.9
      —-
      1.9^0.5 > 0.5^1.9
      —-
      —-
      1.9^0.6 > 0.6^1.9
      —-
      1.9^0.9 > 0.9^1.9

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