Can You Spot the Pattern?

by on May 15th, 2012

High-school mathematics courses often leave the impression that math involves nothing more than the recall of formulas. Nothing could be farther from the truth in real mathematics and on the real GMAT. In fact, the most advanced GMAT problems require you to think on your feet. Often, you’ll need to break down a complex problem into simpler components and apply some clever insight to spot an emerging pattern. Take this problem for instance:

What is the remainder when dividing 2^21 by 3?

The most straightforward approach is to first compute 2^21, then divide that value by 3 to find the remainder. Any decent scientific calculator will be able to handle such a trivial computation in a few milliseconds—the only problem, of course, is that you won’t have such a calculator on the Quant section. So instead, we’ll need a method that will avoid such brute force calculations.

Rather than tackling this problem head-on, let’s try some simplified calculations to see if we can spot an emerging pattern. Let’s calculate the remainder when different powers of 2 are divided by 3:

  • Remainder of 2^3/38/3 is two with a remainder of 2.
  • Remainder of 2^4/316/3 is five with a remainder of 1.
  • Remainder of 2^5/332/3 is ten with a remainder of 2.
  • Remainder of 2^6/364/3 is twenty-one with a remainder of 1.

Can you spot the pattern? Whenever the exponent is odd, the remainder is 2; whenever the exponent is even, the remainder is 1. Returning to our original problem, we want to find 2^21 ÷ 3: since the exponent is odd, the remainder is 2.

Spotting the pattern is a powerful technique that will work for a wide variety of complicated problems, especially those with tricky wording. Patterns often emerge when you start with the simplest case possible, then work your way up:

In a certain sequence of numbers, a_1,a_2,a_3,...,a_n, the average (arithmetic mean) of the first m consecutive terms starting with a_1 is m, for any positive integer m. If a_1=1, what is a_10?

The meaning behind this problem has been intentionally obfuscated, but the pattern becomes obvious once you start plugging-in a few numbers. Let’s work out the first few terms of the sequence:

We’ll first assume that m = 2. Using this plug-in, we find that the average (arithmetic mean) of the first m = 2 consecutive terms becomes the average of a_1 = 1 and a_2. Algebraically, this means,

{1+a_2}/2 = 2

Solving, we get a_2 = 3.

Next, let’s solve for the case m = 3. The average of the first m = 3 terms of the sequence is now the average of a_1 = 1, a_2 = 3, and a_3:

{1+3+a_3}/3 = 3

Solving, we get a_3 = 5.

Repeating the process, we find that for m = 4,

{1+3+5+a_4}/3 = 4

a_4 = 7

By now, the pattern should be apparent. Each term in the sequence belongs to the set of consecutive odd integers: 1, 3, 5, 7. Following the pattern, we find that a_10=19.


By using the Spot the Pattern technique, you can now solve complex problems by working out simpler cases and analyzing the resulting trend. To make sure you grasp the material, try your hand at this difficult problem:

Amit and Ian paint a wall in alternating shifts, first Amit paints alone, then Ian paints alone, then Amit paints alone, etc. During his shifts, Amit paints 1/2 of the remaining unpainted area of the wall, while Ian, paints 1/3 of the remaining unpainted area of the wall during his shifts. If Amit goes first, what fraction of the wall’s area will remain unpainted after Amit has completed his 4th shift?

(Hint: The pattern may be more visible if you don’t multiply out all the fractions)

9 comments

  • The other way for finding remainder when 2^21 is divided with 3:
    When 2 is divided with 3, the reminder is 2, which can be written as 3-1. here remainder -1 is same as remainder 2.
    =>reminder of (2^21)/3 is same as (-1)^21 = -1, which is nothing but 2
    If we have even power for example, 2^20, the remainder will be (-1)^20 = 1

  • To find remainder after 4th shift, multiply 1 (job) by the fraction remaining after completion of said shift

    So, A=1/2 and I=2/3 (2/3 will remain after Ian completes 1/3 of remaining job).

    Therefore,

    1/2*2/3*1/2*2/3*1/2*2/3*1/2=

    1/3*3*3*2= 1/54

    • Dear Daren,
      I am bogged down in the following part:

      I supposed that the total unpainted area= 1

      During the first shift Amit paints = 1*1/2 = 1/2 of the wall. That means the remaining unpainted area is= 1-1/2= 1/2.
      As it is said in the question Ian paints, during his first shift, 1/3 of the unpainted area, so Ian should have painted= 1/2*1/3= 1/6 of the wall.

      But how did you get 2/3 for the first shift of Iann.
      Would you break your answer down in details?

      Thanks.

  • Hi Robin,

    Your math is correct. However, your approach adds unnecessary steps.

    First shift: Amit paints 1/2 of remaining wall (1), leaving (1-1/2=1/2) 1/2 of the job. So, the remainder AFTER Amit paints will always be 1/2 of the remaining job prior to his shift.

    Second Shift: Ian paints 1/3 of remaining job. You correctly identified that 1/3 of the 1/2 remaining wall is 1/6. However, you now need to subtract 1/6 from 1/2. Further, you would have to complete this calculation (and likely adjustments to find common denominator) after each of Ian's shifts.

    The tricky part is re-framing the question: what will remain after each shift? Amit's shift will always be equal to the remainder after his shift (1/2 + 1/2 = 1)

    However, Ian's completion of 1/3 of the remaining job will LEAVE 2/3 REMAINING.

    Thus, we may multiple by the 1-rate to find the remainder of the job at the end of each shift.

    A= 1/2
    I= 2/3

    This allows us to quickly cancel out numerous terms

    1*2*1*2*1*2*1/2*3*2*3*2*3*2

    2*2*2*1/2*2*2*2*3*3*3=

    1/2*3*3*3

    • Thanks a lot.

    • Thanks a lot.. this was a really good question

  • Awesome solution. Thanks a ton!

  • Good one..!!

  • Calculate the amount of the UNPAINTED walls that Amit and Ian leave behind once they finish with their shift. Then multiply times the amounts of shifts they each work.

    Amit leaves 1/2 of the wall unpainted each time and works 4 shifts.
    Ian leaves 2/3 of the wall unpainted each time and works 3 shifts.

    Therefore,
    Amit leaves (1/2)^4 = 1/16 unpainted
    Ian leaves (2/3)^3 = 8/27 unpainted

    TOGETHER, they leave the following portion of the wall unpainted:
    (1/16)*(8/27) =
    Simplify by canceling out the 8 (1/2)*(1/27) = 1/54

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