Solve GMAT Quant questions faster by spotting patternsThe 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 problem is, of course, that you won’t have such a calculator on the Quant section. So instead, we'll need a method that will avoid computations that are best suited for calculators. 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 \(\frac{2^3}{3} → \frac{8}{3} = 2\) with a remainder of 2
- Remainder of \(\frac{2^4}{3} → \frac{16}{3} = 5\) with a remainder of 1
- Remainder of \(\frac{2^5}{3} → \frac{32}{3} = 10\) with a remainder of 2
- Remainder of \(\frac{2^6}{3} → \frac{64}{3} = 21\) 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 the remainder of \(\frac{2^{21}}{3}\).
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.
Let's try a more complex example:In a certain sequence of numbers, a1, a2, a3, ... an, the average (arithmetic mean) of the first m consecutive terms starting with a1 is m for any positive integer m. If a1=1, what is a10?
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 a1=1 and a2. Algebraically, this means \(\frac{(1+a2)}{2} = 2\)
Solving, we get a2 = 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 a1= 1, a2 = 3, and a3: \(\frac{(1+ 3 + a3)}{3}=3\)
Solving, we get a3 = 5
Repeating the process, we find that for \(m = 4, \frac{(1 + 3 + 5 + a4) }{ 3} = 4\)
Solving, we get a4 = 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 a10 = 19
By using the
Spot the Pattern technique, you can now solve complex problems by working out simpler cases and analyzing the resulting trend.
Originally posted at Economist GMAT Blog _________________
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