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Solve GMAT Quant questions faster by spotting patterns
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 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:
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
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 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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