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Positive integers
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didieravoaka
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Hi didieravoaka,
This prompt is an example of a 'Symbolism' question - the prompt 'makes up' a math symbol, tells you what it means and asks you to perform a calculation using it. Based on the information in the prompt, we're told that....
32~ is the remainder when (32-1)! is divided by 32.
Now, there's no way that the GMAT would expect you to calculate the value of 31!, so we have to think in terms of what 31! actually is.
Here's an example that's a bit easier to deal with:
4! = (4)(3)(2)(1) = 24
What numbers divide EVENLY into 24?
1, 2, 3, 4, 6, 8, 12 and 24
You can clearly see why 1, 2, 3 and 4 divide in - they're in the 'chain' of numbers that are multiplied together.
6 divides in because (2)(3) = 6 - and you can see the (2) and the (3) in the 'chain'
Similarly, 8, 12 and 24 are also 'combinations' of the numbers in the 'chain', so they divide evenly in too.
31! has LOTS of numbers in it, so it's evenly divisibly by LOTS of different integers. If you were to write out 31!, you would see a (2) and a (16). This means that (2)(16) = 32 divides evenly into 31!, so there will be a remainder of 0.
Final Answer: A
GMAT assassins aren't born, they're made,
Rich
This prompt is an example of a 'Symbolism' question - the prompt 'makes up' a math symbol, tells you what it means and asks you to perform a calculation using it. Based on the information in the prompt, we're told that....
32~ is the remainder when (32-1)! is divided by 32.
Now, there's no way that the GMAT would expect you to calculate the value of 31!, so we have to think in terms of what 31! actually is.
Here's an example that's a bit easier to deal with:
4! = (4)(3)(2)(1) = 24
What numbers divide EVENLY into 24?
1, 2, 3, 4, 6, 8, 12 and 24
You can clearly see why 1, 2, 3 and 4 divide in - they're in the 'chain' of numbers that are multiplied together.
6 divides in because (2)(3) = 6 - and you can see the (2) and the (3) in the 'chain'
Similarly, 8, 12 and 24 are also 'combinations' of the numbers in the 'chain', so they divide evenly in too.
31! has LOTS of numbers in it, so it's evenly divisibly by LOTS of different integers. If you were to write out 31!, you would see a (2) and a (16). This means that (2)(16) = 32 divides evenly into 31!, so there will be a remainder of 0.
Final Answer: A
GMAT assassins aren't born, they're made,
Rich
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DavidG@VeritasPrep
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It might help to see it in slightly more detail. We want the remainder for when 31! is divided by 32. 31! is the product of every integer between 1 and 31, and 32 = 16*2, sojain2016 wrote:Hi All ,
Is there any other approach for this question.
Many thanks in advance.
SJ
31!/32 = (31*30*29*...17*16....*3*2*1)/(16 * 2)
The terms in red will cancel each other out. Because we can eliminate the denominator entirely, we're dealing with an integer value, and the remainder must be 0.
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DavidG@VeritasPrep
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Or think of it this way. 32 = 2^5. Clearly there are more than five 2's in 31!, as every even number will contain at least one. So we know that in 31!/(2^5), the denominator will be completely eliminated, leaving us with no remainder.jain2016 wrote:Hi All ,
Is there any other approach for this question.
Many thanks in advance.
SJ
31*30*29*28*27*26*25*24*23*22..../(2^5)
(And really, we don't have to go that far, as many of the even numbers will contain multiple 2's.)
