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dimitri92
What is the probability of creating a three digit number with exactly two consecutive primes as digits?

a) 3/100
b) 9/250
c) 9/125
d) 21/250
e) 51/500

E,

For numbers between 100- 199

For 1 st digit we just have 1 option

For second digit we have 3 options (2,3,5) and for all these 3 options we have one option ( i.e for 2 we have 3, for 3 we have 5 and for 5 we have 7)

Hence for numbers between 100-199 we have in total 3 options

Now for numbers between 200-299
For 1st digit again we have only one option but now for 2nd we have either 1 option i.e 3 or we have 8 options i.e when 3rd digit is 3 ( so when 3rd digit is 3 we have 8 options for 2nd digit.)
Hence in total 1* 2* 8 ( 2 because we 2 places for placing 3 ) = 16

Now, similarly for numbers starting from 300-399 we will have same 32 options and same goes for other numbers

So now using POE we can clearly see the total numbers will be more than 21 for sure,

Hence E
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question? consecutive prime is 5 - 7? I suppose that makes sense I just interpreted it as consecutive, prime which the only one would be 2,3 - thanks for the clarity

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How come the demoniator is 500 in the suggested answer??

Totally lost. :(
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Hi Bunuel, Can you please help with the above question?
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dimitri92
What is the probability of creating a three digit number with exactly two consecutive primes as digits?

a) 3/100
b) 9/250
c) 9/125
d) 21/250
e) 51/500

seekmba
Hi Bunuel, Can you please help with the above question?

This is not a good question, the wording is very ambiguous. So don't worry about it. What does "with exactly two consecutive primes as digits" even mean?

Though we can guess that the answer is A (well the only answer which makes sense). There are 900 3-digit numbers, which means that 900 is the total # of outcomes and the answer should be (# of favorable outcomes)/(total # of outcomes)=(# of favorable outcomes)/(900), the only answer choice that could be written as such fraction is A - 3/100=27/900 (no other fraction has denominator factor of 900).
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Thanks. that helps.
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Hey guys,

While this question itself isn't written precisely enough to be an actual GMAT question - each of which costs, on average, around $8,000 to create using a team of PhDs if I remember correctly, so don't feel badly if your creations don't quite work perfectly - I think it's a great exercise in thinking within a GMAT mindframe:

1) Bunuel's point was my first thought...there are 900 total 3-digit numbers, so the denominator would have to be a factor of 900, and only A works. If you're thinking that way, you're in great shape - the less math you have to do to guarantee the answer, the better!

2) There are a few different interpretations of the question, but I firmly believe that thinking of all of them is a great way to ensure that you don't make assumptions when you see actual questions, so please think away!

I took "exactly two consecutive primes as digits" to mean that the 3 digits are "exactly" created by two consecutive primes. That way, you'd only have 7 and 11 as possible primes, as you'd need two consecutive primes to total 3 digits - one 2-digit (11) and one 1-digit (7) number.

You could also look at it as meaning that each prime accounts for one digit, in which case your possibilities are 2, 3, 5, and 7 (the one-digit primes). In that case you'd need to consider that 2 only gives you one option (3) for its consecutive digit, but 3 has two (2 and 5) because it has neighboring consecutive primes on either side.

There are other interpretations you could make, too (which is why this question would need to be reworked to make it more precise), but part of the GMAT study process is understanding the different variations that the test could take on each type of problem, so I fully embrace the idea of creating problems like this and then looking at how they could be tweaked to take different forms. Right now we're all "thinking like the test maker", which is a great way to get some insight into the way these questions will try to trick you.
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