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Number Prop DS

Expert replies
Source: — Data Sufficiency |

by pesfunk » Thu Jan 27, 2011 3:26 am
yellowho wrote:Is the prime number q equal to 29 ?
(1) q -1 has 6 factors
2) q+1 has 2 and 3 as prime factors
I'll go with C.

! or 2 independently can lead to diff numbers other than 29. example:

1 - It can be 19 for which 18 has 6 factors - 18 = 2*3*3 = (1+1)(2+1) = 6
2 - It can be 17 which will again server option 2.

If we use both of them together, the answer will lead to 29
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by Anurag@Gurome » Thu Jan 27, 2011 4:06 am
yellowho wrote:Is the prime number q equal to 29 ?
(1) (q - 1) has 6 factors
(2) (q + 1) has 2 and 3 as prime factors
Note that, 29 satisfy both the statements.
Hence, if we can find at least one example that satisfy any of the statement, then the statement will be insufficient.

Statement 1: There can be infinite number of integers with 6 factors. Hence, infinite possible values of q are there.

Not sufficient

Statement 2: There can be infinite number of integers with 2 and 3 as prime factors, i.e. divisible by 6. Hence, infinite possible values of q are there.

Not sufficient

1 & 2 Together: Both statements together also doesn't limit the number of possible values of q. This is because both of them is also not an unique property for a particular integer. For example, q can be equal to 29 as well as 53.

Not sufficient

The correct answer is E.
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by pesfunk » Thu Jan 27, 2011 4:13 am
HI Anurag, is there a quick way to get 53 ?
Anurag@Gurome wrote:
yellowho wrote:Is the prime number q equal to 29 ?
(1) (q - 1) has 6 factors
(2) (q + 1) has 2 and 3 as prime factors
Note that, 29 satisfy both the statements.
Hence, if we can find at least one example that satisfy any of the statement, then the statement will be insufficient.

Statement 1: There can be infinite number of integers with 6 factors. Hence, infinite possible values of q are there.

Not sufficient

Statement 2: There can be infinite number of integers with 2 and 3 as prime factors, i.e. divisible by 6. Hence, infinite possible values of q are there.

Not sufficient

1 & 2 Together: Both statements together also doesn't limit the number of possible values of q. This is because both of them is also not an unique property for a particular integer. For example, q can be equal to 29 as well as 53.

Not sufficient

The correct answer is E.
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by Anurag@Gurome » Thu Jan 27, 2011 5:14 am
pesfunk wrote:HI Anurag, is there a quick way to get 53?
There is some logic I can give you which will show you the way, but it is not definitely quick. In my opinion, this particular problem is not asking for a solution. It is testing whether you have the sense to identify a statement as unique property for an integer or not.

Also knowing the fact that there is no particular formula or relation by which we can uniquely point out prime numbers, is sufficient to answer the problem.

Now, here is the logic I was talking about.
Disclaimer: I am going to make some huge step jumps so if you do not follow do not bother. This is not really necessary.

Note that for statement 1 to be true, the prime factorization of (q - 1) must be of the form a²b, where a and b are prime numbers. And for statement 2 to be true, (q + 1), i.e (a²b + 2) must be divisible by 6, i.e. both by 2 and 3. Now it can be shown that a will be equal to 2 and b will be greater than 3. Therefore, q will be of the form (a²b + 1) = (4b + 1), where b is any prime number greater than 3.

Again it can be shown that, q will of the form (24n + 5) for some positive integer n. (Note that this doesn't mean any integer of the form (24n + 5) will be a possible value of q.)

Now you have to look for prime integers of the form (24n + 5).
For, n = 1 => q = 29
For, n = 2 => q = 53
...

Again, this doesn't mean all integer of the form (24n + 5) will be a possible value of q. For example, n = 3 doesn't give a prime.
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by GMATGuruNY » Thu Jan 27, 2011 6:43 am
yellowho wrote:Is the prime number q equal to 29 ?
(1) q -1 has 6 factors
2) q+1 has 2 and 3 as prime factors
Here is an efficient approach, if you know the following method for counting the number of positive factors of an integer:
1) Prime-factorize the integer
2) Add 1 to each exponent
3) Multiply


Since 200 = 2^3 * 5^2, it has (3+1)*(2+1) = 12 factors.
Statement 1:
If q-1 = 2 * 3^2 = 18 -- which yields 2*3 = 6 factors -- then q=19.
If q-1 = 2^2 * 7 = 28 -- which yields 3*2 = 6 factors -- then q=29.
Insufficient.

Statement 2:
If q+1 = 2*3 = 6 -- so it has 2 and 3 as prime factors -- then q=5.
If q+1 = 2*3*5 = 30 -- so it has 2 and 3 as prime factors -- then q=29.
Insufficient.

Statements 1 and 2 together:
We have already seen that q=29 satisfies both statements.

Out of every 3 consecutive integers, exactly 1 will be a multiple of 3. Thus, if q+1 is a multiple of 3, then q and q-1 cannot be multiples of 3. So to satisfy statement 2, q-1 cannot have 3 as a prime factor.
We also know that q-1 must be even, since prime number q must be odd.
Thus, q-1 cannot have 3 as a prime factor and must have 2 as a prime factor:

If q-1 = 2^2 * 11 = 44 -- which yields 3*2 = 6 factors -- then q=45, which is not prime.
If q-1 = 2^2 * 13 = 52 -- which yields 3*2 = 6 factors -- then q=53 and q+1=54, satisfying all the conditions in the problem.
Since q=29 and q=53 both work, insufficient.

The correct answer is E.
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by tomada » Thu Jan 27, 2011 2:10 pm
What are the 6 factors of the number 28?
I'm really old, but I'll never be too old to become more educated.
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by GMATGuruNY » Thu Jan 27, 2011 2:36 pm
tomada wrote:What are the 6 factors of the number 28?
28 = 2^2 * 7, yielding 3*2 = 6 factors.

The 6 factors are 1, 2, 4, 7, 14, 28.

Please feel free to read the explanation that I posted in the following thread about counting factors:

https://www.beatthegmat.com/factors-of-36-2-t68679.html
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by clock60 » Thu Jan 27, 2011 3:56 pm
hi guys, i have some ideas about above problem, and it would be great that somebody verify my reasoning, at first i read the problem the following way
if q is prime does it equal to 29
(1) q-1 has 6 factors, it means that q-1=a^2*b^1 where a,b are primes (different)
q-1=2^2*3, q-1=4*3, q=13, no
q-1=2^2*7, q-1=28, q=29, yes
(2) q+1 has 2 and 3 as prime factors
i read somewhere that any primes that are greater than 2 or 3 can be written in the form
q=6k+1, or q=6k-1, where q is prime >than 2 or 3 and k is an integer (+ve)
so st 2 only says that
q+1=6k, or q=6k-1, it gives us that q is definitely greater that 2 or 3 and nothing else
both gives us that q is >2,and 3 and that q -1 looks like a^2*b^1 with a,b primes but with no fixed values of a and b
do you think this approach is valid or it is just oversimplification
thanks
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by Anurag@Gurome » Sat Jan 29, 2011 3:28 am
clock60 wrote:....

do you think this approach is valid or it is just oversimplification
thanks
Yes, you're reasoning is good.
We can generalize it more and get a general expression for such primes as I did in one of the above posts.

But, none of this is required (even picking any number), if you know that there is no unique way to represent prime numbers unless they point to a particular prime. Therefore, any bunch of properties can point to more than one prime unless they limit the range.
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by clock60 » Sat Jan 29, 2011 11:51 am
hi Anurag
thank you for reply and validation!
appreciate it
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