If n is a an integer between 10 and 99 is n < 80?
(1) The sum of the two digits of n is a prime number.
(2) Each of the two digits of n is a prime number.
GMAT Forum
5 postsPage 1 of 1
- Misha
I got the answer right on this test so will try to explain:
First thing I did was to reframe the question:
10<n<99, is n<80?
For (2) Each of the two digits of n is a prime number:
I listed some #'s that only have prime #'s for each of the two digits, such as, 22, 33, 55, 77, etc. From this I deduced that the largest number n could be is 77, thus n<80. SUFF
For (1) The sum of the two digits of n is a prime number.
This was a little more tedious, I listed all possible combo's in my own short way:
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
By visualizing this I could use connecting arrows to see what the possibilities are. I could make a combo of a # below 80, such as, if n = 74 -> then 7+4 = 11, a prime number, and thus n<80 (and still maintaining 10<n<99).
Also, I could make a combo of a # greater than 80, such as, if n = 83 -> 8+ 3 = 11, a prime number, and thus n>80. Since we have n<80 and n>80, INSUFF.
Thus answer is B.
First thing I did was to reframe the question:
10<n<99, is n<80?
For (2) Each of the two digits of n is a prime number:
I listed some #'s that only have prime #'s for each of the two digits, such as, 22, 33, 55, 77, etc. From this I deduced that the largest number n could be is 77, thus n<80. SUFF
For (1) The sum of the two digits of n is a prime number.
This was a little more tedious, I listed all possible combo's in my own short way:
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
By visualizing this I could use connecting arrows to see what the possibilities are. I could make a combo of a # below 80, such as, if n = 74 -> then 7+4 = 11, a prime number, and thus n<80 (and still maintaining 10<n<99).
Also, I could make a combo of a # greater than 80, such as, if n = 83 -> 8+ 3 = 11, a prime number, and thus n>80. Since we have n<80 and n>80, INSUFF.
Thus answer is B.
- Khalid
Misha Wrote:I got the answer right on this test so will try to explain:
First thing I did was to reframe the question:
10<n<99, is n<80?
For (2) Each of the two digits of n is a prime number:
I listed some #'s that only have prime #'s for each of the two digits, such as, 22, 33, 55, 77, etc. From this I deduced that the largest number n could be is 77, thus n<80. SUFF
For (1) The sum of the two digits of n is a prime number.
This was a little more tedious, I listed all possible combo's in my own short way:
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
By visualizing this I could use connecting arrows to see what the possibilities are. I could make a combo of a # below 80, such as, if n = 74 -> then 7+4 = 11, a prime number, and thus n<80 (and still maintaining 10<n<99).
Also, I could make a combo of a # greater than 80, such as, if n = 83 -> 8+ 3 = 11, a prime number, and thus n>80. Since we have n<80 and n>80, INSUFF.
Thus answer is B.
I think the answer is E
For statement 2 you could have 32 and 3+2 = 5 which is prime
- RonPurewal
- Students
- Posts: 19744
- Joined: Tue Aug 14, 2007 8:23 am
remember the prime directive of data sufficiency number plugging. here it is:
LAW #1 OF D.S. NUMBER PLUGGING:
TRY FOR "INSUFFICIENT".
this means that you should plug in numbers in a deliberate attempt to create at least one "yes" possibility and at least one "no" possibility. furthermore, once you've generated a "yes" possibility, you should apply all your energies to producing a "no", and vice versa.
ignore this law at your peril.
note that this applies to NUMBER PLUGGING ONLY. when you're using theory, sometimes the most efficient approaches will be directed at "sufficient" instead. but with number plugging, the only way to get "sufficient" is to go through EVERY possibility, whereas "insufficient" demands nothing more than one "yes" and one "no".
--
reply one:
well, sure.
but 32 is less than 80. that's a "yes" possibility.
to get "insufficient" for statement 2, you'd also have to be able to generate a number that's MORE than 80, which would be a "no" to the problem. see law #1 above.
there is no such number, since the prime tens digit must be 7 or less.
--
reply two:
if you follow law #1 above, this doesn't have to be "tedious".
instead, just realize that you need "insufficient". this means that you need to find a possibility at least 80 as well as another possibility under 80.
to get the former, you should shoot for relatively big numbers.
let's take, say, a sum of 13, which is prime, and somewhat big.
this could be 9+4 --> 94 (more than 80), or 4+9 --> 49 (less than 80).
done.
no need to make such a huge matrix of two columns of 1 through 9, especially because you aren't combining them in any obviously linear sort of way. instead, just follow the law above: "go for insufficient".
LAW #1 OF D.S. NUMBER PLUGGING:
TRY FOR "INSUFFICIENT".
this means that you should plug in numbers in a deliberate attempt to create at least one "yes" possibility and at least one "no" possibility. furthermore, once you've generated a "yes" possibility, you should apply all your energies to producing a "no", and vice versa.
ignore this law at your peril.
note that this applies to NUMBER PLUGGING ONLY. when you're using theory, sometimes the most efficient approaches will be directed at "sufficient" instead. but with number plugging, the only way to get "sufficient" is to go through EVERY possibility, whereas "insufficient" demands nothing more than one "yes" and one "no".
--
reply one:
I think the answer is E
For statement 2 you could have 32 and 3+2 = 5 which is prime
well, sure.
but 32 is less than 80. that's a "yes" possibility.
to get "insufficient" for statement 2, you'd also have to be able to generate a number that's MORE than 80, which would be a "no" to the problem. see law #1 above.
there is no such number, since the prime tens digit must be 7 or less.
--
reply two:
For (1) The sum of the two digits of n is a prime number.
This was a little more tedious
if you follow law #1 above, this doesn't have to be "tedious".
instead, just realize that you need "insufficient". this means that you need to find a possibility at least 80 as well as another possibility under 80.
to get the former, you should shoot for relatively big numbers.
let's take, say, a sum of 13, which is prime, and somewhat big.
this could be 9+4 --> 94 (more than 80), or 4+9 --> 49 (less than 80).
done.
no need to make such a huge matrix of two columns of 1 through 9, especially because you aren't combining them in any obviously linear sort of way. instead, just follow the law above: "go for insufficient".
5 posts Page 1 of 1