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Data Sufficiency: Number Properties

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Data Sufficiency: Number Properties

by saracuse » Fri Sep 18, 2009 12:06 pm
Having issue with wording here. can someone help?

#132.

If the integer n>1, is n =2?

1) n has exactly two positive factors

2) the Difference of any two distinct positive factors of n is odd.

the Answer is B. #2 is sufficient
However .. see below

Answer:

1) n is a prime number, but n could be = to 2, 3, 5, etc. Not sufficient.

2) if n is 5.. factors are 5 and 1.. so difference is even. So could be 2. 2-1 = odd number

BUT, what about 6 -- 6-1. odd number. what about 3-2 = odd number. N and 1 are still factors.

Am i missing something in the language in Assumption #2? In #2, we stll dont know that N is a prime number. But the answer is B....
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Re: Data Sufficiency: Number Properties

by Ian Stewart » Fri Sep 18, 2009 4:57 pm
saracuse wrote:Having issue with wording here. can someone help?

#132.

If the integer n>1, is n =2?

1) n has exactly two positive factors

2) the Difference of any two distinct positive factors of n is odd.

the Answer is B. #2 is sufficient
However .. see below

Answer:

1) n is a prime number, but n could be = to 2, 3, 5, etc. Not sufficient.

2) if n is 5.. factors are 5 and 1.. so difference is even. So could be 2. 2-1 = odd number

BUT, what about 6 -- 6-1. odd number. what about 3-2 = odd number. N and 1 are still factors.

Am i missing something in the language in Assumption #2? In #2, we stll dont know that N is a prime number. But the answer is B....
There are four positive factors of 6: 1, 2, 3 and 6. From these numbers, we can make five different positive differences (we can get 1 in two different ways):

6 - 3 = 3
6 - 2 = 4
6 - 1 = 5
3 -2 = 1
3 -1 = 2
2 - 1 = 1

Some of these differences are odd, some even.

In general, if n is even, and n > 2, then n has itself and 2 as factors, and the difference n - 2 will certainly be even. If n is odd, and n > 1, then n has itself and 1 as factors, and n - 1 will be even. So the only possible value of n, given Statement 2, is n = 2.
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Can somebody help me with this

by jiteshch » Fri Sep 18, 2009 8:10 pm
DS question ::
Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

I am finding difficulties in inequalties and somebody suggest me how i can get extra pratice in these kinda sums...
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by adamsmith2009 » Sat Sep 19, 2009 5:05 pm
There are four positive factors of 6: 1, 2, 3 and 6. From these numbers, we can make five different positive differences (we can get 1 in two different ways):

6 - 3 = 3
6 - 2 = 4
6 - 1 = 5
3 -2 = 1
3 -1 = 2
2 - 1 = 1

Some of these differences are odd, some even.

In general, if n is even, and n > 2, then n has itself and 2 as factors, and the difference n - 2 will certainly be even. If n is odd, and n > 1, then n has itself and 1 as factors, and n - 1 will be even. So the only possible value of n, given Statement 2, is n = 2.
Ian,

Statement 2 says that the difference between any two distinct factors of n is odd. So taking 6 as an example, there are many ways for this to occur. E.g., 6 - 3 = 3, 6 - 1 = 5, 2 - 1 = 1. How/why did you choose 2 - 1 = 1 and know that would be sufficient to answer whether n = 2? I don't understand. Thanks.
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by Ian Stewart » Sat Sep 19, 2009 7:12 pm
adamsmith2009 wrote: Ian,

Statement 2 says that the difference between any two distinct factors of n is odd. So taking 6 as an example, there are many ways for this to occur. E.g., 6 - 3 = 3, 6 - 1 = 5, 2 - 1 = 1. How/why did you choose 2 - 1 = 1 and know that would be sufficient to answer whether n = 2? I don't understand. Thanks.
You're only demonstrating that the difference between *some* pairs of factors of 6 is odd, not that the difference between *any* (or, if it's more clear, *every*) pair of factors of 6 is odd. Since 6 is divisible by itself and by 2, there is a pair of factors with an even difference: 6 -2 = 4. The same will be true for any even number larger than 2. Similarly, any odd number - 9, for example - will have itself and 1 as a factor, and therefore two factors with an even difference: 9-1 = 8.
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