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Re: data sufficiency question II [#permalink]
21 Dec 2009, 07:12

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A.. SI tells us that the tens digit is a factor of units digit so possible values of units digit are 4,6,8,9.....all would be even nos except no ending with digit 9, which will be 39... so all r composite.... SII.. no can be any of 21,22,23.....29..ie mix of composite and prime

Re: data sufficiency question II [#permalink]
21 Dec 2009, 07:53

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Expert's post

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An integer grater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite?

Given: \(n>20\) --> two digit integer can be written as follows: \(n=10b+a>20\) --> \(2\leq{b}\leq{9}\), \(0\leq{a}\leq{9}\).

(1) The tens digit of n is a factor of the units digit of n. This statement implies that \(a=kb\), (\(0\leq{k}\leq{4}\)) --> \(n=10b+a=10b+kb=b(10+k)\) --> as \(b\geq{2}\), \(n\) will always be composite and factor of \(b\). Sufficient

(2) The tens digit of n is 2. This statement implies that \(b=2\), but \(n\) can be for instance composite 25 or prime 29. Not sufficient.

Answer: A.

chetan2u wrote:

A.. SI tells us that the tens digit is a factor of units digit so possible values of units digit are 4,6,8,9.....all would be even nos except no ending with digit 9, which will be 39... so all r composite.... SII.. no can be any of 21,22,23.....29..ie mix of composite and prime

Actually units digit of \(n\) can be any digit but 1, meaning that 0, 2, 3, 5, 7 too. And for 9: 99 also fits: eg . 30, 40, 50, 60, 70, 80, 90, 22, 33, 55, 77, 99. _________________

Re: An integer greater than 1 that is not prime is called compos [#permalink]
19 Dec 2010, 15:20

Expert's post

ajit257 wrote:

Q13: An integer greater than 1 that is not prime is called composite. If the two-digit integer n is greater than 20, is n composite?

(1) The tens digit of n is a factor of the units digit of n. (2) The tens digit of n is 2.

Not sure about the ans

The moment I hear two digit prime number, I think of numbers having 1/3/7/9 as unit's digit. A two digit number ending in 0/2/4/5/6/8 cannot be prime since they are either even or a multiple of 5 (if it is a single digit, 5 is prime)

Stmnt 1: Ten's digit is a factor of the unit's digit. It is easy to see that there are many such composite numbers e.g. 24, 26 etc.

I will try to find if there are any such prime numbers: The only factor of 1 is 1; of 3 are 1 and 3; of 7 are 1 and 7; of 9 are 1, 3 and 9. 11, 13, 17 and 19 are not allowed since the number should be greater than 20. 33, 77, 39 and 99 are all composite. So there is no such prime number. All such number will be composite so n is composite. Hence sufficient.

Stmnt 2: The tens digit of n is 2. 21 is not prime while 23 is. Hence not sufficient.

An integer greater than 1 that is not [#permalink]
09 Jan 2013, 04:29

The tricky thing about this question was for me a very simple thing, i.e. pace. Because I was in a hurry I did not wrote down n > 20 and therefore 11 was in my range. So my advice is be calm. Because 11 is a prime where the tens digit is a factor of the units digit. This is correct isn't it?

Re: An integer grater than 1 that is not prime is called [#permalink]
01 Jun 2014, 19:00

Hi Bunuel,

I got the answer by other way. However, i would like to understand your explanation on (1) where you stated that a=kb, (0<=k<=4) --> n=10b+a=10b+kb=b(10+k) --> as b>=2, n will always be composite and factor of b.

Can you please explain in detail this a=kb part? why (0<=k<=4)?

Re: An integer grater than 1 that is not prime is called [#permalink]
02 Jun 2014, 00:33

Expert's post

mrvora wrote:

Hi Bunuel,

I got the answer by other way. However, i would like to understand your explanation on (1) where you stated that a=kb, (0<=k<=4) --> n=10b+a=10b+kb=b(10+k) --> as b>=2, n will always be composite and factor of b.

Can you please explain in detail this a=kb part? why (0<=k<=4)?

Regards, Mitesh

(1) says that the tens digit of n, which is b, is a factor of the units digit of n, which is a. So, b is a factor of a --> \(a=kb\), for some integer k.

As for \(0\leq{k}\leq{4}\): the least value of b is 2, thus if k is 5 or more, then a becomes 10 or more, which is not possible since we know that a is a digit (\(0\leq{k}\leq{4}\)).

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