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Re: DS- Integers m and p [#permalink]
18 Jun 2007, 19:38

LM wrote:

The integers m and p are such that 2 < m < p and m is not a factor of p. If r is the remainder when p is divided by m, is r > 1?

( i ) The greatest common factor of m and p is 2.

( ii ) the least common multiple of m and p is 30 .

I got A

(1) Greatest common factor of m & p is 2. We know that both p and m must be even. Given 2<m<p, the minimum difference between the two integers is 2. It is also given that m isn't a factor of p; thus, r must be greater than 1. SUFFICIENT.

(2) p = (30^x) * integer
m = (30^x) * integer
In this case, if you plug in the intergers, you can get anything. INSUFFICIENT.

Re: DS- Integers m and p [#permalink]
18 Jun 2007, 19:40

LM wrote:

The integers m and p are such that 2 < m <p> 1?

( i ) The greatest common factor of m and p is 2.

( ii ) the least common multiple of m and p is 30 .

Should be A.

m and p are both greater than 2 (according to the info).

consider (i)

m & p both are even numbers. the smallest of even numbers that satisfy (i) and the given information are 4 & 6. clearly remainder r > 1 for all cases.

consider (ii)

m=3, p=10
or
m=5, p=6
or
m=2, p=15

clearly, r is not greater than 1 in all cases.

please feel free to correct me.

Guys, it's not a poll, votes for A,B,C,D or E won't count. Please provide your answers with explanations so that we can have a healthy discussion. _________________

for every person who doesn't try because he is
afraid of loosing , there is another person who
keeps making mistakes and succeeds..

(i) agree with explantions
(ii) The nos. can only be:
3,10 or 5,6 and not 2,15 as 2<m<p
for both 3,10 and 5,6 remainder is 1 hence less than 2. sufficient

I could be completely wrong for the second statement.

(i) agree with explantions (ii) The nos. can only be: 3,10 or 5,6 and not 2,15 as 2<m<p for both 3,10 and 5,6 remainder is 1 hence less than 2. sufficient

I could be completely wrong for the second statement.

Least common multiple isn't what you described.
Say you have 30 and 40.
30 = 2 *3 *5
40 = 2^3 *5

Stmt2: LCM is 30, p/m could be 6/5, 10/3, 15/2, 30/10...each of them has remainder <= 1
So SUFF

You meant 30/1 right????

This is what I am getting confused about - if the statement says 2<m<p then they have to positive integers greater than 2.
If the LCM is 30 then the only nos. which qualify with all these conditions have to be 5,6 and 3,10. In which case remainder is 1.

I feel it is D. Agree with statement 1. But for 2 we can only have number 5,6 and 10,3 and in either case the remainder is 1 which is not greater than 1. So D