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few quetions which are not clear...

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few quetions which are not clear...

by ritz » Thu Apr 17, 2008 4:24 pm
can some one explain the following...
--------------------------------------
1. A list consist of several integers. is the product of all the integers positive?
(1) product of greatest & smalles interger of the list is positive,
(2) there is an even number of integers in the list.
------------------------------------
2. If N is positive integer & r is remainder when (n-1)(n+1) is divided by 24, what is the value of r
(1) 2 is not a factor of n
(2) 3 is not a factor of n
-------------------------------
Is the integer n odd?
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n
--------------------

Please give the explainations also.

thanks
Ritz
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Re: few quetions which are not clear...

by lunarpower » Thu Apr 24, 2008 2:19 am
ritz wrote:can some one explain the following...
--------------------------------------
1. A list consist of several integers. is the product of all the integers positive?
(1) product of greatest & smalles interger of the list is positive,
(2) there is an even number of integers in the list.
(1)
rephrase in terms of individual signs:
this means that the greatest and smallest number are either both positive or both negative.
...which, because those are the greatest and the smallest numbers, means that all the numbers in the list have the same sign.

this is insufficient, because, if the numbers are all negative, the sign of the product depends on how many numbers there are (if there are an odd # of numbers the product is negative; if there are an even # of numbers the product is positive).

(2)
clearly insufficient, because we know nothing about the signs of any of the numbers.

(together)
if they are all positive, then the product is positive.
if they are all negative, then the product is still positive, because the negative signs cancel in pairs.
sufficient

answer = c


------------------------------------
wrote:2. If N is positive integer & r is remainder when (n-1)(n+1) is divided by 24, what is the value of r
(1) 2 is not a factor of n
(2) 3 is not a factor of n
(1)
pick numbers
n = 3 --> (n-1)(n+1) = 8 --> remainder = 8
n = 5 --> (n-1)(n+1) = 24 --> remainder = 0
insufficient

(2)
pick numbers
n = 5 --> (n-1)(n+1) = 24 --> remainder = 0
n = 6 --> (n-1)(n+1) = 35 --> remainder = 11
insufficient

(together)
if you pick numbers (n = 1, 5, 7, 11, etc.), you'll find that the remainder is 0 every time.
sufficient

if you want the theory, here it is:
if n is not a multiple of 2, then both n-1 and n+1 are multiples of 2. furthermore, one of those two numbers (n-1 and n+1) is a multiple of 4, because every other even number is a multiple of 4. therefore, the product (n-1)(n+1) contains 2 x 4 = 8.
if n is not a multiple of 3, then one of n-1 and n+1 is a multiple of 3, because every third integer is a multiple of 3. therefore, the product (n-1)(n+1) contains 3.
therefore, the product contains 8 x 3 = 24
so the remainder is 0

-------------------------------
wrote:Is the integer n odd?
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n
---
(1)
3 --> yes
6 --> no
insufficient

(2)
for this to be true, every existing factor must yield a NEW factor upon being multiplied by 2. that means that no existing factor can contain 2, because otherwise some of the multiplications would yield factors that already exist.
the only way in which this can happen is if n contains no 2's in its prime factorization, which means it's odd.

(or you could try this for different numbers, and find that it only works for the odd ones)

sufficient

answer = b
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by goelmohit2002 » Sun Aug 23, 2009 11:24 pm
is there a algebraic way to solve this problem rather than relying on number picking ?
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by lunarpower » Mon Aug 24, 2009 12:05 am
goelmohit2002 wrote:is there a algebraic way to solve this problem rather than relying on number picking ?
which problem?

there are 3 posted on this thread.
Ron has been teaching various standardized tests for 20 years.

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by goelmohit2002 » Mon Aug 24, 2009 12:10 am
lunarpower wrote:
goelmohit2002 wrote:is there a algebraic way to solve this problem rather than relying on number picking ?
which problem?

there are 3 posted on this thread.
Is the integer n odd?
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n
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by goelmohit2002 » Mon Aug 24, 2009 12:14 am
goelmohit2002 wrote:
lunarpower wrote:
goelmohit2002 wrote:is there a algebraic way to solve this problem rather than relying on number picking ?
which problem?

there are 3 posted on this thread.


Is the integer n odd?
(1) n is divisible by 3
(2) 2n is divisible by twice as many positive integers as n
Hi Ron,

Basically as you mentioned above how to prove that...why can't an even number when multiplied by two have double the factors....

for this to be true, every existing factor must yield a NEW factor upon being multiplied by 2. that means that no existing factor can contain 2, because otherwise some of the multiplications would yield factors that already exist.
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Re: few quetions which are not clear...

by rish » Mon Aug 24, 2009 3:22 am
lunarpower wrote:
ritz wrote:can some one explain the following...
--------------------------------------
1. A list consist of several integers. is the product of all the integers positive?
(1) product of greatest & smalles interger of the list is positive,
(2) there is an even number of integers in the list.
(1)
rephrase in terms of individual signs:
this means that the greatest and smallest number are either both positive or both negative.
...which, because those are the greatest and the smallest numbers, means that all the numbers in the list have the same sign.

this is insufficient, because, if the numbers are all negative, the sign of the product depends on how many numbers there are (if there are an odd # of numbers the product is negative; if there are an even # of numbers the product is positive).

(2)
clearly insufficient, because we know nothing about the signs of any of the numbers.

(together)
if they are all positive, then the product is positive.
if they are all negative, then the product is still positive, because the negative signs cancel in pairs.
sufficient

answer = c
what if one of the integer in the list is 0 ?
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