ritz wrote:can some one explain the following...
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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
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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
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wrote: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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(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
