If anyone is still (silently) confused, here's an approach:
Let's call our set {a1, a2, a3, ..., aN}, where a1 is our first integer, a2 our second, and aN our nth.
Since the integers are distinct, let's also say that a1 < a2 < a3 < ... < aN. (This is just a way of arranging the integers from least to greatest, as the order of the set hasn't otherwise been stipulated.)
(1) tells us that
(a2 + a3 + a4 + ... + aN)/(n-1) - 1 = (a1 + a2 + a3 + ... + a(N-1))/(n-1)
Multiplying both sides by (n-1), we get
(a2 + a3 + ... + aN) - (n-1) = (a1 + a2 + ... + a(N-1))
Subtracting all the terms from a2 to a(N-1) from both sides, we get
aN - (n-1) = a1
or
aN - a1 = (n-1)
This tells us that the distance between a1 and aN is (n-1); in other words, that aN is (n-1) integers to the right of a1 on the number line. Since every term in between a1 and aN is distinct, and there are are (n-2) terms between 1 and n, this means we have ALL the integers from 1 to n; SUFFICIENT.
(2) This gives us the system of equations
aN - a1 < n
aN - a2 < n
aN - a3 < n
...
aN - a(N-1) < n
But the first one is the only one we need. Since a1 < a2 < ... < aN, we know that if the integers ARE consecutive, the distance between a1 and aN will be (n - 1). Since n is an integer, and our distance is LESS THAN n, it must be at most (n - 1). Further, if our distance is LESS THAN (n - 1), all the integers can't be distinct -- we must have more than one of at least one of them -- but that would violate our condition. So the distance from a1 to aN on the number line must be (n - 1), and as in S1 this is enough to show that we have all the integers from a1 to aN; SUFFICIENT.
Not the easiest question in the world, and one on which you're probably advised to guess & check, experiment, wing it, and/or live with being 70% sure of your answer on test day.
Let's call our set {a1, a2, a3, ..., aN}, where a1 is our first integer, a2 our second, and aN our nth.
Since the integers are distinct, let's also say that a1 < a2 < a3 < ... < aN. (This is just a way of arranging the integers from least to greatest, as the order of the set hasn't otherwise been stipulated.)
(1) tells us that
(a2 + a3 + a4 + ... + aN)/(n-1) - 1 = (a1 + a2 + a3 + ... + a(N-1))/(n-1)
Multiplying both sides by (n-1), we get
(a2 + a3 + ... + aN) - (n-1) = (a1 + a2 + ... + a(N-1))
Subtracting all the terms from a2 to a(N-1) from both sides, we get
aN - (n-1) = a1
or
aN - a1 = (n-1)
This tells us that the distance between a1 and aN is (n-1); in other words, that aN is (n-1) integers to the right of a1 on the number line. Since every term in between a1 and aN is distinct, and there are are (n-2) terms between 1 and n, this means we have ALL the integers from 1 to n; SUFFICIENT.
(2) This gives us the system of equations
aN - a1 < n
aN - a2 < n
aN - a3 < n
...
aN - a(N-1) < n
But the first one is the only one we need. Since a1 < a2 < ... < aN, we know that if the integers ARE consecutive, the distance between a1 and aN will be (n - 1). Since n is an integer, and our distance is LESS THAN n, it must be at most (n - 1). Further, if our distance is LESS THAN (n - 1), all the integers can't be distinct -- we must have more than one of at least one of them -- but that would violate our condition. So the distance from a1 to aN on the number line must be (n - 1), and as in S1 this is enough to show that we have all the integers from a1 to aN; SUFFICIENT.
Not the easiest question in the world, and one on which you're probably advised to guess & check, experiment, wing it, and/or live with being 70% sure of your answer on test day.
sidceg wrote:Hi! This was from one of the MGMAT CATs. Can some one please explain?
A list contains n distinct integers. Are all n integers consecutive?
(1) The average (arithmetic mean) of the list with the lowest number removed is 1 more than the average (arithmetic mean) of the list with the highest number removed.
(2) The positive difference between any two numbers in the list is always less than n.
OA is D
