pappueshwar wrote:If the sum of four consecutive positive integers a three digit multiple of 50, the mean of the these integers must be one of x possible values, where x=

(A) 7
(B) 8
(C) 9
(D) 10
(E) more than 10

OA C

The rule offered above need to be qualified.
The sum of n consecutive integers will always be a multiple of n IF N IS ODD.
But IF N IS EVEN -- as it is here -- then the sum of the n consecutive integers WILL NOT BE a multiple of n.

Consider the 4 consecutive integers {1,2,3,4}.
Sum = 1+2+3+4 = 10, which is NOT a multiple of 4.

In the problem at hand:
Since the sum of 4 consecutive integers CANNOT be a multiple of 4, any multiple of 50 that is a multiple of 4 CANNOT be the sum of 4 consecutive integers.
Thus, NO multiple of 100 can be the sum of 4 consecutive integers.

Quite the opposite is true:
Any multiple of 50 that is NOT a multiple of 100 will be the sum of 4 consecutive integers.
The reason is that the MEAN -- which is also the MEDIAN -- will always be HALFWAY between two consecutive integers.

If the sum of the 4 integers = 150:
Average = median = 150/4 = 37.5, which is halfway between 37 and 38.
Thus, the 4 integers are 36, 37, 38, and 39.
Sum = 36+37+38+39 = 150.

If the sum of the 4 integers = 250:
Average = median = 250/4 = 62.5, which is halfway between 62 and 63.
Thus, the 4 integers are 61, 62, 63, and 64.
Sum = 61+62+63+64 = 250.

Thus, each of the following 3-digit multiples of 50 is the sum of 4 consecutive integers:
150, 250, 350, 450, 550, 650, 750, 850, 950.
Since there are 9 possible sums, there are 9 possible means.

The correct answer is C.

For any set of consecutive integers:
The number of integers = biggest - smallest + 1.
The average = the median = (biggest+smallest)/2.
Sum = number*average.
If the number of integers is ODD, then the average (and thus the median) will be an integer.
If the number of integers is EVEN, then the average (and thus the median) will NOT be an integer.