Mrs. Walter gave an exam in a mathematics class of five students. She

Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71,76,80,82, and 91.

What was the last score Mrs. Walters entered?

Let the last score entered by Mrs Walters be x.

(71+76+80+82+91+x)/6 is an integer
(400+x)/6 is an integer

(A) 71 : 471/6 is NOT an integer
(B) 76: 476/6 is NOT an integer
(C) 80: 480/6 is an integer
(D) 82: 482/6 is NOT an integer
(E) 91: 491/6 is NOT an integer

IMO C

Scores: 71, 76, 80, 82, 91.

After each score entered, the average was an integer.

(1) 1st number added: Whatever the first number is, the average would always be an integer (the same integer)

(2) 2nd number added: for the average to be an integer, the sum of the 2 numbers entered has to be even.

Cases:

a) 71, 91
b) 76,80
c) 80, 82
d) 76, 84

(3) 3rd number added: for the average to be an integer, the sum of the 3 numbers entered has to be a multiple of 3.

Let's explore the possibilities with the 4 cases above.

a) 71, 91. Sum 162 = 3k. For the sum after adding the next number to still be a multiple of 3, we need to add a number which is a multiple of 3. But none of the 5 numbers given (71, 76, 80, 82, 91) are multiples of 3. Thus, these two numbers are not the first two numbers.

b) 76, 80. Sum 156 = 3k. For the sum after adding the next number to still be a multiple of 3, we need to add a number which is a multiple of 3. But none of the 5 numbers given (71, 76, 80, 82, 91) are multiples of 3. Thus, these two numbers are not the first two numbers.

c) 80, 82. Sum 162 = 3k. For the sum after adding the next number to still be a multiple of 3, we need to add a number which is a multiple of 3. But none of the 5 numbers given (71, 76, 80, 82, 91) are multiples of 3. Thus, these two numbers are not the first two numbers.

d) 76, 82. Sum 158 = 3k+2. For the sum after adding the next number to be a multiple of 3, we need to add a number of the form 3k+1. Among the remaining 3 numbers, 91 fits.

Hence,

First 3 numbers: (76, 82), and 91.

Now,

(4) 4th number added: for the average to be an integer, the sum of the 4 numbers entered has to be a multiple of 4.

Sum of 1st 3 numbers = 76 + 82 + 91 = 249.

The 4th number has to be 71 (if we take 80 as the 4th number, the sum 249 + 80 is not even an even number, it is not divisible by 4).

So, the first 4 numbers -> (76,82), 91, 71.

Hence, the last number entered = 80. Choice C.

Hope this helps.
Harsha