2000 AMC 12 Problems/Problem 9
- The following problem is from both the 2000 AMC 12 #9 and 2000 AMC 10 #14, so both problems redirect to this page.
Contents
Problem
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 , , , , and . What was the last score Mrs. Walters entered?
Solutions
Solution 1
The first number is divisible by .
The sum of the first two numbers is even.
The sum of the first three numbers is divisible by
The sum of the first four numbers is divisible by
The sum of the first five numbers is
Since is divisible by the last score must also be divisible by Therefore, the last score is either or
Case 1: is the last number entered.
Since , the fourth number must be divisible by but none of the scores are divisible by
Case 2: is the last number entered.
Since , the fourth number must be . The only number which satisfies this is . The next number must be since the sum of the first two numbers is even. So the only arrangement of the scores or
Solution 2
We know the first sum of the first three numbers must be divisible by so we write out all numbers , which gives respectively. Clearly, the only way to get a number divisible by by adding three of these is by adding the three ones. So those must go first. Now we have an odd sum, and since the next average must be divisible by must be next. That leaves for last, so the answer is .
Solution 3
we know that the average of the scores is an integer
so that means s1+s2+s3+s4 must be an even number divisible by 4
we have 3 even scores and 2 odd scores
which means that the last score cannot be odd because otherwise, we would get an odd number divided by an even number in the denominator.
so we have answers that are even
76,80,82
We see 3 cases where 76 is the last score, 80 is the last score, and 82 is the last score 76= 1 mod(5) which means 80+82+71+91= 0 mod(4) 80+82+71+91= 4 mod(5)
80= 0 mod(5) 76+82+71+91= 0 mod(5) 76+82+71+91= 0 mod(4)
82= 2 mod(5) 76+80+71+91= 0 mod(4) 76+80+71+91= 3 mod(5)
Case 1: 76 324 is divisible by 4 324 divided by 5 is 1 which means 76 is not the last number
case 2: 324-4=320 320 is divisible by 4 320 is divisible by 5 which means this case is true.
case 3: 320-2=318 318 is not divisible by 4 which makes it incorrect even though
it has a remainder of 3 when divided by 5
Solution 4
the test numbers we have include 71, 76, 80, 82, 91 the answers are same numbers: 71, 76, 80, 82, 91
you could either subrtact eveyrthing by 66 to make things easier
5,10,14,16,25
gettigs 70 when adding up now add 5,10,13,16,25 and it easier to see which one is divisable by 6
or you could do his perform the following action test number minus 6[test number/6]
5,4,2,4,1 add them up to get 6+6+4 now we have to add another 2 for it to be divisable by 6
before the operation 2 was actually 80 thus 80 is the answer
Video Solution
https://www.youtube.com/watch?v=IJ4xXPEfrzc
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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