Difference between revisions of "2000 AMC 12 Problems/Problem 9"

Revision as of 20:11, 22 March 2023

The following problem is from both the 2000 AMC 12 #9 and 2000 AMC 10 #14, so both problems redirect to this page.

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 $71$ , $76$ , $80$ , $82$ , and $91$ . What was the last score Mrs. Walters entered?

$\textbf{(A)} \ 71 \qquad \textbf{(B)} \ 76 \qquad \textbf{(C)} \ 80 \qquad \textbf{(D)} \ 82 \qquad \textbf{(E)} \ 91$

Solutions

Solution 1

The first number is divisible by $1$ .

The sum of the first two numbers is even.

The sum of the first three numbers is divisible by $3.$

The sum of the first four numbers is divisible by $4.$

The sum of the first five numbers is $400.$

Since $400$ is divisible by $4,$ the last score must also be divisible by $4.$ Therefore, the last score is either $76$ or $80.$

Case 1: $76$ is the last number entered.

Since $400\equiv 76\equiv 1\pmod{3}$ , the fourth number must be divisible by $3,$ but none of the scores are divisible by $3.$

Case 2: $80$ is the last number entered.

Since $80\equiv 2\pmod{3}$ , the fourth number must be $2\pmod{3}$ . The only number which satisfies this is $71$ . The next number must be $91$ since the sum of the first two numbers is even. So the only arrangement of the scores $76, 82, 91, 71, 80$ or $82, 76, 91, 71, 80$ $\Rightarrow \text{(C)}$

Solution 2

We know the first sum of the first three numbers must be divisible by $3,$ so we write out all $5$ numbers $\pmod{3}$ , which gives $2,1,2,1,1,$ respectively. Clearly, the only way to get a number divisible by $3$ 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 $4, 71$ must be next. That leaves $80$ for last, so the answer is $\mathrm{C}$ .

Solution 3

we know that the scores avarge is a integer

so that means s1+s2+s3+s4 must be a even number divisble by 4

we have 3 even scores and 2 odd scores

which means that the last score cannot be a odd because outwise we get odd number divided by a even number in the demointor.

so we have the answers that are even

76,80,82

We see 3 cases where 76 is 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 divisble by 4 324 divided by 5 is 1 which means 76 is not the last number

case 2: 324-4=320 320 is divisble by 4 320 is divisble by 5 which measn this case true.

case 3: 320-2=318 318 is not disvible by 4 which makes it inccorect even though

it has a remiander of 3 when divided by 5

Video Solution

https://www.youtube.com/watch?v=IJ4xXPEfrzc

@@ Line 31: / Line 31: @@
 We know the first sum of the first three numbers must be divisible by <math>3,</math> so we write out all <math>5</math> numbers <math>\pmod{3}</math>, which gives <math>2,1,2,1,1,</math> respectively. Clearly, the only way to get a number divisible by <math>3</math> 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 <math>4, 71</math> must be next. That leaves <math>80</math> for last, so the answer is <math>\mathrm{C}</math>.
-=== Solution 2 ===
+=== Solution 3 ===
 we know that the scores avarge is a integer

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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