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

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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 <math>71</math>, <math>76</math>, <math>80</math>, <math>82</math>, and <math>91</math>. What was the last score Mrs. Walters entered?
 
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 <math>71</math>, <math>76</math>, <math>80</math>, <math>82</math>, and <math>91</math>. What was the last score Mrs. Walters entered?
  
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<math>\text{(A)} \ 71 \qquad \text{(B)} \ 76 \qquad \text{(C)} \ 80 \qquad \text{(D)} \ 82 \qquad \text{(E)} \ 91</math>
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<math>\textbf{(A)} \ 71 \qquad \textbf{(B)} \ 76 \qquad \textbf{(C)} \ 80 \qquad \textbf{(D)} \ 82 \qquad \textbf{(E)} \ 91</math>
  
 
== Solutions ==
 
== Solutions ==
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Case 2: <math>80</math> is the last number entered.
 
Case 2: <math>80</math> is the last number entered.
  
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Since <math>80\equiv 2\pmod{3}</math>, the fourth number must be <math>2\pmod{3}</math>. That number is <math>71</math> and <math>71</math> only. The next number must be <math>91</math> since the sum of the first two numbers is even. So the only arrangement of the scores <math>76, 82, 91, 71, 80</math> <math>\Rightarrow \text{(C)}</math>
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Since <math>80\equiv 2\pmod{3}</math>, the fourth number must be <math>2\pmod{3}</math>. The only number which satisfies this is <math>71</math>. The next number must be <math>91</math> since the sum of the first two numbers is even. So the only arrangement of the scores <math>76, 82, 91, 71, 80</math> or <math>82, 76, 91, 71, 80</math> <math>\Rightarrow \text{(C)}</math>
  
 
=== Solution 2 ===
 
=== Solution 2 ===
 
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>.
 
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 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 
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76= 1 mod(5)
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which means
 +
80+82+71+91= 0 mod(4)
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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
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324 divided by 5 is 1
 +
which means 76 is not the last number
 +
 +
case 2:
 +
324-4=320
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320 is divisible by 4
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320 is divisible by 5
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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 ===
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 +
We can simply use the answer choices. Since we know after the sixth test, the sum of the scores he received on the test is divisible by 6, and the sum of the scores he already has is 400, we can simply add the answer choices to 400 to see if it is divisible by 6 or not. By doing this, we can see that 80 is our only answer, so the answer is <math>\mathrm{C}</math>.
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== Solution 5 ==
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 +
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the test numbers we have include 71, 76, 80, 82, 91
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the answers are same numbers: 71, 76, 80, 82, 91
 +
 +
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you could either subtract eveyrthing by 66 to make things easier
 +
5,10,14,16,25
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getting 70 when adding up now add 5,10,13,16,25 and it easier to see which one is divisible by 6
 +
 +
 +
 +
or you could do logma balls all bit
 +
perform the following action test number minus 6[test number/6]
 +
 +
5,4,2,4,1
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add them up to get
 +
6+6+4
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now we have to add another 2 for it to be divisible by 6
 +
 +
before the operation 2 was actually 80
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thus 80 is the answer
  
 
== Video Solution ==
 
== Video Solution ==
 
https://www.youtube.com/watch?v=IJ4xXPEfrzc
 
https://www.youtube.com/watch?v=IJ4xXPEfrzc
  
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== See also ==
+
==Video Solution by Daily Dose of Math==
 +
 
 +
https://youtu.be/Tz-duTUTj78?si=0fm2msZ9V2AAUlFI
 +
 
 +
~Thesmartgreekmathdude
 +
 
 +
==See Also==
 +
 
 
{{AMC12 box|year=2000|num-b=8|num-a=10}}
 
{{AMC12 box|year=2000|num-b=8|num-a=10}}
 
{{AMC10 box|year=2000|num-b=13|num-a=15}}
 
{{AMC10 box|year=2000|num-b=13|num-a=15}}
−
 
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
−
[[Category:Introductory Number Theory Problems]]
 
−
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:43, 8 October 2024

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 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

We can simply use the answer choices. Since we know after the sixth test, the sum of the scores he received on the test is divisible by 6, and the sum of the scores he already has is 400, we can simply add the answer choices to 400 to see if it is divisible by 6 or not. By doing this, we can see that 80 is our only answer, so the answer is $\mathrm{C}$.

Solution 5

the test numbers we have include 71, 76, 80, 82, 91 the answers are same numbers: 71, 76, 80, 82, 91


you could either subtract eveyrthing by 66 to make things easier 5,10,14,16,25 getting 70 when adding up now add 5,10,13,16,25 and it easier to see which one is divisible by 6


or you could do logma balls all bit 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 divisible by 6

before the operation 2 was actually 80 thus 80 is the answer

Video Solution

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

Video Solution by Daily Dose of Math

https://youtu.be/Tz-duTUTj78?si=0fm2msZ9V2AAUlFI

~Thesmartgreekmathdude

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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