Difference between revisions of "2017 AMC 10B Problems/Problem 25"

(Solution 2 (Cheap Solution))
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By inspection, the sequences <math>91,93,92,96,98,100,95</math> and <math>93,91,92,96,98,100,95</math> work, so the answer is <math>\boxed{\textbf{(E) } 100}</math>.
 
By inspection, the sequences <math>91,93,92,96,98,100,95</math> and <math>93,91,92,96,98,100,95</math> work, so the answer is <math>\boxed{\textbf{(E) } 100}</math>.
Note: A method of finding this "cheap" solution is to create a "mod chart", basically list out the residues of 91-100 modulo 1-7 and then finding the two sequences should be made substantially easier.
+
Note: A method of finding this "cheap" solution is to create a "mod chart", basically list out the residues of 91-100 modulo 1-7 and then finding the two sequences should be made substantially easier.FUCK YOU BITCHES
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=24|after=Last Problem}}
 
{{AMC10 box|year=2017|ab=B|num-b=24|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:18, 14 March 2017

Problem

Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100$, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95$. What was her score on the sixth test?

$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 94\qquad\textbf{(C)}\ 96\qquad\textbf{(D)}\ 98\qquad\textbf{(E)}\ 100$

Solution 1

Let the sum of the scores of Isabella's first $6$ tests be $S$. Since the mean of her first $7$ scores is an integer, then $S + 95 \equiv 0 \text{ (mod 7)}$, or $S \equiv 3 \text{ (mod 7)}$. Also, $S \equiv 0 \text{ (mod 6)}$, so by CRT, $S \equiv 24 \text{ (mod 42)}$. We also know that $91 \cdot 6 \leq S \leq 100 \cdot 6$, so by inspection, $S = 570$. However, we also have that the mean of the first $5$ integers must be an integer, so the sum of the first $5$ test scores must be an multiple of $5$, which implies that the $6$th test score is $\boxed{\textbf{(E) } 100}$.

Solution 2 (Cheap Solution)

By inspection, the sequences $91,93,92,96,98,100,95$ and $93,91,92,96,98,100,95$ work, so the answer is $\boxed{\textbf{(E) } 100}$. Note: A method of finding this "cheap" solution is to create a "mod chart", basically list out the residues of 91-100 modulo 1-7 and then finding the two sequences should be made substantially easier.FUCK YOU BITCHES

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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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