Difference between revisions of "2016 AMC 10B Problems/Problem 4"

(Solution 1)
 
(8 intermediate revisions by 6 users not shown)
Line 5: Line 5:
 
<math>\textbf{(A)}\ \text{Sunday}\qquad\textbf{(B)}\ \text{Monday}\qquad\textbf{(C)}\ \text{Wednesday}\qquad\textbf{(D)}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}</math>
 
<math>\textbf{(A)}\ \text{Sunday}\qquad\textbf{(B)}\ \text{Monday}\qquad\textbf{(C)}\ \text{Wednesday}\qquad\textbf{(D)}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}</math>
  
==Solution==
+
==Solution 1==
 
The process took <math>1+2+3+\ldots+13+14+15=120</math> days, so the last day was <math>119</math> days after the first day.
 
The process took <math>1+2+3+\ldots+13+14+15=120</math> days, so the last day was <math>119</math> days after the first day.
Since <math>119</math> is divisible by <math>7</math>, both must have been the same day of the week, so the answer is <math>\textbf{(B)}\ \text{Monday}</math>.
+
Since <math>119</math> is divisible by <math>7</math>, both must have been the same day of the week, so the answer is <math>\boxed{\textbf{(B)}\ \text{Monday}}</math>.
 +
 
 +
==Solution 2==
 +
Similar to solution 1, the process took 120 days. <math>120 \equiv 1 \mod 7</math>. Since Zoey finished the first book on Monday and the second book (after three days) on Wednesday, we conclude that the modulus must correspond to the day (e.g., <math>1\mod 7</math> corresponds to Monday, <math>4\mod 7</math> corresponds to Thursday, <math>0\mod 7</math> corresponds to Sunday, etc.). The solution is therefore <math>\boxed{\textbf{(B)}\ \text{Monday}}</math>.
 +
 
 +
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/XG5fR4xl1o4
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
 
 +
==Video Solution==
 +
https://youtu.be/8_xEaEIJZ24
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=3|num-a=5}}
 
{{AMC10 box|year=2016|ab=B|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:22, 30 December 2023

Problem

Zoey read $15$ books, one at a time. The first book took her $1$ day to read, the second book took her $2$ days to read, the third book took her $3$ days to read, and so on, with each book taking her $1$ more day to read than the previous book. Zoey finished the first book on a Monday, and the second on a Wednesday. On what day the week did she finish her $15$th book?

$\textbf{(A)}\ \text{Sunday}\qquad\textbf{(B)}\ \text{Monday}\qquad\textbf{(C)}\ \text{Wednesday}\qquad\textbf{(D)}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}$

Solution 1

The process took $1+2+3+\ldots+13+14+15=120$ days, so the last day was $119$ days after the first day. Since $119$ is divisible by $7$, both must have been the same day of the week, so the answer is $\boxed{\textbf{(B)}\ \text{Monday}}$.

Solution 2

Similar to solution 1, the process took 120 days. $120 \equiv 1 \mod 7$. Since Zoey finished the first book on Monday and the second book (after three days) on Wednesday, we conclude that the modulus must correspond to the day (e.g., $1\mod 7$ corresponds to Monday, $4\mod 7$ corresponds to Thursday, $0\mod 7$ corresponds to Sunday, etc.). The solution is therefore $\boxed{\textbf{(B)}\ \text{Monday}}$.

Video Solution (CREATIVE THINKING)

https://youtu.be/XG5fR4xl1o4

~Education, the Study of Everything


Video Solution

https://youtu.be/8_xEaEIJZ24

~savannahsolver

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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