Answer:
Step-by-step explanation:
2005 AMC 8 Problems/Problem 20
Problem
Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$
Solution
Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$, is a multiple of $12$. Since this number must be a multiple of $12$, as stated in the previous sentence, $14$ has a factor $2$, $k$ must have a factor of $6$. The smallest number of turns that is a multiple of $6$ is $\boxed{\textbf{(A)}\ 6}$.
See Also
2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 19 Followed by
Problem 21
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All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
The total cost for a $1,200 washing machine is $1,200. In the real world it would be 1,200 + tax.
Answer:
17 /6 t
Step-by-step explanation:
3t2+4t3
=32t+43t
Combine Like Terms:
=32t+43t
=(32t+43t)
=176t
Answer:
1364
Step-by-step explanation:
1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364
a1+a2 = a3, a2+a3=a4 etcetera..
1+3 =4
3+4 =7
4+7=11
.
.
a13+a14 = a15
521+843 = 1364
so, 1364 is the answer
Answer:
165 different teams of 3 students can be formed for competitions
Step-by-step explanation:
Combinations of m elements taken from n in n (m≥n) are called all possible groupings that can be made with the m elements so that:
- Not all items fit
- No matter the order
- Elements are not repeated
That is, a combination is an arrangement of elements where the place or position they occupy within the arrangement does not matter. In a combination it is interesting to form groups and their content.
To calculate the number of combinations, the following expression is applied:
It indicates the combinations of m objects taken from among n objects, where the term "n!" is called "factorial of n" and is the multiplication of all the numbers that go from "n" to 1.
In this case:
Replacing:
Solving:
being:
- 3!=3*2*1=6
- 8!=8*7*6*5*4*3*2*1=40,320
- 11!=39,916,800
So:
C= 165
<u><em>165 different teams of 3 students can be formed for competitions</em></u>
