Answer: B
Step-by-step explanation:
Answer:
The number of deserters is 34.
Step-by-step explanation:
We have to calculate the number of desertors in a group of 1500 soldiers.
The sergeant divides in groups of different numbers and count the lefts over.
If he divide in groups of 5, he has on left over. The amount of soldiers grouped has to end in 5 or 0, so the total amount of soldiers has to end in 1 or 6.
If he divide in groups of 7, there are three left over. If we take 3, the number of soldiers gruoped in 7 has to end in 8 or 3. The only numbers bigger than 1400 that end in 8 or 3 and have 7 as common divider are 1428 and 1463.
If we add the 3 soldiers left over, we have 1431 and 1466 as the only possible amount of soldiers applying to the two conditions stated until now.
If he divide in groups of 11, there are three left over. We can test with the 2 numbers we stay:
As only 1466 gives a possible result (no decimals), this is the amount of soldiers left.
The deserters are 34:
Answer: 12 x 15= 180 cuadrados
juas juas juas 180 + 180 = 360
15 x 15 cm = 360
Step-by-step explanation:
988 i believe all i did was add the foreign only up
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.
