2010 AMC 12A Problems and Solution

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Hence for we get paths, for we get

paths, and for

paths. This gives us

hence the total number of paths is

we get

paths through the second quadrant,

.

Combinatorial Solution 2

Each path that goes through the second quadrant must pass through exactly one of the points

,

, and

.

There is exactly path of the first kind, paths of the second kind, and

paths of the third type. The conclusion remains the same.

Problem 19

Each of 2010 boxes in a line contains a single red marble, and for box in the

, the

position also contains white marbles. Isabella begins at the first box

be the probability that Isabella

and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let

stops after drawing exactly marbles. What is the smallest value of for which

?

Solution

The probability of drawing a white marble from box is drawing a red marble from box is

.

. The probability of

The probability of drawing a red marble at box is therefore

It is then easy to see that the lowest integer value of that satisfies the inequality is

.

Problem 20

Arithmetic sequences value of ?

and and

have integer terms with

for some . What is the largest possible

Solution Solution 1

Since

and

have integer terms with

, we can write the terms of

each sequence as

where and ( Since

) are the common differences of each, respectively.

it is easy to see tha

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